Inverse, Exponential, and Logarithmic Functions

Home , Common logarithm, Constant function, Inverse function

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Inverse, Exponential, and 4 Logarithmic Functions

The magnitudes of earthquakes, the loudness of sounds, and the growth or decay of some populations are examples of quantities that are described by exponential functions and their inverses, logarithmic functions.

4.1 Inverse Functions

4.2 Exponential Functions

4.3 Logarithmic Functions

Summary Exercises on Inverse, Exponential, and Logarithmic Functions

4.4 Evaluating Logarithms and the Change-of-Base Theorem Chapter 4 Quiz

4.5 Exponential and Logarithmic Equations

4.6 Applications and Models of Exponential Growth and Decay

Summary Exercises on Functions: Domains and Defining Equations

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4.1 Inverse Functions

■ One-to-One Functions One-to-One Functions Suppose we define the following function F. ■ Inverse Functions F 2, 2 , 1, 1 , 0, 0 , 1, 3 , 2, 5 ■ Equations of Inverses = - - ■ An Application of (We have defined F so51 that each2 1 second2 component1 2 1 is2 used1 only26 once.) We can Inverse Functions to form another set of ordered pairs from F by interchanging the x- and y-values of Cryptography each pair in F. We call this set G. G = 2, -2 , 1, -1 , 0, 0 , 3, 1 , 5, 2 G is the inverse of F. Function F was defined with each second component Domain Range 51 2 1 2 1 2 1 2 1 26 f used only once, so set G will also be a function. (Each first component must 1 6 be used only once.) In order for a function to have an inverse that is also a func- 2 7 3 tion, it must exhibit this one-to-one relationship. 8 4 In a one-to-one function, each x-value corresponds to only one y-value, and 5 9 each y-value corresponds to only one x-value.

Not One-to-One The function ƒ shown in Figure 1 is not one-to-one because the y-value 7 corre- Figure 1 sponds to two x-values, 2 and 3. That is, the ordered pairs 2, 7 and 3, 7 both belong to the function. The function ƒ in Figure 2 is one-to-one. 1 2 1 2

Domain Range One-to-One Function f 1 5 A function ƒ is a one-to-one function if, for elements a and b in the domain 2 6 of ƒ, 3 7 a 3 b implies ƒ a 3 ƒ b . 4 8 That is, different values of the domain correspond1 2 1to2 different values of the One-to-One range. Figure 2 Using the concept of the contrapositive from the study of logic, the boldface statement in the preceding box is equivalent to ƒ a ƒ b implies a b.

This means that if two range1 2 values1 2are equal, then their corresponding domain values are equal. We use this statement to show that a function ƒ is one-to-one in Example 1(a).

Classroom Example 1 Example 1 Deciding Whether Functions Are One-to-One Determine whether each function is one-to-one. Determine whether each function is one-to-one. (a) ƒ x = -3x + 7 (a) ƒ x = -4x + 12 (b) ƒ x = 25 - x2 (b) ƒ x = 49 - x2 2 1 2 Solut1 io2 n 1 2 Answers: 2 1 2 (a) We can determine that the function ƒ x 4x 12 is one-to-one by (a) It is one-to-one. = - + showing that ƒ a = ƒ b leads to the result a = b. (b) It is not one-to-one. 1 2 1 2 ƒ1 a2 = ƒ b -4a +1122 = -14b2 + 12 ƒ x = -4x + 12 -4a = -4b Subtract1 2 12. a = b Divide by -4. By the definition, ƒ x = -4x + 12 is one-to-one. Copyright Pearson. All Rights Reserved. 1 2

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(b) We can determine that the function ƒ x = 25 - x2 is not one-to-one by showing that different values of the domain correspond2 to the same value of the range. If we choose a = 3 and b =1 -23, then 3 -3, but

ƒ 3 = 25 - 32 = 25 - 9 = 16 = 4 2 2 2 and ƒ -132 = 25 - -3 2 = 25 - 9 = 4. 2 2 Here, even though1 32 -3, ƒ 3 1= ƒ2-3 = 4. By the definition, ƒ is not a one-to-one function. 1 2 1 2 ■✔ Now Try Exercises 17 and 19.

As illustrated in Example 1(b), a way to show that a function is not one- to-one is to produce a pair of different domain elements that lead to the same function value. There is a useful graphical test for this, the horizontal line test.

y Horizontal Line Test

2 5 f(x) = Ë25 – x A function is one-to-one if every horizontal line intersects the graph of the (–3, 4)(3, 4) function at most once.

x 0 –5 5 Note In Example 1(b), the graph of the function is a semicircle, as shown in Figure 3. Because there is at least one horizontal line that intersects the graph in more than one point, this function is not one-to-one. Figure 3

Classroom Example 2 Example 2 Using the Horizontal Line Test Determine whether each graph is the graph of a one-to-one function. Determine whether each graph is the graph of a one-to-one function. (a) y (a) y (b) y

4 y1 (x2, y1) (x1, y1) (x3, y1) x –20 2 y1 y 2 x 0 3 x –4 x x1 x2 x3 x1 x2 0 y3 (b) y

4 Solution –2 2 x 0 (a) Each point where the horizontal line intersects the graph has the same value of y but a different value of x. Because more than one different value of x (here three) lead to the same value of y, the function is not one-to-one. Answers: (a) It is one-to-one. (b) Every horizontal line will intersect the graph at exactly one point, so this function is one-to-one. (b) It is not one-to-one. ■✔ Now Try Exercises 11 and 13.

The function graphed in Example 2(b) decreases on its entire domain. In general, a function that is either increasing or decreasing on its 3 1 entire domain, such as ƒ x x, g x x , and h x x , must be one-to-one. Copyright Pearson. All Rights Reserved. 1 2 1 2 1 2

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Tests to Determine Whether a Function Is One-to-One

1. Show that ƒ a = ƒ b implies a = b. This means that ƒ is one-to-one. (See Example 1(a).) 1 2 1 2 2. In a one-to-one function, every y-value corresponds to no more than one x-value. To show that a function is not one-to-one, find at least two x-values that produce the same y-value. (See Example 1(b).) 3. Sketch the graph and use the horizontal line test. (See Example 2.) 4. If the function either increases or decreases on its entire domain, then it is one-to-one. A sketch is helpful here, too. (See Example 2(b).)

Teaching Tip This is a good Inverse Functions Certain pairs of one-to-one functions “undo” each place to review composition of other. For example, consider the functions functions. 1 5 g x = 8x + 5 and ƒ x = x - . 8 8 We choose an arbitrary1 2element from the domain1 2 of g, say 10. Evaluate g 10 .

g x = 8x + 5 Given function 1 2 g 101 2 = 8 # 10 + 5 Let x = 10. g1102 = 85 Multiply and then add. Now, we evaluate ƒ 185 2. 1 2 1 5 ƒ x = x - Given function 8 8 1 2 1 5 ƒ 85 = 85 - Let x = 85. 8 8 1 2 851 25 ƒ 85 = - Multiply. 8 8 ƒ1852 = 10 Subtract and then divide. Starting with 10, we1 “applied”2 function g and then “applied” function ƒ to the result, which returned the number 10. See Figure 4.

Function Function These functions 10 85 10 contain inverse g(x) = 8x + 5 f(x) = 1x – 5 operations that 8 8 “undo” each other.

Figure 4

As further examples, confirm the following.

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1 5 In particular, for the pair of functions g x = 8x + 5 and ƒ x = 8 x - 8 , ƒ g 2 = 2 and1 2 g ƒ 2 = 2. 1 2 In fact, for any value of 1x, 1 22 1 1 22

ƒ g x = x and g ƒ x = x. Using the notation for composition1 1 22 of functions,1 1 22 these two equations can be written as follows.

ƒ ∘ g x = x and g ∘ ƒ x = x The result is the identity function. Because1 the21 compositions2 of 1ƒ and21 g yield2 the identity function, they are inverses of each other.

Inverse Function

Let ƒ be a one-to-one function. Then g is the inverse function of ƒ if

ƒ g x x for every x in the domain of g, and 1g ƒ2 1x2 x for every x in the domain of ƒ. 1 2 1 2

The condition that f is one-to-one in the definition of inverse function is essential. Otherwise, g will not define a function. Classroom Example 3 Let functions ƒ and g be defined respectively by Example 3 Determining Whether Two Functions Are Inverses ƒ x = 2x + 5 and Let functions ƒ and g be defined respectively by 1 2 1 3 3 g x = x - 5. ƒ x x 1 and g x x 1. 2 = - = + 2 Is g the inverse1 2 function of ƒ? Is g the inverse function1 2 of ƒ? 1 2

Answer: no Solution As shown in Figure 5, the horizontal line test applied to the graph indicates that ƒ is one-to-one, so the function has an inverse. Because it is one- to-one, we now find ƒ ∘ g x and g ∘ ƒ x . y ƒ ∘ g1 x 21 2 1 21g2∘ ƒ x

1 = ƒ21g2x 1 = g21ƒ2x 3 3 3 x = 1 1x 22+ 1 - 1 = 1 1x322- 1 + 1 ƒ(x) = x3 – 1 2 2 x 1 1 3 x3 = A + - B = 1 2 = x = x2

Figure 5 Since ƒ ∘ g x = x and g ∘ ƒ x = x, function g is the inverse of function ƒ. 1 21 2 1 21 2 ■✔ Now Try Exercise 41.

A special notation is used for inverse functions: If g is the inverse of a function ƒ, then g is written as ƒ1 (read “ƒ-inverse”).

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Caution Do not confuse the 1 in ƒ1 with a negative exponent. -1 1 The symbol ƒ x represents the inverse function of ƒ, not ƒ x . 1 2 1 2

Teaching Tip Remind students By the definition of inverse function, the domain of ƒ is the range of ƒ1, that composition of functions and the range of ƒ is the domain of ƒ1. See Figure 6. is not commutative. Therefore, ƒ ∘ g x and g ∘ ƒ x must X both be checked. Use an example Y such1 as21 2 1 21 2 Domain of f f Range of f ƒ x = x2 and g x = x x y Range of f –1 2 Domain of f –1 to illustrate1 2 the necessity1 2 of f –1 checking both compositions if you do not verify the functions are one-to-one. Figure 6

Example 4 Finding Inverses of One-to-One Functions

Number of Find the inverse of each function that is one-to-one. Year Hurricanes (a) F = -2, 1 , -1, 0 , 0, 1 , 1, 2 , 2, 2 2009 3 (b) G = 3, 1 , 0, 2 , 2, 3 , 4, 0 2010 12 51 2 1 2 1 2 1 2 1 26 2011 7 (c) The table51 in2 the1 margin2 1 shows2 1 the26 number of hurricanes recorded in the North Atlantic during the years 2009–2013. Let ƒ be the function defined in 2012 10 the table, with the years forming the domain and the numbers of hurricanes 2013 2 forming the range. Source: www.wunderground.com Solution (a) Each x-value in F corresponds to just one y-value. However, the y-value 2 Classroom Example 4 corresponds to two x-values, 1 and 2. Also, the y-value 1 corresponds to Find the inverse of each function both -2 and 0. Because at least one y-value corresponds to more than one that is one-to-one. x-value, F is not one-to-one and does not have an inverse. (a) F = -2, -8 , -1, -1 , 0, 0 , 1, 1 , 2, 8 (b) Every x-value in G corresponds to only one y-value, and every y-value cor- 51 2 1 2 (b) G 2, 5 , 1, 2 , responds to only one x-value, so G is a one-to-one function. The inverse = 1 - 2 1 -2 1 26 0, 1 , 1, 2 , 2, 5 function is found by interchanging the x- and y-values in each ordered pair. 51 2 1 2 (c) Let h be the function defined 1 1 2 1 2 1 26 G- = 1, 3 , 2, 0 , 3, 2 , 0, 4 by the table in Example 4(c) if the number of hurricanes for Notice how the domain and51 range2 1 of 2G1 become2 1 the26 range and domain, 2009 is decreased by 1. respectively, of G-1. Answers: (c) Each x-value in ƒ corresponds to only one y-value, and each y-value corre- (a) F -1 = -8, -2 , -1, -1 , sponds to only one x-value, so ƒ is a one-to-one function. The inverse function 0, 0 , 1, 1 , 8, 2 51 2 1 2 is found by interchanging the x- and y-values in the table. (b) G is not one-to-one. 1 2 1 2 1 26 (c) h is not one-to-one. ƒ-1 x = 3, 2009 , 12, 2010 , 7, 2011 , 10, 2012 , 2, 2013 The domain1 2 and51 range of2 1ƒ become 2the1 range and2 1 domain of2 ƒ1-1. 26

■✔ Now Try Exercises 37, 51, and 53.

Equations of Inverses The inverse of a one-to-one function is found by interchanging the x- and y-values of each of its ordered pairs. The equation of the inverse of a function defined by y = ƒ x is found in the same way. Copyright Pearson. All Rights Reserved.1 2

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Finding the Equation of the Inverse of y f x

For a one-to-one function ƒ defined by an equation1 y2 = ƒ x , find the defining equation of the inverse as follows. (If necessary, replace ƒ x with y first. Any restrictions on x and y should be considered.) 1 2 1 2 Step 1 Interchange x and y. Step 2 Solve for y. Step 3 Replace y with ƒ-1 x .

1 2

Classroom Example 5 Example 5 Finding Equations of Inverses Determine whether each function defines a one-to-one function. Determine whether each equation defines a one-to-one function. If so, find the If so, find the equation of the equation of the inverse. inverse. (a) ƒ x = 2x + 5 (b) y = x2 + 2 (c) ƒ x = x - 2 3 (a) ƒ x = x (b) g x = 4x - 7 Solut1 io2 n 1 2 1 2 1 2 0 0 (c) h x = x3 + 5 1 2 (a) The graph of y = 2x + 5 is a nonhorizontal line, so by the horizontal line Answers:1 2 test, ƒ is a one-to-one function. Find the equation of the inverse as follows. (a) ƒ is not one-to-one. ƒ x 2x 5 Given function (b) g is one-to-one; = + -1 1 7 y 2x 5 Let y ƒ x . g x = 4 x + 4 . 1 2 = + = (c) h is1 one-to-one;2 Step 1 x = 2y + 5 Interchange1 2x and y. h-1 x 3 x 5. (+ = - Step 2 x 5 2y Subtract 5. - = 1

2 )+ 1 2 x - 5 Divide by 2. Solve for y.

y 1

= * 2 Rewrite. Replace y with ƒ-1 x . 1 1 5 Step 3 ƒ- x x a b b = - - 1 a 2 2 c = c - c 1 2 1 2 x Teaching Tip To emphasize -1 - 5 1 5 Thus, the equation ƒ x = 2 A= B2 x - 2 represents a linear function. In the relationship between inverse the function y = 2x + 5, the value of y is found by starting with a value of x, functions, for ƒ x = 2x + 5 and 1 2 x = 3, find multiplying by 2, and adding 5. x 1 2 The equation ƒ-1 x - 5 for the inverse subtracts 5 and then divides y = 2 # 3 + 5 = 11. = 2 by 2. An inverse is used to “undo” what a function does to the variable x. -1 1 5 1 2 Now, for ƒ x = 2 x - 2 and x = 11, find (b) The equation y = x2 + 2 has a parabola opening up as its graph, so some 1 2 1 5 horizontal lines will intersect the graph at two points. For example, both y = 11 - = 3. 2 2 x = 3 and x = -3 correspond to y = 11. Because of the presence of the 2 1 2 x -term, there are many pairs of x-values that correspond to the same This shows numerically that when y-value. This means that the function defined by y x2 2 is not one-to- we substitute the original y-value = + in ƒ-1 x , we obtain the original one and does not have an inverse. x-value. Proceeding with the steps for finding the equation of an inverse leads to 1 2 y = x2 + 2 x = y2 + 2 Interchange x and y. x 2 y2 Solve for y. Remember - = both roots. { x - 2 = y. Square root property 2 The last equation shows that there are two y-values for each choice of x greater than 2, indicating that this is not a function.

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y f(x) = (x – 2)3 (c) Figure 7 shows that the horizontal line test assures us that this horizontal translation of the graph of the cubing function is one-to-one. 8 ƒ x = x - 2 3 Given function 3 1 y2 = 1x - 22 Replace ƒ x with y. Step 1 x = y - 2 3 Interchange1 2x and y.

1 2 (++ 1 3 3 3 x Step 2 x = 1 y -2 2 Take the cube root on each side. 1

0 2 )++ 23 2 3 3 Solve for y. x = y -1 2 2 a = a 2 3 2 1 This graph x + 2 = y Add 2. * passes the 2 3 horizontal Step 3 ƒ-1 x x 2 Replace y with ƒ-1 x . Rewrite. line test. = + 2 –8 1 2 ■✔ Now Try Exercise1 s2 59(a), 63(a), and 65(a). Figure 7 Example 6 Finding the Equation of the Inverse of a Rational Function Classroom Example 6 The following rational function is one-to-one. Find its inverse. The following rational function is one-to-one. Find its inverse. 2x + 3 ƒ x = , x 4 -3x + 1 x - 4 ƒ x = , x 5 x - 5 1 2 2x + 3 1 2 5x 1 Solution ƒ x , x 4 Given function Answer: ƒ-1 x + , = = x + 3 x - 4 x -3 1 2 1 2 2x + 3 y = Replace ƒ x with y. x - 4 1 2 2y + 3 Step 1 x = , y 4 Interchange x and y. y - 4 ( 11

Step 2 x y - 4 = 2y + 3 Multiply by y - 4. ++++)++++ x1y - 4x2 = 2y + 3 Distributive property Pay close attention here. xy 2y 4x 3 Add 4x and -2y. - = + Solve for y. y x - 2 = 4x + 3 Factor out y. 4x + 3 1 2 11 y = , x 2 Divide by x - 2. x - 2 * In the final line, we give the condition x 2. (Note that 2 is not in the range of ƒ, so it is not in the domain of ƒ-1.)

4x + 3 Step 3 ƒ-1 x = , x 2 Replace y with ƒ-1 x . x - 2 1 2 ■✔ Now Try1 2Exercise 71(a).

One way to graph the inverse of a function ƒ whose equation is known follows. Step 1 Find some ordered pairs that are on the graph of ƒ. Step 2 Interchange x and y to find ordered pairs that are on the graph of ƒ-1. Step 3 Plot those points, and sketch the graph of ƒ-1 through them. Another way is to select points on the graph of ƒ and use symmetry to find corresponding points on the graph of ƒ-1.

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y For example, suppose the point a, b shown in Figure 8 is on the graph of (a, b) a one-to-one function ƒ. Then the point b, a is on the graph of ƒ-1. The line b segment connecting a, b and b, a1 is perpendicular2 to, and cut in half by, the y = x line y = x. The points a, b and b, a are1 “mirror2 images” of each other with respect to y = x. 1 2 1 2 1 2 1 2 Thus, we can find the graph of ƒ1 from the graph of ƒ by locating the (b, a) a mirror image of each point in ƒ with respect to the line y x. x 0 a b Figure 8 Example 7 Graphing f 1 Given the Graph of f

In each set of axes in Figure 9, the graph of a one-to-one function ƒ is shown in -1 Classroom Example 7 blue. Graph ƒ in red. Determine whether functions ƒ Solution In Figure 9, the graphs of two functions ƒ shown in blue are given and g graphed are inverses of with their inverses shown in red. In each case, the graph of ƒ-1 is a reflection of each other. the graph of ƒ with respect to the line y = x. y y = g(x) y = x y y f –1 f y = x x 5 y = x y = f(x) (1, 3) (3, 1) 5 f (2, 4) (–4, 0) x 0 –1 Answer: They are not inverses. –2 5 f –2 (4, 2) 0 (1, 1) x (0, –4) –2 5 –2 Figure 9 ■✔ Now Try Exercises 77 and 81.

Classroom Example 8 Example 8 Finding the Inverse of a Function (Restricted Domain) Let ƒ x = x2 + 4, x … 0. Find ƒ-1 x . Let ƒ x = x + 5, x Ú-5. Find ƒ-1 x . 1 2 Answer: 2 1 2 Solut1 2ion The domain of ƒ is restricted1 2to the interval -5, ∞ . Function ƒ is ƒ-1 x = - x - 4, x Ú 4 one-to-one because it is an increasing function and thus has an inverse function. 2 Now we find the equation of the inverse. 3 2 1 2

ƒ x = x + 5, x Ú-5 Given function 2 1 2 y = x + 5, x Ú-5 Replace ƒ x with y. 2 Step 1 x = y + 5, y Ú-5 Interchange1 2x and y.

2 ( 2 2 1 Step 2 x = y + 5 Square each side. + 11

2 2 ) x2 y 5 a a for a 0 Solve for y. = + = Ú 1111 A B 2 y x2 5 Subtract 5. Rewrite. * = - A B However, we cannot define ƒ-1 x as x2 - 5. The domain of ƒ is -5, ∞ , and its range is 0, ∞ . The range of ƒ is the domain of ƒ-1, so ƒ-1 must be defined as follows. 1 2 3 2 3 2 Step 3 ƒ-1 x = x2 - 5, x Ú 0 As a check, the range of ƒ-1, -1 5,2 ∞ , is the domain of ƒ. Copyright Pearson. All Rights Reserved.3 2

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Graphs of ƒ and ƒ-1 are shown in Figures 10 and 11. The line y = x is included on the graphs to show that the graphs of ƒ and ƒ-1 are mirror images with respect to this line.

y f(x) = Ίx+5, x ≥ −5 5 f(x) = Ίx + 5, x Ն – 5 10 y = x

x –5 0 5 −16.1 16.1 y = x

−10 –5 f –1(x) = x2 – 5, x Ն 0 f –1(x) = x2 − 5, x ≥ 0 Figure 10 Figure 11

■✔ Now Try Exercise 75.

Important Facts about Inverses

1. If ƒ is one-to-one, then ƒ-1 exists. 2 y = x 2. The domain of ƒ is the range of ƒ-1, and the range of ƒ is the domain of ƒ-1.

-1 4.1 3. If the point a, b lies on the graph of ƒ, then b, a lies on the graph of ƒ . The graphs of ƒ and ƒ-1 are reflections of each other across the line y = x. 1 2 1 2 4. To find the equation for ƒ-1, replace ƒ x with y, interchange x and y, 6.6 −6.6 and solve for y. This gives ƒ-1 x . 1 2 1 2 −4.1

x = y2 Some graphing calculators have the capability of “drawing” the reflection Despite the fact that y = x2 is not of a graph across the line y = x. This feature does not require that the function be one-to-one, the calculator will draw its one-to-one, however, so the resulting figure may not be the graph of a function. 2 “inverse,” x = y . See Figure 12. It is necessary to understand the mathematics to interpret Figure 12 results correctly. ■

An Application of Inverse Functions to Cryptography A one-to-one function and its inverse can be used to make information secure. The function is used to encode a message, and its inverse is used to decode the coded message. In practice, complicated functions are used. Classroom Example 9 The function Example 9 Using Functions to Encode and Decode a Message ƒ x = 3x - 1 Use the one-to-one function ƒ x 3x 1 and the following numerical values was used to1 encode2 a message as = + assigned to each letter of the alphabet to encode and decode the message BE 11 44 74 44 62 MY FACEBOOK FRIEND. 1 2 59 68 14 14 59 Find the inverse function and A 1 H 8 O 15 V 22 decode the message. (Use the B 2 I 9 P 16 W 23 same numerical values for the C 3 J 10 Q 17 X 24 letters as in Example 9.) D 4 K 11 R 18 Y 25 E 5 L 12 S 19 Z 26 -1 1 1 Answer: ƒ x = 3 x + 3 ; F 6 M 13 T 20 DO YOU TWEET G 7 N 14 U 21 1 2 Copyright Pearson. All Rights Reserved.

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A 1 N 14 Solution The message BE MY FACEBOOK FRIEND would be encoded as B 2 O 15 C 3 P 16 7 16 40 76 19 4 10 16 7 D 4 Q 17 46 46 34 19 55 28 16 43 13 E 5 R 18 because F 6 S 19 B corresponds to 2 and ƒ 2 3 2 1 7, G 7 T 20 = + = H 8 U 21 E corresponds to 5 and ƒ152 = 3152 + 1 = 16, and so on. I 9 V 22 J 10 W 23 -1 1 1 1 2 1 2 Using the inverse ƒ x = 3 x - 3 to decode yields K 11 X 24 L 12 Y 25 1 1 2 1 ƒ-1 7 = 7 - = 2, which corresponds to B, M 13 Z 26 3 3 1 2 1 1 2 1 ƒ-1 16 = 16 - = 5, which corresponds to E, and so on. 3 3 1 2 1 2 ■✔ Now Try Exercise 97.

4.1 Exercises

1. one-to-one 2. not one-to-one Concept Preview Determine whether the function represented in each table is one-to-one. 3. one-to-one 4. x; g ∘ ƒ x 5. range; domain 1. The table shows the number of registered passenger cars in the United States for the 6. b, a 7. 3 1x 21 2 years 2008–2012. 8. y = x 9. -23 1 2 10. does not; it is not one-to-one Registered Passenger Cars Year (in thousands) 11. one-to-one 12. one-to-one 2008 137,080 13. not one-to-one 14. not one-to-one 2009 134,880 15. one-to-one 16. one-to-one 2010 139,892 17. one-to-one 18. one-to-one 2011 125,657 19. not one-to-one 20. not one-to-one 2012 111,290 21. one-to-one 22. one-to-one Source: U.S. Federal Highway Administration. 23. one-to-one 24. one-to-one 25. not one-to-one 2. The table gives the number of representatives currently in Congress from each of 26. not one-to-one five New England states. 27. one-to-one 28. one-to-one Number of 29. no 30. no State Representatives 31. untying your shoelaces Connecticut 5 32. stopping a car Maine 2 33. leaving a room 34. descending the stairs Massachusetts 9 35. unscrewing a light bulb New Hampshire 2 36. emptying a cup Vermont 1 37. inverses 38. not inverses Source: www.house.gov 39. not inverses 40. inverses Concept Preview Fill in the blank(s) to correctly complete each sentence. 41. inverses 42. inverses 43. not inverses 44. not inverses 3. For a function to have an inverse, it must be . 45. inverses 46. inverses 4. If two functions ƒ and g are inverses, then ƒ ∘ g x = and = x. 47. not inverses 48. not inverses 49. inverses 50. inverses 5. The domain of ƒ is equal to the 1 of ƒ-211, and2 the range of ƒ is equal to the of ƒ-1. 51. 6, -3 , 1, 2 , 8, 5 2 6. If the point a, b lies on the graph of ƒ, and ƒ has an inverse, then the point 52. 51-1, 32, 10, 52, 15, 026, 3 , 4 lies on the graph of ƒ-1. Copyright Pearson.1 All Rights2 Reserved. E 1 2 1 2 1 2 A B F

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53. not one-to-one 7. If ƒ x = x3, then ƒ-1 x = . 54. not one-to-one 8. If a 1function2 ƒ has an1 inverse,2 then the graph of ƒ-1 may be obtained by reflecting 55. inverses 56. inverses the graph of ƒ across the line with equation . 57. not inverses 58. not inverses 9. If a function ƒ has an inverse and ƒ -3 = 6, then ƒ-1 6 = . -1 1 4 59. (a) ƒ x = 3 x + 3 10. If ƒ -4 = 16 and ƒ 4 = 16, then1 ƒ 2 1 2 have an inverse because (b) y . (does/does not) 1 2 ƒ 1 2 1 2 2 ƒ –1 Determine whether each function graphed or defined is one-to-one. See Examples 1 x –4 0 2 and 2.

–4 11. y 12. y (c) Domains and ranges of both ƒ and ƒ-1 are -∞, ∞ . 60. (a) ƒ-1 x 1 x 5 = 4 +1 4 2 x x 0 0 (b) y 1 2 f

f –1 x –5 0 13. y 14. y –5 (c) Domains and ranges of both ƒ and ƒ-1 are -∞, ∞ . x x -1 1 3 0 0 61. (a) ƒ x = - 4 x1 + 4 2 (b) y 1 2

f –1 x 15. y 16. y –2 0 3

f (c) Domains and ranges of both ƒ and ƒ-1 are , . x x -∞ ∞ 0 0 -1 1 4 62. (a) ƒ x = - 6 x1 - 3 2 (b) y 1 2 ƒ 4 4 0 x 17. y 2x 8 18. y 4x 20 19. y 36 x2 –8 = - = + = - ƒ –1 2 20. y 100 x2 21. y 2x3 1 22. y 3x3 6 –8 = - - = - = - 2 (c) Domains and ranges of both -1 4 2 1 23. y = 24. y = 25. y = 2 x + 1 - 6 ƒ and ƒ- are -∞, ∞ . x + 2 x - 8 -1 3 63. (a) ƒ x = x - 1 3 1 3 2 1 2 26. y 3 x 6 2 8 27. y x 1 3 28. y x 2 8 (b) y 2 = - - + = + - = - + - 1 2 2 2 ƒ 1 2 ƒ –1 Concept Check Answer each question. x 0 29. Can a constant function, such as ƒ x = 3, defined over the set of real numbers, be one-to-one? 1 2 (c) Domains and ranges of both 30. Can a polynomial function of even degree defined over the set of real numbers have ƒ and ƒ-1 are -∞, ∞ . an inverse? 1 3 64. (a) ƒ- x = -1x - 2 2 (b) y 2 Concept Check An everyday activity is described. Keeping in mind that an inverse 1 2 f operation “undoes” what an operation does, describe each inverse activity. f –1 31. tying your shoelaces 32. starting a car x –2 0 33. entering a room 34. climbing the stairs –2

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65. not one-to-one Determine whether the given functions are inverses. See Example 4. 66. not one-to-one 37. x ƒ x x g x 38. x ƒ x x g x -1 1 67. (a) ƒ x = x , x 0 (b) y 3 -1 42 -4 13 2 -2 -1 82 8 -1 22 1 2 2 -6 -6 2 -1 -1 1 -1 5 8 8 5 0 0 0 0 1 1 9 9 1 1 1 1 1 x - 0 1 4 3 3 4 2 8 -8 2 ƒ = ƒ –1 39. ƒ = 2, 5 , 3, 5 , 4, 5 ; g = 5, 2 (c) Domains and ranges of both ƒ and ƒ-1 are 40. ƒ = 511, 12, 13, 32, 15, 526; g = 511, 126, 3, 3 , 5, 5 -∞, 0 ´ 0, ∞ . 51 2 1 2 1 26 51 2 1 2 1 26 -1 4 68. (a) 1ƒ x 2= x , 1x 02 (b) y Use the definition of inverses to determine whether f and g are inverses. See Example 3. 1 2 1 1 41. ƒ x = 2x + 4, g x = x - 2 42. ƒ x = 3x + 9, g x = x - 3 2 2 3 x 0 2 –1 1 2 1 2 1 2 1 2 f = f 1 1 43. ƒ x = -3x + 12, g x = - x - 12 44. ƒ x = -4x + 2, g x = - x - 2 3 4 (c) Domains and ranges 1 2 1 2 1 2 1 2 of both ƒ and ƒ-1 are x + 1 2x + 1 x - 3 4x + 3 45. ƒ x = , g x = 46. ƒ x = , g x = -∞, 0 ´ 0, ∞ . x - 2 x - 1 x + 4 1 - x x 69. (a) ƒ-1 x 1 + 3 , x 0 1 2 1 2 1 2 1 2 1 2= 1x 2 2 6x + 2 -1 1 - x 47. ƒ x , g x 48. ƒ x , g x (b) y = = = = 1 2 ƒ x + 6 x x + 1 x –1 ƒ –1 ƒ 1 2 1 2 1 2 1 2 49. ƒ x = x2 + 3, x Ú 0; g x = x - 3, x Ú 3 0 x 2 50. ƒ1x2 x 8, x 8;1 2g x x2 8, x 0 ƒ = + Ú- = - Ú 2 (c) Domain of ƒ = range of 1 2 1 2 ƒ-1 = -∞, 3 ´ 3, ∞ . Domain of ƒ-1 = range of Find the inverse of each function that is one-to-one. See Example 4. 1 2 1 2 ƒ = -∞, 0 ´ 0, ∞ . 2 51. 3, 6 , 2, 1 , 5, 8 52. 3, 1 , 5, 0 , 0, 5 , 4, -1 1 - 2x - - 70. (a) ƒ x1 = 2x ,1 x 02 3 (b) y 51 2 1 2 1 26 e 1 2 1 2 1 2 a b f 1 2 53. 1, -3 , 2, -7 , 4, -3 , 5, -5 54. 6, -8 , 3, -4 , 0, -8 , 5, -4 ƒ 51 2 1 2 1 2 1 26 51 2 1 2 1 2 1 26 x 0 –1 –1 ƒ ƒ Determine whether each pair of functions graphed are inverses. See Example 7. ƒ (c) Domain of ƒ = range 55. y 56. y of ƒ-1 = -∞, -2 ´ -2, ∞ . 4 y = x 4 1 1 2 Domain of ƒ- = range of 3 ƒ1 = -2∞, 0 ´ 0, ∞ . x 71. (a) ƒ-1 x 3 + 1 , x 1 x x 1 = x 2- 1 1 2 0 0 3 4 y = x 4 (b) y 1 2

5 ƒ –1 ƒ –1 ƒ 57. y 58. y 0 x –1 3

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-1 x + 2 For each function that is one-to-one, (a) write an equation for the inverse function, 72. (a) ƒ x = x 1 , x 1 - (b) graph ƒ and ƒ-1 on the same axes, and (c) give the domain and range of both f and (b) y 1 2 ƒ-1. If the function is not one-to-one, say so. See Examples 5–8. ƒ = ƒ –1 4 2 59. ƒ x = 3x - 4 60. ƒ x = 4x - 5 61. ƒ x = -4x + 3 0 x 3 3 –2 2 62. ƒ1x2 = -6x - 8 63. ƒ1x2 = x + 1 64. ƒ1x2 = -x - 2 ƒ = ƒ –1 1 2 1 2 1 2 1 65. ƒ x = x2 + 8 66. ƒ x = -x2 + 2 67. ƒ x = , x 0 (c) Domain of ƒ = range of x ƒ-1 = -∞, 1 ´ 1, ∞ . 1 2 1 2 1 2 1 4 1 1 Domain of ƒ- = range of 68. ƒ x = , x 0 69. ƒ x = , x 3 70. ƒ x = , x -2 x x 3 x 2 ƒ = -1∞, 1 ´2 1,1 ∞ .2 - + 3x 6 1 2 1 2 1 2 73. (a) ƒ-1 x + , x 2 x + 1 x + 2 1 = x 2- 2 1 2 71. ƒ x = , x 3 72. ƒ x = , x 1 (b) y x - 3 x - 1 1 2 1 2 1 2 2x + 6 -3x + 12 ƒ –1 73. ƒ x = , x 3 74. ƒ x = , x 6 –1 x - 3 x - 6 ƒ ƒ 2 x 1 2 1 2 2 3 75. ƒ x = x + 6, x Ú-6 76. ƒ x = - x - 16, x Ú 4 ƒ 2 2 Graph1 the2 inverse of each one-to-one function. See Example1 2 7. (c) Domain of ƒ = range of -1 ƒ = -∞, 3 ´ 3, ∞ . 77. y 78. y 79. y Domain of ƒ-1 = range of ƒ = -1∞, 2 ´2 2,1 ∞ .2 x 74. (a) ƒ-1 x 6 + 12 , x 3 1 = x2+ 3 1 2- (b) –1 y ƒ x x x ƒ 1 2 0 0 0

6 4 ƒ –1 x Ϫ3 0 46 Ϫ3 ƒ 80. y 81. y 82. y

(c) Domain of ƒ = range of ƒ-1 = -∞, 6 ´ 6, ∞ . -1 Domain of ƒ = range of x x x ƒ = -1∞, -32 ´ 1-3, 2∞ . 0 0 0 -1 2 75. (a) ƒ x1 = x -26, x1Ú 0 2 (b) y 1 2 ƒ –1 ƒ x Concept Check The graph of a function ƒ is shown in the figure. Use the graph to find 0 –6 each value. y

–6 83. ƒ-1 4 84. ƒ-1 2 4

(c) Domain of ƒ = range of 1 1 85. ƒ- 102 86. ƒ- 1-22 2 ƒ-1 = -6, ∞ . -1 -1 -1 x Domain of ƒ = range of 87. ƒ 1-23 88. ƒ 1-42 –4 –2 0 2 4 ƒ = 0,3 ∞ . 2 1 2 1 2 –2 76. (a) ƒ-1 x3 = 2 x2 + 16, –4 x … 0 2 (b) 1 2y f –1 Concept Check Answer each of the following. 4 89. Suppose ƒ x is the number of cars that can be built for x dollars. What does 4 x 1 0 ƒ- 1000 represent? f 1 2 90. Suppose1 ƒ2 r is the volume (in cubic inches) of a sphere of radius r inches. What does ƒ-1 5 represent? (c) Domain of ƒ = range of 1 2 -1 ƒ = 4, ∞ . 91. If a line has1 2 slope a, what is the slope of its reflection across the line y = x? Domain of ƒ-1 = range of 1 ƒ = -3∞, 0 2. 92. For a one-to-one function ƒ, find ƒ- ∘ ƒ 2 , where ƒ 2 = 3. 1 4 Copyright Pearson. All Rights1 Reserved.21 2 1 2

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77. y 78. y Use a graphing calculator to graph each function defined as follows, using the given ƒ viewing window. Use the graph to decide which functions are one-to-one. If a function is ƒ –1 ƒ one-to-one, give the equation of its inverse. x y = x x 0 0 y = x 3 2 4 2 ƒ –1 93. ƒ x = 6x + 11x - 6; 94. ƒ x = x - 5x ; -3, 2 by -10, 10 -3, 3 by -8, 8 79. y 80. y 1 2 1 2 ƒ 3 4x - 35 4 3 4 -x3 4 95. ƒ x , x 3; 96. ƒ x , x 4; ƒ –1 = - = ƒ –1 y = x x + 3 x - 4 x x y = x 0 0 ƒ 1-8,2 8 by -6, 8 1-1,2 8 by -6, 6 3 4 3 4 3 4 3 4 81. y 82. y Use the following alphabet coding assignment to work each problem. See Example 9. ƒ ƒ –1 A 1 H 8 O 15 V 22 y = x x x B 2 I 9 P 16 W 23 0 ƒ 0 C 3 J 10 Q 17 X 24 –1 y = x ƒ D 4 K 11 R 18 Y 25 E 5 L 12 S 19 Z 26 83. 4 84. 3 85. 2 F 6 M 13 T 20 G 7 N 14 U 21 86. 0 87. -2 88. -4 97. The function ƒ x 3x 2 was used to encode a message as 89. It represents the cost, in = - dollars, of building 1000 cars. 37 25 191 2 61 13 34 22 1 55 1 52 52 25 64 13 10. 90. It represents the radius of a sphere with volume 5 in.3. Find the inverse function and determine the message. 91. 1 92. 2 a 98. The function ƒ x = 2x - 9 was used to encode a message as 93. not one-to-one 5 9 15 2 5 9 27 15 29 1 21 19 31 3 27 41. 94. not one-to-one - - - 95. one-to-one; Find the inverse function and determine the message. 5 3x ƒ-1 x = - - , x 1 x - 1 99. Encode the message SEND HELP, using the one-to-one function 96. one-to-one; 1 2 4x ƒ-1 x , x 1 3 = x + 1 - ƒ x = x - 1.

-11 2 1 2 97. ƒ x = 3 x + 3 ; Give the inverse function that the decoder1 2 will need when the message is received. MIGUEL HAS ARRIVED 1 2 100. Encode the message SAILOR BEWARE, using the one-to-one function -1 1 9 98. ƒ x = 2 x + 2 ; 3 BIG GIRLS DONT CRY ƒ x = x + 1 . 1 2 99. 6858 124 2743 63 511 124 1727 4095; Give the inverse function that the decoder1 2 1 will need2 when the message is received. 3 ƒ-1 x = x + 1 100. 8000 82 1000 2197 40961 2 6859 27 216 13824 8 6859 216; 3 ƒ-1 x = x - 1 2 1 2 4.2 Exponential Functions

■ Exponents and Exponents and Properties Recall the definition of am/n: If a is a real Properties n number, m is an integer, n is a positive integer, and a is a real number, then ■ Exponential Functions ■ Exponential Equations n m 2 am/n a . ■ Compound Interest ■ The Number e 3 3/4 4 ! 3 and Continuous For example, 16 = 16A =B 2 = 8, Compounding 2 ■ 1 1 1 1 1 1 Exponential Models 27-1/3 = = = A , andB 64-1/2 = = = . 271/3 3 27 3 641/2 64 8 2 2 Copyright Pearson. All Rights Reserved.

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Teaching Tip Use this opportu- In this section, we extend the definition of ar to include all real (not just nity to emphasize the distinction rational) values of the exponent r. Consider the graphs of y = 2x for different between exact and approximate domains in Figure 13. values. y y y

8 8 8

6 6 6

4 4 4 2Ί3 2 2 2 2 x x x –2 2 –2 2 –2 Ί3

x x y = 2 ; y = 2x; y = 2 ; integers as domain selected rational numbers real numbers as domain as domain Figure 13

The equations that use just integers or selected rational numbers as domain in Figure 13 leave holes in the graphs. In order for the graph to be continuous, we must extend the domain to include irrational numbers such as 3. We might evaluate 2 3 by approximating the exponent with the rational2 numbers 1.7, 2 1.73, 1.732, and so on. Because these values approach the value of 3 more and more closely, it is reasonable that 2 3 should be approximated 2more and more closely by the numbers 21.7, 21.73, 212.732, and so on. These expressions can be evaluated using rational exponents as follows.

10 17 21.7 = 217/10 = 2 ≈ 3.249009585 2 Because any irrational numberQ may beR approximated more and more closely using rational numbers, we can extend the definition of ar to include all real number exponents and apply all previous theorems for exponents. In addition to the rules for exponents presented earlier, we use several new properties in this chapter.

Additional Properties of Exponents

For any real number a 7 0, a 1, the following statements hold. Property Description (a) ax is a unique real number y = ax can be considered a function for all real numbers x. ƒ x = ax with domain -∞, ∞ . b c x (b) a a if and only if b c. The1 2 function ƒ x = a 1is one-to-one.2 3 4 (c) if a + 1 and m * n, then Example: 2 1622 a 7 1 m n a * a . Increasing the exponent leads to a greater number. The function1 2 ƒ x = 2x is an increasing function. 1 2 1 3 (d) if 0 * a * 1 and m * n, Example:1 2 2 7 2 0 6 a 6 1 m n then a + a . Increasing the exponent leads1 to 2 a lesser number.A B TheA Bfunction 1 x ƒ x = 2 is a decreasing function. 1 2 Copyright Pearson. All Rights Reserved. A B

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Exponential Functions We now define a function ƒ x = ax whose domain is the set of all real numbers. Notice how the independent variable x appears in the exponent in this function. In earlier chapters,1 2this was not the case.

Exponential Function

If a 7 0 and a 1, then the exponential function with base a is ƒ x ax.

1 2

Note The restrictions on a in the definition of an exponential function are important. Consider the outcome of breaking each restriction. 1 1 1/2 If a 6 0, say a = -2, and we let x = 2 , then ƒ 2 = -2 = -2, which is not a real number. 2 1 2 If a = 1, then the function becomes the constant functionA B ƒ x = 1x = 1, which is not an exponential function. 1 2

Classroom Example 1 Example 1 Evaluating an Exponential Function For ƒ x 4x, find each of the = x following. For ƒ x = 2 , find each of the following. 1 2 (a) ƒ -2 (b) ƒ 5 5 (a) ƒ1 -21 (b) ƒ 3 (c) ƒ (d) ƒ 4.92 2 2 (c) ƒ1 2 (d) ƒ12.152 3 Solut1 io2n 1 2 a b 1 2 Answers:a b 1 2 1 3 1 (a) (b) 1024 (c) 2 2 -1 3 16 (a) ƒ -1 = 2 = Replace x with -1. (b) ƒ 3 = 2 = 8 2 (d) 19.69831061 2 1 2 1 2 5 (c) ƒ = 25/2 = 25 1/2 = 321/2 = 32 = 16 # 2 = 4 2 2 2 2 2 a b 1 2 (d) ƒ 4.92 = 24.92 ≈ 30.2738447 Use a calculator. 1 2 ■✔ Now Try Exercises 13, 19, and 23.

We repeat the final graph of y = 2x (with real num- y bers as domain) from Figure 13 and summarize impor- x 8 f(x) = 2 tant details of the function ƒ x = 2x here. 6 • The y-intercept is 0, 1 .1 2 x x • Because 2 7 0 for1 all2 x and 2 S 0 as x S -∞, 4 (2, 4) the x-axis is a horizontal asymptote. 2 (1, 2) –1, 1 • As the graph suggests, the domain of the function is Q 2R (0, 1) ∞, ∞ and the range is 0, ∞ . x - –22 The function is increasing on its entire domain. • 1 2 1 2 Graph of ƒ x 2x with Therefore, it is one-to-one. domain H, H 1 2 These observations lead to the following generaliza- Figure1 13 2 tions about the graphs of exponential functions. (repeated)

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Exponential Function f x ax

Domain: -∞, 1∞ 2 Range: 0, ∞ For ƒ x = 2x: 1 2 1 f(2x) = ax, a > 1 f(x) = a x, a > 1 x1 2 ƒ x y 2 11 2 - 4 -1 1 (1, a) 2 –1, 1 0 1 ( a) (0, 1) x 1 2 0 2 4 3 8

This is the general behavior seen on a calculator graph for any base a, for a + 1. Figure 14 x • ƒ x a , for a + 1, is increasing and continuous on its entire domain, -∞, ∞ . 1 2 • The1 x-axis2 is a horizontal asymptote as x S -∞. 1 • The graph passes through the points -1, a , 0, 1 , and 1, a .

1 x 1 2 1 2 For ƒ x = : A B 2 f(x) = ax, 0 < a < 1 x ƒ x y 1 2 A B x -3 18 2 f(x) = a , 0 < a < 1 2 4 1 - (–1, a ) 1 2 (1, a) - (0, 1) 0 1 x 1 0 1 2 1 2 4

This is the general behavior seen on a calculator graph for any base a, for 0 * a * 1. Figure 15 x • ƒ x a , for 0 * a * 1, is decreasing and continuous on its entire domain, -∞, ∞ . 1 2 • The x-axis1 is a horizontal2 asymptote as x S ∞. 1 • The graph passes through the points -1, a , 0, 1 , and 1, a .

A B 1 2 1 2 Recall that the graph of y = ƒ -x is the graph of y = ƒ x reflected across the y-axis. Thus, we have the following. 1 2 1 2 1 x If ƒ x = 2x, then ƒ -x = 2-x = 2-1# x = 2-1 x = . 2

This is supported1 2 by the graphs in1 Figures2 14 and 15. 1 2 a b The graph of ƒ x = 2x is typical of graphs of ƒ x = ax where a 7 1. For larger values of a, the graphs rise more steeply, but the general shape is similar to the graph in Figure1 2 14. When 0 6 a 6 1, the graph1 2 decreases in a manner 1 x similar to the graph of ƒ x = 2 in Figure 15. Copyright Pearson. All Rights Reserved. 1 2 A B

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In Figure 16, the graphs of several typical exponential functions illustrate these facts. y

1 x f(x) = 5 x Teaching Tip The point 0, 1 For 0 < a < 1, (10 ) f(x) = 10 For a > 1, x x the function 1 f(x) = 3 the function lies on the graph of f(x) = ( 3 ) 4 1 2 is decreasing. x x is increasing. x 1 f(x) = 2 ƒ x = a f(x) = ( 2 ) 3 for all a. We can1 2 informally consider it “home plate” when graphing translations and reflections, as seen later in The x-axis is Example 3. x a horizontal 0 –3 –2 –1 123 asymptote. f(x) = ax Domain: (–∞, ∞); Range: (0, ∞)

Figure 16

In summary, the graph of a function of the form ƒ x = ax has the following features. 1 2

Teaching Tip Encourage x students to learn the characteristics Characteristics of the Graph of f x a of the graph of 1 1 2 x 1. The points -1, , 0, 1 , and 1, a are on the graph. ƒ x = a . a 2. If a 7 1, then ƒ is an increasing function. Ask them how1 the2 coordinates A B 1 2 1 2 0, 1 and 1, a are affected by If 0 6 a 6 1, then ƒ is a decreasing function. different variations of the graph of 1 2 1 2 3. The x-axis is a horizontal asymptote. ƒ x = -ax-h + k. 4. The domain is -∞, ∞ , and the range is 0, ∞ . 1 2 1 2 1 2

Classroom Example 2 Example 2 Graphing an Exponential Function x 1 x Graph ƒ x = . Give the 1 2 Graph ƒ x = 5 . Give the domain and range. domain and1 2 range.a b Solutio1 n2 The y-intercept is 0, 1 , and the x-axis is a horizontal asymptote. Answer: y Plot a few orderedA B pairs, and draw a smooth curve through them as shown in Figure 17. 1 2

x ( 1 ) ؍ (f(x 4 2 x ƒ x y

1 x -2 251 2 0 (–2, 25) 25 –4 –2 42 -1 5 1 x 20 f(x) = ( ) -∞, ∞ ; 0, ∞ 0 1 5 1 15 1 5 1 2 1 2 10 1 1, 1 2 ( 5) 25 (–1, 5) 5 (0, 1) x –1–2 0 12

Figure 17

This function has domain -∞, ∞ , range 0, ∞ , and is one-to-one. It is decreasing on its entire domain. 1 2 1 2 ■✔ Now Try Exercise 29.

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Classroom Example 3 Example 3 Graphing Reflections and Translations Graph each function. Give the x domain and range. Graph each function. Show the graph of y = 2 for comparison. Give the domain and range. (a) ƒ x = -3x (b) ƒ x = 3x-2 (c) ƒ x = 3x+2 - 2 (a) ƒ x 2x (b) ƒ x 2x+3 (c) ƒ x 2x-2 1 1 2 1 2 = - = = - Answers: 1 2 Solut1 io2 n In each graph, we show1 2 in particular how the1 point2 0, 1 on the (a) y x x graph of y = 2 has been translated. f(x) = –3 1 2 0 x (a) The graph of ƒ x 2x is that of ƒ x 2x reflected across the x-axis. See –1 2 = - = Figure 18. The domain is -∞, ∞ , and the range is -∞, 0 . 1 2 1 2 –5 (b) The graph of ƒ x = 2x+3 1is the graph2 of ƒ x = 2x translated1 2 3 units to the left, as shown in Figure 19. The domain is -∞, ∞ , and the range is 0, ∞ . 1 2 1 2 (c) The graph of ƒ x 2x-2 1 is that of ƒ x 2x translated 2 units to the -∞, ∞ ; -∞, 0 = - 1 = 2 1 2 right and 1 unit down. See Figure 20. The domain is ∞, ∞ , and the range (b) y - 1 f2(x)1 = 3x – 2 2 is -1, ∞ . 1 2 1 2 1 2 x؊2 y 2 y y f(x) = 2 ؊ 1 1 y = 2x

y = 2x 8 1 x 2 0 2 6 –2 (0, 1) 4 0 x -∞, ∞ ; 0, ∞ f(x) = 2x+3 4 ؊1) 2 x ,0) y y = 2 (c) –2 f(x) = –2x 2 (0, 1) f1(x) = 3x2 + 2 ؊ 2 (؊3, 1) (2, 0) 4 2 1 (0, 1) x 7 –4 x 0 0 –6 –3 2 y ϭ Ϫ1

2 Figure 18 Figure 19 Figure 20 ؊2 x 0 2 ■✔ Now Try Exercises 39, 41, and 47. y = ؊2 -∞, ∞ ; -2, ∞ Exponential Equations Because the graph of ƒ x ax is that of a 1 2 1 2 = x x one-to-one function, to solve a 1 = a 2, we need only show that x1 = x2. This property is used to solve an exponential equation, which1 is2 an equation with a variable as exponent.

Classroom Example 4 Example 4 Solving an Exponential Equation 1 Solve 5x = . 1 x 125 Solve 3 = 81. Answer: -3 Solution Write each side of the equation using a common base. A B 5 6 1 x = 81 3 100 a3-1b x = 81 Definition of negative exponent -x m n mn 1 32 = 81 a = a −5 1 3-x = 34 Write1 2 81 as a power of 3. -x = 4 Set exponents equal (Property (b) given earlier). −100 x = -4 Multiply by -1. The x-intercept of the graph of 1 x Check by substituting -4 for x in the original equation. The solution set is -4 . y = 3 - 81 can be used to verify the solution in Example 4. ■✔ Now Try Exercise5 73.6 A B Copyright Pearson. All Rights Reserved.

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Classroom Example 5 Example 5 Solving an Exponential Equation Solve 3x+1 = 9x-3. Solve 2x+4 = 8x-6. Answer: 7 Solution Write each side of the equation using a common base. 5 6 2x+4 = 8x-6 Teaching Tip Now is a good time to review the properties of 2x+4 = 23 x-6 Write 8 as a power of 2. exponents. x+4 3x-18 m n mn 2 = 12 2 a = a x + 4 = 3x - 18 Set1 2exponents equal (Property (b)). -2x = -22 Subtract 3x and 4. x = 11 Divide by -2. Teaching Tip Give an example such as Check by substituting 11 for x in the original equation. The solution set is 11 . 16x-2 = 91-x ■✔ Now Try Exercise5 81.6 that cannot be solved using the methods of this section. Show how Later in this chapter, we describe a general method for solving exponential a graphing calculator can be used equations where the approach used in Examples 4 and 5 is not possible. For to find the approximate solution instance, the above method could not be used to solve an equation like 1.558. 7x = 12 because it is not easy to express both sides as exponential expressions with the same base. In Example 6, we solve an equation that has the variable as the base of an exponential expression.

Classroom Example 6 Example 6 Solving an Equation with a Fractional Exponent Solve x2/3 = 25. Solve x4/3 = 81. Answer: {125 Solution Notice that the variable is in the base rather than in the exponent. 5 6 x4/3 = 81 Teaching Tip Caution students 3 4 that in solving equations like the x = 81 Radical notation for am/n one in Example 6, they must consider both even roots. 2 3 Take fourth roots on each side. x = 3 A B { Remember to use {. 2 x = {27 Cube each side. Check both solutions in the original equation. Both check, so the solution set is {27 . Alternative5 6 Method There may be more than one way to solve an exponential equation, as shown here.

x4/3 = 81 x4/3 3 = 813 Cube each side.

4 4 3 4 1 x2 = 3 Write 81 as 3 . 4 12 m n mn x = 13 2 a = a 4 x = { 312 Take1 2 fourth roots on each side. 23 x = {3 Simplify the radical. x = {27 Apply the exponent. The same solution set, {27 , results. ■✔ Now Try Exercise 83. Copyright Pearson. All Rights5 Reserved.6

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Compound Interest Recall the formula for simple interest, I = Prt, where P is principal (amount deposited), r is annual rate of interest expressed as a decimal, and t is time in years that the principal earns interest. Suppose t = 1 yr. Then at the end of the year, the amount has grown to the following. P + Pr = P 1 + r Original principal plus interest If this balance earns interest1 at the2 same interest rate for another year, the balance at the end of that year will increase as follows.

P 1 + r + P 1 + r r = P 1 + r 1 + r Factor.

2 2 3 1 24 3 1 24 = P3 1 + r 241 2 a # a = a After the third year, the balance will grow1 in a similar2 pattern.

P 1 + r 2 + P 1 + r 2 r = P 1 + r 2 1 + r Factor.

3 2 3 3 1 2 4 3 1 2 4 = P3 1 + r 2 41 2 a # a = a Continuing in this way produces a formula1 for interest2 compounded annually. A P 1 r t

The general formula for compound 1interest2 can be derived in the same way.

Compound Interest

If P dollars are deposited in an account paying an annual rate of interest r compounded (paid) n times per year, then after t years the account will contain A dollars, according to the following formula.

r tn A P 1 n a b

Classroom Example 7 Example 7 Using the Compound Interest Formula Repeat Example 7 if $2500 is deposited in an account paying Suppose $1000 is deposited in an account paying 4% interest per year com- 3% per year compounded pounded quarterly (four times per year). semi-annually (twice per year). (a) Find the amount in the account after 10 yr with no withdrawals. Answers: (b) How much interest is earned over the 10-yr period? (a) $3367.14 (b) $867.14 Solution r t n (a) A P 1 Compound interest formula = + n a b 0.04 10 4 A = 1000 1 + Let P = 1000, r = 0.04, n = 4, and t = 10. 4 1 2 A = 1000a1 + 0.01 40b Simplify. A = 1488.861 2 Round to the nearest cent. Thus, $1488.86 is in the account after 10 yr.

(b) The interest earned for that period is $1488.86 - $1000 = $488.86. ■✔ Now Try Exercise 97(a).

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In the formula for compound interest

r t n A P 1 , = + n A is sometimes called the future valuea and P bthe present value. A is also called the compound amount and is the balance after interest has been earned.

Classroom Example 8 Example 8 Finding Present Value Brendan must pay a lump sum of $15,000 in 8 yr. Becky must pay a lump sum of $6000 in 5 yr. (a) What amount deposited (a) What amount deposited today (present value) at 3.1% compounded annually today (present value) at 2.4% will grow to $6000 in 5 yr? compounded annually will give $15,000 in 8 yr? (b) If only $5000 is available to deposit now, what annual interest rate is neces- (b) If only $12,025 is available sary for the money to increase to $6000 in 5 yr? to deposit now, what annual Solution interest rate is necessary for the money to increase to r t n (a) A P 1 Compound interest formula $15,000 in 8 yr? = + n Answers: a 0.031b 5 1 (a) $12,407.71 (b) 2.80% 6000 = P 1 + Let A = 6000, r = 0.031, n = 1, and t = 5. 1 1 2 6000 = P a1.031 5 b Simplify. 6000 P = 1 2 Divide by 1.031 5 to solve for P. 1.031 5 1 2 P ≈ 15150.602 Use a calculator. If Becky leaves $5150.60 for 5 yr in an account paying 3.1% compounded annually, she will have $6000 when she needs it. Thus, $5150.60 is the present value of $6000 if interest of 3.1% is compounded annually for 5 yr. r t n (b) A P 1 Compound interest formula = + n 5 6000 = 5000a 1 +br Let A = 6000, P = 5000, n = 1, and t = 5. 6 = 1 +1r 5 2 Divide by 5000. 5 6 1/5 1 2 = 1 + r Take the fifth root on each side. 5 6 a1/5 b - 1 = r Subtract 1. 5 a b r ≈ 0.0371 Use a calculator. An interest rate of 3.71% will produce enough interest to increase the $5000 to $6000 by the end of 5 yr.

■✔ Now Try Exercises 99 and 103.

Caution When performing the computations in problems like those in Examples 7 and 8, do not round off during intermediate steps. Keep all calculator digits and round at the end of the process.

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1 n n 1 The Number e and Continuous Compounding The more often inter- n est is compounded within a given time period, the more interest will be earned. a(rounded)b Surprisingly, however, there is a limit on the amount of interest, no matter how 1 2 often it is compounded. 2 2.25 Suppose that $1 is invested at 100% interest per year, compounded n times 5 2.48832 per year. Then the interest rate (in decimal form) is 1.00, and the interest rate per 10 2.59374 1 100 2.70481 period is n . According to the formula (with P = 1), the compound amount at the 1000 2.71692 end of 1 yr will be 10,000 2.71815 1 n 1,000,000 2.71828 A 1 . = + n A calculator gives the results in the margina forb various values of n. The table sug- 1 n gests that as n increases, the value of 1 + n gets closer and closer to some fixed number. This is indeed the case. This fixed number is called e. (In mathematics, e is a real number and not a variable.)A B

Value of e y y = 3x e ? 2.718281828459045 8 y = e x 7 6 Figure 21 shows graphs of the functions 5 x x x 4 y = 2x y = 2 , y = 3 , and y = e . 3 x 2 Because 2 6 e 6 3, the graph of y = e lies “between” the other two graphs. 1 As mentioned above, the amount of interest earned increases with the fre- x n 0 1 –1 12 quency of compounding, but the value of the expression 1 + n approaches e as n gets larger. Consequently, the formula for compound interest approaches Figure 21 a limit as well, called the compound amount from continuousA compounding.B

Continuous Compounding Teaching Tip Tell students that as n increases without bound, If P dollars are deposited at a rate of interest r compounded continuously 1 n 1 + n approaches e. for t years, then the compound amount A in dollars on deposit is given by the following formula. A B A Pert

Classroom Example 9 Example 9 Solving a Continuous Compounding Problem Suppose $8000 is deposited in an account paying 5% interest Suppose $5000 is deposited in an account paying 3% interest compounded con- compounded continuously for 6 yr. tinuously for 5 yr. Find the total amount on deposit at the end of 5 yr. Find the total amount on deposit at Solution the end of 6 yr. rt Answer: $10,798.87 A = Pe Continuous compounding formula 0.03 5 A = 5000e Let P = 5000, r = 0.03, and t = 5. 1 2 A = 5000e0.15 Multiply exponents. A ≈ 5809.17 or $5809.17 Use a calculator. Check that daily compounding would have produced a compound amount about $0.03 less. ■✔ Now Try Exercise 97(b).

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Classroom Example 10 Example 10 Comparing Interest Earned as Compounding Is More In Classroom Example 7, we Frequent found the total amount on deposit after 10 yr in an account paying In Example 7, we found that $1000 invested at 4% compounded quarterly for 3% semiannually in which $2500 10 yr grew to $1488.86. Compare this same investment compounded annually, was invested. Find the amounts semiannually, monthly, daily, and continuously. from the same investment for Solution interest compounded quarterly, Substitute 0.04 for r, 10 for t, and the appropriate number of com- monthly, daily, and continuously. pounding periods for n into the formulas Answer: r t n A P 1 Compound interest formula quarterly: $3370.87; = + n monthly: $3373.38; daily: $3374.61; and A = Peart. b Continuous compounding formula continuously: $3374.65 The results for amounts of $1 and $1000 are given in the table.

Compounded $1 $1000 Annually 1 + 0.04 10 ≈ 1.48024 $1480.24 1 0.04 1022 Semiannually 1 + ≈ 1.48595 $1485.95 2 1 2 a 0.04 b 10 4 Quarterly 1 + ≈ 1.48886 $1488.86 4 1 2 a 0.04 b10 12 Monthly 1 + ≈ 1.49083 $1490.83 12 1 2 a 0.04 b10 365 Daily 1 + ≈ 1.49179 $1491.79 365 1 2 Continuously a eb10 0.04 ≈ 1.49182 $1491.82 1 2

Comparing the results for a $1000 investment, we notice the following. • Compounding semiannually rather than annually increases the value of the account after 10 yr by $5.71. • Quarterly compounding grows to $2.91 more than semiannual compounding after 10 yr. • Daily compounding yields only $0.96 more than monthly compounding. • Continuous compounding yields only $0.03 more than daily compounding. Looking Ahead To Calculus In calculus, the derivative allows us Each increase in compounding frequency earns less additional interest. to determine the slope of a tangent ■✔ Now Try Exercise 105. line to the graph of a function. For the function ƒ x = ex, Exponential Models The number e is important as the base of an expo- the derivative is1 the2 function ƒ itself: nential function in many practical applications. In situations involving growth or decay of a quantity, the amount or number present at time t often can be closely ƒ x ex. ′ = modeled by a function of the form Therefore, in calculus1 2 the exponential kt function with base e is much easier to y y0e , work with than exponential functions where y is the amount or number present at time t 0 and k is a constant. having other bases. 0 = Exponential functions are used to model the growth of microorganisms in a culture, the growth of certain populations, and the decay of radioactive material.

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Classroom Example 11 Example 11 Using Data to Model Exponential Growth Refer to the model given in Example 11. Data from recent years indicate that future amounts of carbon dioxide in the (a) What will be the atmospheric atmosphere may grow according to the table. Amounts are given in parts per carbon dioxide level in parts million. per million in 2015? (a) Make a scatter diagram of the data. Do Year Carbon Dioxide (ppm) (b) Use a graph of this model the carbon dioxide levels appear to grow 1990 353 to estimate when the carbon exponentially? dioxide level will be double 2000 375 the level that it was in 2000. (b) One model for the data is the function 2075 590 Answers: y = 0.001942e0.00609x, 2175 1090 (a) 415 ppm (b) by 2112 2275 2000 where x is the year and 1990 … x … 2275. Use a graph of this model to estimate Source: International Panel on Climate when future levels of carbon dioxide will Change (IPCC).

2100 double and triple over the preindustrial level of 280 ppm. Solution

(a) We show a calculator graph for the data in Figure 22(a). The data appear to 1975 2300 300 resemble the graph of an increasing exponential function. (a) (b) A graph of y = 0.001942e0.00609x in Figure 22(b) shows that it is very close to the data points. We graph y 2 280 560 in Figure 23(a) and y = 0.001942e0.00609x 2 = # = y2 = 3 # 280 = 840 in Figure 23(b) on the same coordinate axes as the given 2100 function, and we use the calculator to find the intersection points.

0.00609x 0.00609x y1 = 0.001942e y1 = 0.001942e

1975 2300 300 2100 2100 (b) Figure 22

1975 2300 1975 2300

−500 −500

y2 = 560 y2 = 840 (a) (b) Figure 23

The graph of the function intersects the horizontal lines at x-values of approximately 2064.4 and 2130.9. According to this model, carbon dioxide levels will have doubled during 2064 and tripled by 2131. (a) ■✔ Now Try Exercise 107. x y1 = 0.001923(1.006109)

2100 Graphing calculators are capable of fitting exponential curves to scatter diagrams like the one found in Example 11. The TI-84 Plus displays another (different) equation in Figure 24(a) for the atmospheric carbon dioxide example, approximated as follows.

1975 2300 x 300 y = 0.001923 1.006109

(b) This calculator form differs from the model1 in Example2 11. Figure 24(b) shows Figure 24 the data points and the graph of this exponential regression equation. ■

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4.2 Exercises

1 Concept Preview Fill in the blank(s) to correctly complete each sentence. 1. 16; 16 2. rises 3. falls 1. If ƒ x = 4x, then ƒ 2 = and ƒ -2 = . 4. -∞, ∞ ; 0, ∞ x 1 2. If a1721, then the graph1 2 of ƒ x = a 1 2 from left to right. 5. 18 ; 1; 8 2 1 6.2 x; 4; 5 (rises/falls) 1 2 7. -3 8. {216 3. If 0 6 a 6 1, then the graph of ƒ x = ax from left to right. 9. 2540.22 10. 0.03 (rises/falls) 5 6 5 6 1 2 4. The domain of ƒ x 4x is and the range is . 5 6 51 6 1 = 11. 9 12. 27 13. 9 14. 27 1 1 5. The graph of ƒ x1 2= 8x passes through the points (-1, ), (0, ), and (1, ). 15. 16 16. 64 17. 16 18. 64 x 4 x 19. 3 3 20. 3 6. The graph of 1ƒ2x 1 + 5 is that of ƒ x 1 reflected across the 227 = - 3 - = 3 21. 1 2 22. 32 -axis, translated units to the left and units down. 8 1 2 1 2 23. 13.076 24. 0.158 A B A B 25. 10.267 26. 0.039 Concept Preview Solve each equation. Round answers to the nearest hundredth 27. y 28. y as needed. x 9 1 8 7. = 64 8. x2/3 = 36 x 4 f(x) = 3 4 a b 1 x x f(x) = 4 8 4 0 0.03 25 2 x 9. A = 2000 1 + 10. 10,000 = 5000 1 + r –10 1 4 1 2 29. y 30. y a b 1 2

4 x x 3 1 x 1 x 1 f(x) = f(x) = For ƒ x 3 and g x , find each of the following. Round answers to the nearest ( 3 ) ( 4) = = 4 1 1 x x –10 1 –10 1 thousandth as needed. See Example 1. 1 2 1 2 A B 11. ƒ 2 12. ƒ 3 13. ƒ -2 14. ƒ -3 31. y 32. y 15. g122 16. g132 17. g1-22 18. g1-32 1 32 1 2 5 1 3 2 1 52 19. ƒ 20. ƒ - 21. g 22. g - 5 x x 3 f(x) = 5 2 2 2 2 f(x) = 3 ( 3 ) ( 2 ) 1 1 x a b a b a b a b x 0 1–1 23. ƒ 2.34 24. ƒ -1.68 25. g -1.68 26. g 2.34 0 3

33. y 34. y 1 2 1 2 1 2 1 2 10 Graph each function. See Example 2.

6 x x x 1 1 –x 27. ƒ x 3 28. ƒ x 4 29. ƒ x f(x) = = = = ؊x ( 6) 3 1 f(x) = 1 (10) x 0 1 1–1 1 2 x 1 2 x 1 2 a b x x 1 3 5 –2 –1 0 12 30. ƒ x = 31. ƒ x = 32. ƒ x = 4 2 3 35. y 36. y 1 2 a 1b -x 1 2 a 1 b -x 1 2 a b –x 10 x f(x) = 4 33. ƒ x = 34. ƒ x = 35. ƒ x = 4- 10 6 4 –1 1 2 a -x b 1 2 a͉x͉ b 1 2 -͉x͉ x –x 36. ƒ x = 10 37. ƒ x = 2 38. ƒ x = 2 0 f(x) = 10 1 x –20 2 1 2 1 2 1 2 Graph each function. Give the domain and range. See Example 3. 37. y 38. y 39. ƒ x = 2x + 1 40. ƒ x = 2x - 4 41. ƒ x = 2x+1 8 1 f(x) = 2–͉ x͉ 42. ƒ x = 2x-4 43. ƒ x = -2x+2 44. ƒ x = -2x-3 1 1 2 1 2 1 2 2 1 -x -x x-1 x 45. ƒ1x2 = 2 46. ƒ1x2 = -2 47. ƒ1x2 = 2 + 2 0 x 3 0 x –2 2 f(x) = 2͉ ͉ 48. ƒ1x2 = 2x+3 + 1 49. ƒ1x2 = 2x+2 - 4 50. ƒ1x2 = 2x-3 - 1

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39. y x 40. y Graph each function. Give the domain and range. See Example 3. f(x) = 2 + 1 f(x) = 2x – 4 x x x+2 8 1 1 1 2 51. ƒ x = - 2 52. ƒ x = + 4 53. ƒ x = x 3 3 3 0 2 1 2 a b 1 2 a b 1 2 a b 2 y = 1 –3 1 x-4 1 -x+1 1 -x-2 x y = –4 54. ƒ x = 55. ƒ x = 56. ƒ x = –2 0 2 3 3 3 ∞, ∞ ; 1, ∞ ∞, ∞ ; 4, ∞ 1 2 a b 1 2 a b 1 2 a b - - - 1 -x 1 -x 1 x-2 57. ƒ x = 58. ƒ x = - 59. ƒ x = + 2 41.1 y2 1 2 42.1 y 2 1 2 3 3 3 f(x) = 2x – 4 1 2 a b x 1 1 2 a xb2 1 2 a b x 3 8 8 1 - 1 + 1 + 60. ƒ x = + 3 61. ƒ x = - 1 62. ƒ x = - 2 3 3 3 x + 1 2 f(x) = 2 2 1 2 a b 1 2 a b 1 2 a b x x Connecting Graphs with Equations Write an equation for the graph given. Each repre- 0 0 4 4 sents an exponential function ƒ with base 2 or 3, translated and/or reflected. -∞, ∞ ; 0, ∞ -∞, ∞ ; 0, ∞ 63. y 64. y 65. y 43.1 y 2 1 2 44.1 y 2 1 2 (4, 9) –2 5 (2, 7) x 0 x (0, 7) 0 2 –4 –4 f(x) = –2x + 2 (3, 3) (–2, 1) x – 3 (1, 1) (–3, 0) f(x) = –2 x 0 (2, 1) x (0, –1) x 0 -∞, ∞ ; -∞, 0 -∞, ∞ ; -∞, 0 0 y = –1 y = –2 45.1 y 2 1 2 46.1 y 2 1 2 y y y –2 2 66. 67. 68. 0 x –1 (–2, 2) 4 –x –x f(x) = –2 y = 3 f(x) = 2 –4 (1, 7) x 0 (–2, 1) –4 2 (–1, 4) (0, 5) (–1, 1) 0 x x 0 -∞, ∞ ; 0, ∞ -∞, ∞ ; -∞, 0 y = 3 (0, –1) (–1, –1) (0, –2)

y x – 1 y x y = –3 2 1 2 0 f(2x) =1 2 2؉ 2 48.1 .147

4 y y 2 y = 1 69. 70. y = 2 y = 5 x x ؊2 0 2 21 0 x + 3 (0, 4) (f(x) = 2 ؉ 1 (–1, 3 (–1, 4) 4 , ; 2, ∞, ∞ ; 1, ∞ 1, -∞ ∞ ∞ - (0, 2) ( 3) y = 1 x x 0 49.1 2 1 2 50.1 2 1 2 0 y x – 3 (y f(x) = 2 ؊ 1 (–3, –3

3 2 ؊2 1 3 x 0 x Solve each equation. See Examples 4–6. 0 2 ؊2 y = ؊1 x x x x 5 4 2 9 y = 4 71. 4 = 2 72. 125 = 5 73. = 74. = ؊ 2 25 3 4 f(x) = 2x + 2 ؊ 4 a b a b -∞, ∞ ; -4, ∞ -∞, ∞ ; -1, ∞ 75. 23-2x = 8 76. 52+2x = 25 77. e4x-1 = e2 x 78. e3-x = e3 -x 1 2 1 2 1 2 1 2 79. 274x 9x+1 80. 322x 16x-1 81. 4x-2 23x+3 82. 26-3x 8x+1 51. y 52. y = = = 1 2 =1 2 2/3 2/5 5/2 3/2 9 10 83. x = 4 84. x = 16 85. x = 32 86. x = 27

5 1 1 -6 -4 5/3 7/5 3 y = 4 87. x = 88. x = 89. x = -243 90. x = -128 3 64 256 0 x y = –2 x 0 3 -x x+1 x+1 1 x 1 x 1 1 1 x 4 f(x) = ( ) – 2 f(x) = + 4 x 1 + x 3 ( 3 ) 91. = 92. e - = 93. 2 = 4 e e2 e4 -∞, ∞ ; -2, ∞ -∞, ∞ ; 4, ∞ 2 a b a b x+2 a b 3 -x 1 1 A1 B 1 2 1 2 1 2 1 2 94. 5 = 95. = x-3 96. = x-5 5 27 32 2 a b A B Copyright Pearson. All Rights Reserved.

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53. y 54. y Solve each problem. See Examples 7–9. f(x) = 1 x + 2 ( 3 ) 9 8 f(x) = 1 x – 4 97. Future Value Find the future value and interest earned if $8906.54 is invested for ( 3 ) 4 9 yr at 3% compounded 3 (a) semiannually (b) continuously. 0 x x 4 0 –2 2 98. Future Value Find the future value and interest earned if $56,780 is invested at -∞, ∞ ; 0, ∞ -∞, ∞ ; 0, ∞ 2.8% compounded 55.1 y 2 1 2 56.1 y 2 1 2 (a) quarterly for 23 quarters (b) continuously for 15 yr. 9 9 99. Present Value Find the present value that will grow to $25,000 if interest is 3.2% 6 compounded quarterly for 11 quarters. 3

x x 100. Present Value Find the present value that will grow to $45,000 if interest is 3.6% 0 3 0 –2 –3–2 1 –x + 1 compounded monthly for 1 yr. f(x) = f(x) = 1 –x – 2 ( 3 ) ( 3 ) 101. Present Value Find the present value that will grow to $5000 if interest is 3.5% , ; 0, , ; 0, -∞ ∞ ∞ -∞ ∞ ∞ compounded quarterly for 10 yr. 57. y 58. y 1 2 1 2 1 2 1 2 102. Interest Rate Find the required annual interest rate to the nearest tenth of a percent 9 2 0 –x x for $65,000 to grow to $65,783.91 if interest is compounded monthly for 6 months. f(x) = 1 ( 3 ) –3 3 –3 –x 3 f(x) = – 1 103. Interest Rate Find the required annual interest rate to the nearest tenth of a percent ( 3 ) x 0 for $1200 to grow to $1500 if interest is compounded quarterly for 9 yr. 3 –9 -∞, ∞ ; 0, ∞ -∞, ∞ ; -∞, 0 104. Interest Rate Find the required annual interest rate to the nearest tenth of a percent for $5000 to grow to $6200 if interest is compounded quarterly for 8 yr. 59.1 2 1 2 160. y 2 1 2 y 1 x – 2 f(x) = ؉ 2 1 x – 1 .f(x) = ؉ 3 Solve each problem. See Example 10 ( 3 ) ( 3 ) 6 105. Comparing Loans Bank A is lending money at 6.4% interest compounded annu- 3 ally. The rate at Bank B is 6.3% compounded monthly, and the rate at Bank C is 2 y = 3 y = 2 x 6.35% compounded quarterly. At which bank will we pay the least interest? x 0 0 2 2 106. Future Value Suppose $10,000 is invested at an annual rate of 2.4% for 10 yr. -∞, ∞ ; 2, ∞ -∞, ∞ ; 3, ∞ Find the future value if interest is compounded as follows. 61.1 2y 1 2 1 2 1 2 (a) annually (b) quarterly (c) monthly (d) daily (365 days)

8 (Modeling) Solve each problem. See Example 11. x + 2 f(x) = 1 ؊ 1 ( 3 ) 107. Atmospheric Pressure The atmospheric pressure (in millibars) at a given altitude (in meters) is shown in the table. ؊2 2 0 x Altitude Pressure Altitude Pressure ؊1 y = ؊1 0 1013 6000 472 -∞, ∞ ; -1, ∞ 1000 899 7000 411 162. 2 y1 2 2000 795 8000 357 7 3000 701 9000 308

x + 3 4000 617 10,000 265 f(x) = 1 ؊ 2 ( 3 ) 5000 541 x ؊5 0 Source: Miller, A. and J. Thompson, Elements of y = ؊2 Meteorology, Fourth Edition, Charles E. Merrill Publishing Company, Columbus, Ohio. -∞, ∞ ; -2, ∞ (a) Use a graphing calculator to make a scatter diagram of the data for atmospheric 63.1 ƒ x2 =1 3x - 2 pressure P at altitude x. 64. ƒ x = 3x-2 (b) Would a linear or an exponential function fit the data better? 1 2 65. ƒ x = 2x+3 - 1 1 2 (c) The following function approximates the data. 66. ƒ x = 2x+1 + 3 1 2 P x = 1013e-0.0001341x 67. ƒ x = -2x+2 + 3 1 2 68. ƒ x = 2-x - 3 Use a graphing calculator to graph1 2 P and the data on the same coordinate axes. 1 2 69. ƒ x = 3-x + 1 (d) Use P to predict the pressures at 1500 m and 11,000 m, and compare them to 1 2 70. ƒ x = -2-x + 5 the actual values of 846 millibars and 227 millibars, respectively. 1 2 1 2 Copyright Pearson. All Rights Reserved.

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71. 1 72. 1 108. World Population Growth World population in millions closely fits the exponen- 2 3 tial function 73. -2 74. -2 0.0120x 75. E0F 76. E0F ƒ x = 6084e , 5 6 5 6 1 3 where x is the number of years since1 2 2000. (Source: U.S. Census Bureau.) 77. 52 6 78. 5 -6 2 (a) The world population was about 6853 million in 2010. How closely does the 79. 1 80. 2 E 5 F E - 3 F function approximate this value? 81. 7 82. 1 (b) Use this model to predict world population in 2020 and 2030. E -F E 2 F 83. 5{86 84. {1024 109. Deer Population The exponential 85. 4 86. E9F growth of the deer population in Mas- 5 6 5 6 sachusetts can be approximated using the 87. {2 88. {4 5 6 5 6 model 89. -27 90. -32 5 6 5 6 2 3 ƒ x = 50,000 1 + 0.06 x, 91. 5 - 3 6 92. 5 - 5 6 where 50,000 is the initial deer population 93. 4 94. 3 1 2 1 2 E 3 F E - F and 0.06 is the rate of growth. ƒ x is the 95. 3 96. 52 6 total population after x years have passed. E F Find each value to the nearest thousand.1 2 97. (a)5 6 $11,643.88; $2737.345 6 (b) $11,667.25; $2760.71 (a) Predict the total population after 4 yr. 98. (a) $66,661.21; $9881.21 (b) If the initial population was 30,000 and (b) $86,416.98; $29,636.98 the growth rate was 0.12, how many 99. $22,902.04 100. $43,411.15 deer would be present after 3 yr? 101. $3528.81 102. 2.4% (c) How many additional deer can we expect in 5 yr if the initial population is 45,000 103. 2.5% 104. 2.7% and the current growth rate is 0.08? 105. Bank A (even though it has 110. Employee Training A person learning certain skills involving repetition tends the greatest stated rate) to learn quickly at first. Then learning tapers off and skill acquisition approaches 106. (a) $12,676.51 some upper limit. Suppose the number of symbols per minute that a person using a (b) $12,703.38 keyboard can type is given by (c) $12,709.44 -0.5t (d) $12,712.39 ƒ t = 250 - 120 2.8 , where t is the number of months the operator has been in training. Find each value 107. (a) 1200 1 2 1 2 to the nearest whole number. (a) ƒ 2 (b) ƒ 4 (c) ƒ 10 (d) What happens to the number of symbols per minute after several months of 1 2 1 2 1 2 −1000 11,000 training? 0 (b) exponential Use a graphing calculator to find the solution set of each equation. Approximate the (c) P(x) = 1013e−0.0001341x 1200 solution(s) to the nearest tenth. 111. 5e3x = 75 112. 6-x = 1 - x 113. 3x + 2 = 4x 114. x = 2x 115. A function of the form ƒ x xr, where r is a constant, is a power function. Discuss the difference between an exponential function and a power function. −1000 11,000 0 1 2 116. Concept Check If ƒ x ax and ƒ 3 27, determine each function value. (d) P 1500 ≈ 828 mb; = = P 11,000 ≈ 232 mb (a) ƒ 1 (b) ƒ -1 (c) ƒ 2 (d) ƒ 0 1 2 1 2 1 2 108. (a) The function gives 1 2 1 2 1 2 1 2 1 2 approximately Concept Check Give an equation of the form ƒ x = ax to define the exponential func- 6860 million, which tion whose graph contains the given point. differs by 7 million from 1 2 the actual value. 117. 3, 8 118. 3, 125 119. -3, 64 120. -2, 36 (b) 7734 million; 8720 million 1 2 1 2 1 2 1 2 Concept Check Use properties of exponents to write each function in the form 109. (a) 63,000 (b) 42,000 ƒ t = kat, where k is a constant. (Hint: Recall that ax+y = ax # ay.) (c) 21,000 1 2t 1 2t 110. (a) 207 (b) 235 (c) 249 1 2 1 - 1 - 121. ƒ t = 32t+3 122. ƒ t = 23t+2 123. ƒ t = 124. ƒ t = (d) The number of symbols 3 2 approaches 250. 1 2 1 2 1 2 a b 1 2 a b Copyright Pearson. All Rights Reserved.

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111. 0.9 112. 0, 0.7 In calculus, the following can be shown.

113. -0.5, 1.3 114. ∅ 2 3 4 5 5 6 5 6 x x x x x e = 1 + x + + + + + g 115. The5 variable6 is located in 2 # 1 3 # 2 # 1 4 # 3 # 2 # 1 5 # 4 # 3 # 2 # 1 the base of a power function Using more terms, one can obtain a more accurate approximation for ex. and in the exponent of an exponential function. 125. Use the terms shown, and replace x with 1 to approximate e1 = e to three decimal 1 places. Check the result with a calculator. 116. (a) 3 (b) 3 (c) 9 (d) 1 126. Use the terms shown, and replace x with -0.05 to approximate e-0.05 to four deci- 117. ƒ x = 2x 118. ƒ x = 5x mal places. Check the result with a calculator. 1 x 1 x 119. ƒ1x2 = 4 120. ƒ1x2 = 6 1 2 1 2 121. ƒ t = 27A B# 9t A B Relating Concepts 122. ƒ t = 4 # 8t 1 2 1 t 1 t For individual or collaborative investigation (Exercises 127—132) 123. ƒ1t2 = 3 9 124. ƒ t = 2 4 Consider ƒ x = ax, where a 7 1. Work these exercises in order. 1 2 1 2 125. 2.717 A B A B 127. Is ƒ a one-to-one function? If so, what kind of related function exists for ƒ? (A calculator gives 2.718.) 1 2 126. 0.9512 128. If ƒ has an inverse function ƒ-1, sketch ƒ and ƒ-1 on the same set of axes. (A calculator gives 0.9512.) 129. If ƒ-1 exists, find an equation for y ƒ-1 x . (You need not solve for y.) 127. yes; an inverse function = 128. y -1 y = f(x) 130. If a = 10, what is the equation for y = ƒ 1 2x ? (You need not solve for y.) 1 (0, 1) 131. If a = e, what is the equation for y = ƒ- x1 ?2 (You need not solve for y.) 0 x (1, 0) 132. If the point p, q is on the graph of ƒ, then1 2the point is on the graph -1 –1 of ƒ . y = x y = f (x) 1 2 129. x = ay 130. x = 10y 131. x = ey 132. q, p 1 2

4.3 Logarithmic Functions

■ Logarithms Logarithms The previous section dealt with exponential functions of the ■ Logarithmic Equations form y = ax for all positive values of a, where a 1. The horizontal line test ■ Logarithmic Functions shows that exponential functions are one-to-one and thus have inverse functions. ■ Properties of The equation defining the inverse of a function is found by interchanging x and y Logarithms in the equation that defines the function. Starting with y = ax and interchanging x and y yields x = ay. Here y is the exponent to which a must be raised in order to obtain x. We call this exponent a logarithm, symbolized by the abbreviation “log.” The expression log a x represents the logarithm in this discussion. The number a is the base of the logarithm, and x is the argument of the expression. It is read “logarithm with base a of x,” or “logarithm of x with base a,” or “base a logarithm of x.”

Logarithm

For all real numbers y and all positive numbers a and x, where a 1, y y log a x is equivalent to x a .

The expression log a x represents the exponent to which the base a must be raised in order to obtain x.

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Classroom Example 1 Example 1 Writing Equivalent Logarithmic and Exponential Forms Complete the missing equivalent forms in the table. The table shows several pairs of equivalent statements, written in both logarithmic and exponential forms. Logarithmic Exponential Form Form Solution To remember the relationships among (a) 34 81 = a, x, and y in the two equivalent forms (b) log 8 3 Logarithmic Form Exponential Form y 1/2 = - y = loga x and x = a , refer to these 3 3 diagrams. (c) 10 = 1000 log2 8 = 3 2 = 8 1 A logarithm is an exponent. 1 -4 (d) log5 125 = -3 log1/2 16 = -4 2 = 16 Exponent 1 (e) 12 = 12 5 log10 100,000 = 5 10 = 100,000 (f) log6 1 = 0 A B Logarithmic form: y = loga x 1 -4 1 log3 81 = -4 3 = 81 Answers: Base 1 -3 1 log5 5 1 5 5 Exponent (a) log3 81 = 4 (b) 2 = 8 = = 1 3 0 (c) log 1000 3 (d) 5-3 log 1 0 1 y 10 = A B = 125 3/4 = 4 = Exponential form: a = x 0 (e) log12 12 = 1 (f) 6 = 1 A B Base ■✔ Now Try Exercises 11, 13, 15, and 17. Teaching Tip Have students verbalize the statements in the left Logarithmic Equations The definition of logarithm can be used to solve column of the table in Example 1. a logarithmic equation, which is an equation with a logarithm in at least one term. For instance, “log2 8 is the exponent to which 2 must be raised in order to obtain 8.” Example 2 Solving Logarithmic Equations Solve each equation. Classroom Example 2 8 5 3 Solve each equation. (a) logx = 3 (b) log4 x = (c) log49 7 = x 27 2 16 2 (a) logx = -2 9 Solution Many logarithmic equations can be solved by first writing the 3 equation in exponential form. (b) log16 x = 4 8 (a) logx = 3 (c) log36 6 = x 27 Answers:2 3 8 3 1 x = Write in exponential form. (a) 4 (b) 8 (c) 4 27 3 5 6 3 2 8 2 3 E F E F x = = 3 27 3 2a b A B x = Take cube roots. 3

8 Teaching Tip Point out that Check logx = 3 Original equation Example 2 illustrates three 27 different locations for the 8 ≟ 2 variable: the base, the argument, log2/3 3 Let x = 3 . and the exponent. Remind students 27 of the restrictions on x in parts (a) 2 3 8 and (b), and that there is no ≟ Write in exponential form. restriction on x in part (c). 3 27 a b8 8 = ✓ True 27 27

2 The solution set is 3 .

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5 3 (b) log4 x = (c) log49 7 x 2 = Write in exponential 2 3 Write in exponential 45/2 x 49x 7 = form. = form. 1/2 5 amn am n 2 Write with the same 4 = x = 72 x 71/3 = base. 5 2 x 41/2 = 122 21/2 = 2 1 2 = 2x 1/3 Power rule for 1 72 = 7 exponents 32 = x Apply 1the2 exponent. 1 2x = Set exponents equal. 5 3 Check log 32 ≟ Let x = 32. 4 2 1 5/2 x = Divide by 2. 4 ≟ 32 6 5 5 2 ≟ 32 45/2 = 4 = 25 A check shows that the solution set 2 32 = 32 ✓ True 1 A B is 6 . The solution set is 32 . ■✔ Now Try Exercises 19, 29, and 35. E F 5 6 Logarithmic Functions We define the logarithmic function with base a.

Logarithmic Function

If a 7 0, a 1, and x 7 0, then the logarithmic function with base a is

ƒ x log a x.

1 2 Exponential and logarithmic functions are inverses of each other. To show this, we use the three steps for finding the inverse of a function.

ƒ x = 2x Exponential function with base 2 x 1 y2 = 2 Let y = ƒ x . Step 1 x = 2y Interchange1 2x and y.

Step 2 y = log2 x Solve for y by writing in equivalent logarithmic form.

-1 -1 Step 3 ƒ x = log2 x Replace y with ƒ x . The graph1 of2 ƒ x = 2x has the x-axis as1 horizontal2 asymptote and is shown -1 in red in Figure 25. Its inverse, ƒ x = log2 x, has the y-axis as vertical asymptote and is shown in1 blue.2 The graphs are reflections of each other across the line y = x. As a result, their domains and1 2 ranges are interchanged.

f(x) = 2 x Domain: (– , ) x 1 x ƒ x 2 x ƒ x log2 x y Range: (0, ) y = x 1 21 1 1 2 8 -2 4 4 -2 1 1 6 -1 2 2 -1 (1, 2) 0 1 1 0 4 f –1(x) = log x 1 2 2 1 (0, 1) 2 Domain: (0, ) 2 4 4 2 –1, 1 ( 2) (2, 1) Range: (– , ) x 0 –2 (1, 0) 468 –2 1 , –1 (2 ) Figure 25

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The domain of an exponential function is the set of all real numbers, so the range of a logarithmic function also will be the set of all real numbers. In the same way, both the range of an exponential function and the domain of a logarithmic function are the set of all positive real numbers.

Thus, logarithms can be found for positive numbers only.

Logarithmic Function f x loga x

Domain: 0, ∞ 1 2 Range: -∞, ∞ For ƒ x = log2 x: 1 2 1 2 y x 1ƒ 2x f(x) = loga x, a > 1 f(x) = log x, a > 1 1 1 2 a 4 -2 1 (a, 1) 2 -1 1 0 x 2 1 0 (1, 0) 4 2 1 ( a , –1) 8 3

This is the general behavior seen on a calculator graph for any base a, for a + 1. Figure 26

• ƒ x log a x, for a + 1, is increasing and continuous on its entire domain, 0, ∞ . 1 2 • The y-axis1 is a2 vertical asymptote as x S 0 from the right. 1 • The graph passes through the points a , -1 , 1, 0 , and a, 1 .

A B 1 2 1 2 For ƒ x = log1/2 x:

y x 1ƒ 2x f(x) = loga x, 0 < a < 1

1 1 2 4 2 (a, 1) 1 1 2 (1, 0) x 1 0 0 2 1 - 1 4 -2 ( a , –1) 8 3 - f(x) = log a x, 0 < a < 1

This is the general behavior seen on a calculator graph for any base a, for 0 * a * 1. Figure 27

• ƒ x log a x, for 0 * a * 1, is decreasing and continuous on its entire domain, 0, ∞ . 1 2 • The y-axis1 is a2 vertical asymptote as x S 0 from the right. 1 • The graph passes through the points a , -1 , 1, 0 , and a, 1 .

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Calculator graphs of logarithmic functions sometimes do not give an accurate picture of the behavior of the graphs near the vertical asymptotes. While it may seem as if the graph has an endpoint, this is not the case. The resolution of the calculator screen is not precise enough to indicate that the graph approaches the vertical asymptote as the value of x gets closer to it. Do not draw incorrect conclusions just because the calculator does not show this behavior. ■

The graphs in Figures 26 and 27 and the information with them suggest the following generalizations about the graphs of logarithmic functions of the form ƒ x = loga x. 1 2 Teaching Tip Encourage students to learn the characteristics Characteristics of the Graph of f x loga x of the graph of 1. The points 1 , 1 , 1, 0 , and a, 11 are2 on the graph. ƒ x = loga x. a - Ask them how1 2 the coordinates 2. If a 7 1, then ƒ is an1 increasing2 1function.2 1 , 1 , 1, 0 , and a, 1 are A B a - If 0 6 a 6 1, then ƒ is a decreasing function. affected by different translations 1 2 1 2 ofA the graph.B Also, emphasize 3. The y-axis is a vertical asymptote. that horizontal shifts affect the 4. The domain is 0, ∞ , and the range is -∞, ∞ . domains of logarithmic functions. 1 2 1 2

Classroom Example 3 Example 3 Graphing Logarithmic Functions Graph each function. Graph each function. (a) ƒ x = log1/3 x (b) ƒ x = log5 x (a) ƒ x log1/2 x (b) ƒ x log3 x 1 2 = = Answers:1 2 Solut1 io2 n 1 2 (a) (b) 1 x y y (a) One approach is to first graph y = 2 , which defines the inverse function of ƒ, by plotting points. Some ordered pairs are given in the table with the graph f(x) = log1/3x f(x) = log5 x 1 1 shown in red in Figure 28. A B x x 1 x 0 1 9 0 –2 5 The graph of ƒ x = log1/2 x is the reflection of the graph of y = 2 across the line y = x. The ordered pairs for y = log1/2 x are found by inter- 1 2 1 x A B changing the x- and y-values in the ordered pairs for y = 2 . See the graph in blue in Figure 28. A B y 1 x 1 x y x ƒ x log x y = y = x x ƒ x log3 x x y 2 1/2 ( 2 ) 2 4 4 2 1 1 2 - 1 2 - 3 -1 -1 2 A B 2 -1 1 0 4 3 f(x) = log x 0 1 1 0 3 1 3 1 1 1 1 9 2 2 2 x 0 1 1 x 39 2 2 0 y 4 4 4 Think: x = 3 –2 1 1 –2 4 16 16 4 f(x) = log1/2 x

Figure 28 Figure 29

(b) Another way to graph a logarithmic function is to write ƒ x = y = log3 x in exponential form as x = 3y, and then select y-values and calculate corresponding x-values. Several selected ordered pairs are shown1 in2 the table for the graph in Figure 29. ■✔ Now Try Exercise 55.

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Caution If we write a logarithmic function in exponential form in order to graph it, as in Example 3(b), we start first with y-values to calculate corresponding x-values. Be careful to write the values in the ordered pairs in the correct order.

More general logarithmic functions can be obtained by forming the composition of ƒ x = loga x with a function g x . For example, if ƒ x = log2 x and g x = x - 1, then 1 2 1 2 1 2 1 2 ƒ ∘ g x = ƒ g x = log2 x - 1 . The next example shows1 how21 2to graph1 1 such22 functions.1 2

Classroom Example 4 Example 4 Graphing Translated Logarithmic Functions Graph each function. Give the domain and range. Graph each function. Give the domain and range.

(a) ƒ x = log3 x + 2 (a) ƒ x = log2 x - 1 (b) ƒ x = log3 x - 1 (b) ƒ x = log4 x + 2 1 2 1 2 (c) ƒ1x2 = log4 1x + 22 + 1 1 2 1 2 (c) ƒ x = log2 x - 3 + 2 1 2 1 2 Solution Answers:1 2 1 2 1 2 1 2 (a) y (a) The graph of ƒ x = log2 x - 1 is the graph of g x = log2 x translated f(x) = log (x + 2) x = –2 3 1 unit to the right. The vertical asymptote has equation x 1. Because loga- 2 = x rithms can be found1 2 only for1 positive2 numbers, we solve1 2 x 1 0 to find 0 2 - 7 the domain, 1, ∞ . To determine ordered pairs to plot, use the equivalent -2, ∞ ; -∞, ∞ exponential form of the equation y = log2 x - 1 . 1 2 (b) y 1 2 1 2 y = log2 x - 1 1 2 2 y x x 1 2 Write in exponential form. 0 1 - = 1 2 f(x) = (log x) + 2 4 x = 2y + 1 Add 1. 0, ∞ ; -∞, ∞ (c) y We first choose values for y and then calculate each of the corresponding 1 2 x 1= 3 2 x-values. The range is -∞, ∞ . See Figure 30. 3 1 2 x y y 0 6 x = 1 ؊2 4 f(x) = log (x – 1) 4 2 f(x) = (log3 x) – 1 f(x) = log2 (x ؊ 3) ؉ 2 2 (3, 1) 2 (5, 2) (9, 1) 3, ∞ ; -∞, ∞ (3, 0) x x 0 0 1 2 1 2 4 68 369 (؊1 ,1) 2– (0 ,2) 2–

Figure 30 Figure 31

(b) The function ƒ x = log3 x - 1 has the same graph as g x = log3 x translated 1 unit down. We find ordered pairs to plot by writing the equation 1 2 1 2 1 2 y = log3 x - 1 in exponential form.

1 2 y = log3 x - 1

y + 1 = log1 3 x 2 Add 1. x = 3y+1 Write in exponential form. Again, choose y-values and calculate the corresponding x-values. The graph is shown in Figure 31. The domain is 0, ∞ , and the range is -∞, ∞ . Copyright Pearson. All Rights Reserved.1 2 1 2

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y (c) The graph of ƒ x = log4 x + 2 + 1 is obtained by shifting the graph of 1 2 1 2 y = log4 x to the left 2 units and up 1 unit. The domain is found by solving 4 log4 (x ؉ 2) ؉ 1 ؍ (f(x x + 2 7 0, (؊1, 1) (2, 2) which yields -2, ∞ . The vertical x asymptote has been shifted to the left 0 2 2 units as well,1 and 2it has equation Ϫ2 x = -2. The range is unaffected by the vertical shift and remains -∞, ∞ . x ϭ Ϫ2 See Figure 32. 1 2 Figure 32 ■✔ Now Try Exercises 43, 47, and 61.

Note If we are given a graph such as the one in Figure 31 and asked to find its equation, we could reason as follows: The point 1, 0 on the basic logarithmic graph has been shifted down 1 unit, and the point 3, 0 on the 1 2 given graph is 1 unit lower than 3, 1 , which is on the graph of y = log3 x. Thus, the equation will be 1 2 1 2 y = log3 x - 1. 1 2 Teaching Tip Encourage Properties of Logarithms The properties of logarithms enable us to students to memorize the change the form of logarithmic statements so that products can be converted to sums, properties of logarithms. quotients can be converted to differences, and powers can be converted to products.

Properties of Logarithms

For x 7 0, y 7 0, a 7 0, a 1, and any real number r, the following properties hold. Property Description Product Property The logarithm of the product of two num- Teaching Tip Point out that the log xy log x log y bers is equal to the sum of the logarithms phrase “quotients can be converted a a a of the numbers. to differences” refers to the quotient property. It does NOT Quotient Property The logarithm of the quotient of two num- say that x bers is equal to the difference between the log log x log y log x a a a logarithms of the numbers. a y log x log y = a - a log y a Power Property The logarithm of a number raised to a or log xr r log x power is equal to the exponent multiplied a a log x a by the logarithm of the number. log x y , = a - log y a Logarithm of 1 The base a logarithm of 1 is 0. which are COMMO1N ERR2 ORS log 1 0 a that students make. A quotient or product of logarithms cannot be Base a Logarithm of a The base a logarithm of a is 1. log a 1 simplified using the properties of a logarithms given in the box.

Proof To prove the product property, let m = loga x and n = loga y. m loga x = m means a = x Write in exponential form. n loga y = n means a = y

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Now consider the product xy.

Looking Ahead To Calculus m n xy = a # a x = am and y = an; Substitute. A technique called logarithmic m+n differentiation, which uses the xy = a Product rule for exponents properties of logarithms, can often be loga xy = m + n Write in logarithmic form. used to differentiate complicated functions. loga xy = loga x + loga y Substitute. The last statement is the result we wished to prove. The quotient and power properties are proved similarly and are left as exercises.

Classroom Example 5 Example 5 Using Properties of Logarithms Use the properties of logarithms to rewrite each expression. Assume Use the properties of logarithms to rewrite each expression. Assume all vari- all variables represent positive real ables represent positive real numbers, with a 1 and b 1. numbers, with a 1 and b 1. 15 12 (a) log6 7 # 9 (b) log9 (c) log5 8 (a) log7 8 # 6 (b) log6 7 5 2 1 2 3 5 13 2 5 3 3 mnq n x y (c) log2 9 (d) loga r 2 (d) loga m (e) loga 2 4 (f) logb m 2 2 p t z 2 3 2 2 rs t m r s (e) logb 3 5 (f) loga 4 Solution B u v t 15 (a) log6 7 # 9 (b) log Answers: B 9 Product 7 log 7 log 9 (a) log7 8 + log7 6 =1 6 2 + 6 property Quotient = log9 15 - log9 7 property (b) log6 12 - log6 5 1 3 2 (c) 3 log2 9 (c) log5 8 (d) loga m 3 (d) log r 2 1/2 1/2 2 2/3 n m m/n 5 a = log5 8 a = a = loga m a = a (e) logb r + 2 logb s + logb t - 2 2 1 1 2 2 3 logb u - 5 logb v = log5 8 Power property = loga m Power property 3 2 4 2 3 (f) m loga r + m loga s - m loga t mnq (e) log Use parentheses a p2t4 to avoid errors. 2 4 Product and quotient = loga m + loga n + loga q - loga p + loga t properties = loga m + loga n + loga q - 12 loga p + 4 loga2 t Power property = loga m + loga n + loga q - 12 loga p - 4 loga t 2 Distributive property Be careful 3 5 x y with signs. (f) log n b zm

3 5 1/n B x y n log a a1/n = b = zm 2 1 a x3y5b Teaching Tip Point out that in log Power property = n b m Example 5(e), the product and z quotient properties of logarithms 1 log x3 log y5 log zm Product and quotient properties were applied before the power = n b + b - b property. In Example 5(f), the power property was applied first. 1 1 2 3 log x 5 log y m log z Power property In general, properties of = n b + b - b logarithms should follow the order of operations with respect 3 1 5 m 2 = logb x + logb y - logb z Distributive property to exponents, multiplication, and n n n division. ■✔ Now Try Exercises 71, 73, and 77.

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Classroom Example 6 Example 6 Using Properties of Logarithms Write each expression as a single logarithm with coefficient 1. Write each expression as a single logarithm with coefficient 1. Assume all vari- Assume all variables represent ables represent positive real numbers, with a 1 and b 1. positive real numbers, with a 1 (a) log3 x + 2 + log3 x - log3 2 (b) 2 loga m - 3 loga n and b 1. 1 3 (a) log4 x - log4 y + log4 z 1 2 2 (c) logb m + logb 2n - logb m n (b) 4 logb r - 5 logb s 2 2 1 2 (c) loga x + loga y - loga xy Solution 3 3 (a) log3 x + 2 + log3 x - log3 2 (b) 2 loga m - 3 loga n Answers: Power 2 3 xz r 4 Product and (a) log (b) log 1 2x + 2 x = loga m - loga n property 4 y b s5 quotient = log3 2 3 2 properties m Quotient xy 2 (c) log 1 2 = loga property a 2xy n3 1 3 2 (c) logb m + logb 2n - logb m n 2 2 1/2 3/2 2 = logb m + logb 2n - logb m n Power property 1/2 3/2 m 2n 1 Use2 parentheses = logb Product and quotient properties m2n around 2n. 1 2 23/2n1/2 = logb Rules for exponents m3/2 23n 1/2 Classroom Example 7 = logb Rules for exponents m3 Given that log10 7 ≈ 0.8451, find each logarithm without using a a b 8n calculator. 1/n = logb 3 Definition of a (a) log10 49 (b) log10 70 m ■✔ Now Try Exercises 83, 87, and 89. Answers: B (a) 1.6902 (b) 1.8451 Caution There is no property of logarithms to rewrite a logarithm of a sum or difference. That is why, in Example 6(a),

log3 x + 2 cannot be written as log3 x + log3 2.

The distributive 1property2 does not apply here because log3 x + y is one term. The abbreviation “log” is a function name, not a factor. 1 2

Example 7 Using Properties of Logarithms with Numerical Values Napier’s Rods Given that log10 2 ≈ 0.3010, find each logarithm without using a calculator. The search for ways to make (a) log 4 (b) log 5 calculations easier has been a 10 10 long, ongoing process. Machines Solution built by Charles Babbage and Blaise Pascal, a system of “rods” (a) log10 4 (b) log10 5 used by John Napier, and slide 2 = log10 2 10 rules were the forerunners of = log10 2 today’s calculators and computers. = 2 log10 2 The invention of logarithms by = log10 10 - log10 2 John Napier in the 16th century ≈ 2 0.3010 was a great breakthrough in the ≈ 0.6020 ≈ 1 - 0.3010 search for easier calculation 1 2 methods. ≈ 0.6990 Source: IBM Corporate Archives. ■✔ Now Try Exercises 93 and 95.

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Note The values in Example 7 are approximations of logarithms, so the final digit may differ from the actual 4-decimal-place approximation after properties of logarithms are applied.

Recall that for inverse functions ƒ and g, ƒ ∘ g x = g ∘ ƒ x = x. We can use this property with exponential and logarithmic functions to state two x 1 21 2 1 21 2 more properties. If ƒ x = a and g x = loga x, then

log x x ƒ ∘ g x 1=2a a = x 1and2 g ∘ ƒ x = loga a = x. 1 21 2 1 21 2 1 2 Theorem on Inverses

For a 7 0, a 1, the following properties hold.

Teaching Tip Point out that the log a x x a x for x 0 and log a x theorem on inverses really follows + a from the definition of logarithm. 1 2 Because loga x is the power to log7 10 3 k+1 which a must be raised in order Examples: 7 = 10, log5 5 = 3, and logr r = k + 1 to obtain x, a raised to that power will be x. In practice, x will often The second statement in the theorem will be useful when we solve logarithmic be an algebraic expression. and exponential equations.

4.3 Exercises

1. (a) C (b) A (c) E Concept Preview Match the logarithm in Column I with its value in Column II. (d) B (e) F (f) D Remember that loga x is the exponent to which a must be raised in order to obtain x. 2. (a) F (b) B (c) A I II I II (d) D (e) C (f) E 1. (a) log2 16 A. 0 2. (a) log3 81 A. -2 3 1 1 3. 2 = 8 4. log10 1000 = 3 (b) log3 1 B. (b) log3 B. -1 2 3 4 1 5. 9 6. 6 (c) log10 0.1 C. 4 (c) log10 0.01 C. 0 1 (d) log2 2 D. -3 (d) log 6 D. 7. E Fy E F 6 2 21 2 4 f(x) = log x (e) loge E. -1 5 e2 9 1 (e) loge 1 E. x 2 –2 0 15 (f) log1/2 8 F. -2 –2 3/2 (f) log3 27 F. 4

0, ∞ ; -∞, ∞ Concept Preview Write each equivalent form. 8. y 1 2 1 2 3 3. Write log2 8 = 3 in exponential form. 4. Write 10 = 1000 in logarithmic form. 2 5 x 0 –2 1 Concept Preview Solve each logarithmic equation.

–4 g(x) = log1/5 x 16 3 5. logx = 2 6. log36 6 = x 81 0, ∞ ; -∞, ∞ 2

1 2 1 2 Concept Preview Sketch the graph of each function. Give the domain and range.

7. ƒ x = log5 x 8. g x = log1/5 x

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9. log10 2 + log10 x - log10 7 Concept Preview Use the properties of logarithms to rewrite each expression. As- x3 sume all variables represent positive real numbers. 10. log4 y 5 2x 9. log10 10. 3 log4 x - 5 log4 y 11. log3 81 = 4 12. log2 32 = 5 7 27 13. log2/3 8 = -3 If the statement is in exponential form, write it in an equivalent logarithmic form. If the 14. log10 0.0001 = -4 statement is in logarithmic form, write it in exponential form. See Example 1. 15. 62 = 36 16. 51 = 5 8 1 3 17. 3 = 81 18. 4-3 = 2 - 27 64 11. 34 = 81 12. 25 = 32 13. = 14. 10-4 = 0.0001 2 3 8 19. A 4 B 20. 4 - - a b 1 1 3 15. log 36 2 16. log 5 1 17. log 81 8 18. log 3 6 = 5 = 3 = 4 = - 21. 52 6 22. 54 6 64 2 23. 1 24. 1 E 4 F E 5 F Solve each equation. See Example 2. 25. 8 26. 5 E F E F 1 1 1 27 27. 9 28. 11 19. x = log5 20. x = log3 21. logx = 5 22. logx = 3 5 6 5 6 625 81 32 64 29. 1 30. 1 55 6 54 6 4 5 23. x log 8 24. x log 7 25. x 3log3 8 26. x 12log12 5 31. 64 32. 8 = 8 = 7 = = log 92 log 112 E 2 F E 1 F 27. x = 2 2 28. x = 8 8 29. logx 25 = -2 30. logx 16 = -2 33. 53 6 34. 52 6 3 4 35. 243 36. 128 31. log4 x = 3 32. log2 x = 3 33. x = log4 16 34. x = log5 25 37. E13F 38. E3F 2 2 5 6 5 6 5 7 39. 3 40. 23 35. log9 x = 36. log4 x = 5 6 5 6 2 2 41. 5 42. 6 5 6 5 6 37. log1/2 x + 3 = -4 38. log1/3 x + 6 = -2 43. 5 6 44. 5 6 39. log x 3 6 = 1 40. log x 4 19 = 1 y y +1 2 -1 2 x = –3 41. 3x 1 152 log 1 x 0, x 1 42. 4x 1 242 log 1 x 0, x 1 f(x) = (log2 x) +3 - = x 7 - = x 7 6 3 4 –2 x 1 2 1 2 2 0 25 Graph each function. Give the domain and range. See Example 4. x 0 2468 43. ƒ x = log2 x + 3 44. ƒ x = log2 x + 3 45. ƒ x = log2 x + 3 f(x) = log2 (x + 3) 0, ∞ ; ∞, ∞ 3, ∞ ; ∞, ∞ 1 2 1 2 1 2 1 2 1 2 0 1 2 0 - - - Graph each function. Give the domain and range. See Example 4. 45. 46. 1 2 1 2 1 2 1 2 y y 46. ƒ x = log1/2 x - 2 47. ƒ x = log1/2 x - 2 48. ƒ x = log1/2 x - 2 x = –3

4 x 1 2 1 2 1 2 1 2 1 2 1 2 0 0 0 –2 258 Concept Check In Exercises 49–54, match the function with its graph from choices A–F. x –4 –2 0 2 f(x) = (log x) – 2 1 1/2 49. ƒ x log x 50. ƒ x log 2x 51. ƒ x log = 2 = 2 = 2 x f(x) = ͉log2 (x + 3)͉ 1 2 1 1 2 1 2 3, ; 0, 0, ; , - ∞ ∞ ∞ -∞ ∞ 52. ƒ x = log2 x 53. ƒ x = log2 x - 1 54. ƒ x = log2 -x 2 147. 2 3 2 48.1 2 1 2 y y A. 1 2 a yb B. 1 2y 1 2 C. 1 2y 1 2 x = 2 f(x) = log1/2 (x – 2) 4 2 2 1 1 1 x 1 1 0 x 0 x x x –2 369 0 0 0 –2 369 1

x = 2 f(x) = ͉log1/2 (x – 2)͉ 2, ∞ ; -∞, ∞ 2, ∞ ; 0, ∞ D. y E. y F. y 49.1 E2 1 2 50.1 D2 3 2 51. B 52. C 1 1 1 x x x 53. F 54. A 0 0 0 1 1 2

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55. 56. Graph each function. See Examples 3 and 4. y y 55. ƒ x = log5 x 56. ƒ x = log10 x 57. ƒ x = log5 x + 1 f(x) = log5 x 2 2 1 x x 58. ƒ1x2 log6 x 2 59. ƒ1x2 log1/2 1 x 60. ƒ1x2 log1/31 3 2x 0 525 0 = - = - = - –2 –1 510 f(x) = log x –2 10 61. ƒ1x2 = log3 1x - 1 2+ 2 62. ƒ1x2 = log2 x1 + 2 2- 3 63. ƒ1x2 = log1/2 1x + 3 2- 2 57. y 64. Concept1 2 Check1 To2 graph the function1 2 ƒ 1x 2log5 x 7 1 2 4, reflect1 the graph2 x = –1 = - - - 4 of y = log5 x across the -axis, then shift the graph units to the right and 2 units down. 1 2 1 2 x –2 0 24 –3 Connecting Graphs with Equations Write an equation for the graph given. Each repre- f(x) = log5 (x+1) sents a logarithmic function ƒ with base 2 or 3, translated and/or reflected. See the Note 58. y following Example 4. x = 2 4 65. y 66. y 67. y 2 x = 3 x x = 3 –10 38 x = –1 –3 (7, 1) (7, 0) 0 x x x 0 0 (–1, 0) (1, –1) f(x) = log6 (x–2) (5, 0) (1, –2) (2, –2) (4, –1) 59. y 60. y (0, –3) x = 1

x = 3 2 2 68. y 69. 70. y x 0 y 4 0 x , 1 –4 –2 –3 2 ( 3 ) –2 (–2, 0) f(x) = log1/2 (1 – x) 0 (2, 0) x x (4, 0) f(x) = log (3 – x) x 1/3 (1, –2) 0 0 (–1, –1) (4, –1) (–3, –3) 61. 62. (3, –1) y y x = 1 x = 1 x = –2 x = 5 x = –3 3 3 2 x x 0 0 24 6 Use the properties of logarithms to rewrite each expression. Simplify the result if pos- –3 .؊3 sible. Assume all variables represent positive real numbers. See Example 5 f(x) = log (x ؊ 1) ؉ 2 f(x) = log (x + 2) – 3 3 2 6x 4p 5 7 71. log 72. log 73. log 63. y 64. x; 7; 4 2 y 3 q 5 23

x = –3 2 3 2 74. log 75. log 2x 5y 76. log 7m 3q 0 x 2 4 + 6 + –4 25 25 –2 1 2 1 2 3 5 4 5r 3 m n ab f(x) = log (x+3) – 2 77. log 78. log 79. log 1/2 2 z5 3 t2 2 cd 65. ƒ x log x 1 3 B B = 2 + - xy x # 3 y 3 a # 4 b 66. ƒ x = log2 x - 3 - 1 80. log2 81. log3 82. log4 1 2 1 2 tqr 2w2 2z 2c # 23 d 2 67. ƒ x = log2 -x + 3 - 2 1 2 1 2 2 2 2 68. ƒ x = -log2 x + 3 1 2 1 2 Write each expression as a single logarithm with coefficient 1. Assume all variables rep- 69. ƒ x log x 1 1 2 = - 3 1 - 2 resent positive real numbers, with a 1 and b 1. See Example 6. 70. ƒ x log x 5 1 2 = - 2 1- +2 83. loga x + loga y - loga m 84. logb k + logb m - logb a 1 2 1 2 71. log2 6 + log2 x - log2 y 85. loga m - loga n - loga t 86. logb p - logb q - logb r 72. log3 4 + log3 p - log3 q 1 1 3 1 2 73. 1 log 7 log 3 4 5 2 3 4 4 3 + 2 5 - 5 87. logb x y - logb x y 88. loga p q - loga p q 1 3 4 2 3 74. 1 + 2 log2 3 - log2 5 75. This cannot be simplified. 89. 2 loga z + 1 + loga 3z + 2 90. 5 loga z + 7 + loga 2z + 9 76. This cannot be simplified. 2 1 22 1 1 2 2 3 1 24 2 1 3 2 91. - log5 5m + log5 25m 92. - log3 16p - log3 8p 3 2 4 3

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1 Given that log10 2 ≈ 0.3010 and log10 3 ≈ 0.4771, find each logarithm without using a 77. 2 log2 5 + 3 log2 r - calculator. See Example 7. 51 log2 z 78. 1 5 log m 4 log n 3 2 3 3 2 + 3 - 93. log10 6 94. log10 12 95. log10 96. log10 2 log t 2 9 1 3 79. log a log b log c 2 + 2 - 2 - 9 20 1/3 2 97. log10 98. log10 99. log10 30 100. log10 36 log2 d 4 27 2 80. log2 x + log2 y - log2 t - Solve each problem. log2 q - log2 r 1 1 81. 2 log3 x + 3 log3 y - 101. (Modeling) Interest Rates of Treasury Securities The Time Yield table gives interest rates for various U.S. Treasury Securities 2 log w 1 log z 3-month 0.02% 3 - 2 3 on January 2, 2015. 1 1 6-month 0.10% 82. 3 log4 a + 4 log4 b - (a) Make a scatter diagram of the data. 1 2 2-year 0.66% 2 log4 c - 3 log4 d (b) Which type of function will model this data best: linear, exponential, or logarithmic? 5-year 1.61% xy km 10-year 2.11% 83. loga m 84. logb a m p 30-year 2.60% 85. loga nt 86. logb qr Source:www.federal 87. log x-1/6y11/12 b reserve.gov -7/6 88. loga 1p 2 102. Concept Check Use the graph to estimate each y 2 89. loga z + 1 3z + 2 logarithm. 1 2 1 5 90. loga 31z + 72 12z + 924 (a) log3 0.3 (b) log3 0.8 0.8 51/3 3 5 91. log5 31m1/3 , or l2og15 m 24 x 0.6 1 3 y = 3 92. log3 p5 32 0.4

93. 0.7781 94. 1.0791 0.2 95. 0.1761 96. -0.6532 x 97. 0.3522 98. -0.1303 –2–1.5 –1 –0.5 0 99. 0.7386 100. 0.5187

103. Concept Check Suppose ƒ x = loga x and ƒ 3 = 2. Determine each function 101. (a) 3 value. 1 2 1 2 1 3 (a) ƒ (b) ƒ 27 (c) ƒ 9 (d) ƒ 9 23 −5 35 104. Use propertiesa b of logarithms1 to2 evaluate each expression.1 2 −1 log 3 3 5 log 5 (b) logarithmic (a) 100 10 (b) log10 0.01 (c) log10 0.0001 (d) 1000 10 ¢ ≤ 102. (a) -1.1 (b) -0.2 1 2 1 r t n 2 105. Using the compound interest formula A = P 1 + n , show that the amount of 103. (a) -4 (b) 6 time required for a deposit to double is (c) 4 (d) -1 104. (a) 9 (b) 6 1 A B - . (c) -20 (d) 125 r n log2 1 + n 106. 4 = loga 5 106. Concept Check If 5, 4 is on the graphA of theB logarithmic function with base a, 107. 0.01, 2.38 108. 1.87 which of the following statements is true: 1 2 5 6 5 6 5 = loga 4 or 4 = loga 5?

Use a graphing calculator to find the solution set of each equation. Give solutions to the nearest hundredth.

-x 107. log10 x = x - 2 108. 2 = log10 x x 109. Prove the quotient property of logarithms: loga y = loga x - loga y.

r 110. Prove the power property of logarithms: loga x = r loga x.

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Summary Exercises on Inverse, Exponential, and Logarithmic Functions

1. They are inverses. The following exercises are designed to help solidify your understanding of inverse, 2. They are not inverses. exponential, and logarithmic functions from Sections 4.1–4.3. 3. They are inverses. 4. They are inverses. Determine whether the functions in each pair are inverses of each other. 1 4 1 5. y 6. y 1. ƒ x = 3x - 4, g x = x + 2. ƒ x = 8 - 5x, g x = 8 + x y = x y = x 3 3 5

1 2 1 2 x-1 1 2 x/5 1 2 3. ƒ x 1 log2 x, g x 2 4. ƒ x 3 2, g x 5 log3 x 2 0 x 0 x = + = = - = + 1 2 1 2 1 2 1 2 1 2 Determine whether each function is one-to-one. If it is, then sketch the graph of its 7. It is not one-to-one. inverse function. 8. y y = x 5. y 6. y y = x y = x 0 x

x x 1 9. B 10. D 11. C 12. A

13. The functions in Exercises 9 7. y 8. y and 12 are inverses of one y = x y = x another. The functions in Exercises 10 and 11 are inverses of one another. x x 14. ƒ-1 x = 5x

-11 2 1 15. ƒ x = 3 x + 2; Domains and ranges of both ƒ 1 21 and ƒ- are -∞, ∞ . In Exercises 9–12, match each function with its graph from choices A–D.

x 1 2 9. y = log3 x + 2 10. y = 5 - 2 x 11. y = log2 15 - x2 12. y = 3 - 2 A. y B. y 1 2

1 1 x x 0 0 1 1

C. y D. y

1 x 0 1 1 x 0 1

13. The functions in Exercises 9–12 form two pairs of inverse functions. Determine which functions are inverses of each other.

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3 x For each function that is one-to-one, write an equation for the inverse function. Give the 16. ƒ-1 x = - 1; 2 domain and range of both ƒ and ƒ-1. If the function is not one-to-one, say so. Domains and ranges of both ƒ 1 2 3 and ƒ-1 are -∞, ∞ . 15. ƒ x = 3x - 6 16. ƒ x = 2 x + 1 3 17. ƒ is not one-to-one. 1 2 2x 1 x 1 2 2 1 2 1 - 2 18. ƒ-1 x 5 + 1 ; 17. ƒ x = 3x 18. ƒ x = = 2 + 3x 5 - 3x Domain of ƒ = range of 1 2 1 2 1 2 3 4 2 -1 5 5 19. ƒ x = 5 - x 20. ƒ x = x - 9, x Ú 3 ƒ = -∞, 3 ´ 3 , ∞ . 2 2 Domain of ƒ-1 = range of 1 2 1 2 A 2B A 2 B ƒ = -∞, - 3 ´ - 3 , ∞ . Write an equivalent statement in logarithmic form. 19. ƒ is not one-to-one. -3 4 1 b -1 A B2 A B 21. = 1000 22. a = c 23. 3 = 9 20. ƒ x = x + 9, x Ú 0; 10 Domain of ƒ range of 2 2 = a b -11 2 ƒ = 3, ∞ . 3/2 1 x A 4/3 B 24. 4- = 25. 2 = 32 26. 27 = 81 Domain of ƒ-1 = range of 8 ƒ = 0,3 ∞ .2

21. log1/103 10002 = -3 Solve each equation. 22. loga c = b 1 23. log 9 4 log7 6 3 = 27. 3x = 7 28. x = log10 0.001 29. x = log6 216 21 3 24. log4 8 = - 2 1 25. log2 32 = x 30. logx 5 = 31. log10 0.01 = x 32. logx 3 = -1 4 2 26. log27 81 = 3 3 1 33. logx 1 = 0 34. x = log2 8 35. logx 5 = 27. 2 28. -3 3 29. 3 30. 25 2 2 5 -6 5 6 10 25 1 36. log1/3 x = -5 37. log10 log2 2 = x 38. x = log4/5 31. 5 -26 32. 53 6 16 3 33. 50, 16 1, ∞ 34. 1 2 ´ 2 x 1 x E F 39. 2x - 1 = log6 6 40. x = log1/2 41. 2 = log2 16 16 35. 15 2 1 2 36. 243 E F B 37. 1 38. -2 1 x+1 5 6 5 6 42. log x 2 43. 9x 44. 52x-6 25x-3 39. 1 40. 2 3 = - = = 5 6 5 6 3 1 a b 41. 526 42. 59 6 1 43. 44. ∞, ∞ 5 -63 E-F E F 1 2 4.4 Evaluating Logarithms and the Change-of-Base Theorem

■ Common Logarithms Common Logarithms Two of the most important bases for logarithms ■ Applications and are 10 and e. Base 10 logarithms are common logarithms. The common loga- Models with Common rithm of x is written log x, where the base is understood to be 10. Logarithms ■ Natural Logarithms ■ Applications and Common Logarithm Models with Natural Logarithms For all positive numbers x, ■ Logarithms with Other Bases log x log10 x.

A calculator with a log key can be used to find the base 10 logarithm of any positive number.

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Classroom Example 1 Example 1 Evaluating Common Logarithms with a Calculator Use a calculator to find the values of log 10,000, log 341, and Use a calculator to find the values of log 0.06894. log 1000, log 142, and log 0.005832. Answers: 4; 2.532754379; -1.161528721 Solution Figure 33 shows that the exact value of log 1000 is 3 (because 103 = 1000), and that

log 142 ≈ 2.152288344 and log 0.005832 ≈ -2.234182485. Most common logarithms that appear in calculations are approximations, as seen in the second and third displays. Figure 33

■✔ Now Try Exercises 11, 15, and 17.

For a + 1, base a logarithms of numbers between 0 and 1 are always negative, and base a logarithms of numbers greater than 1 are always positive.

Applications and Models with Common Logarithms In chemistry, the pH of a solution is defined as

pH log H3o ,

+ where H3O is the hydronium ion concentration3 4 in moles* per liter. The pH value is a measure of the acidity or alkalinity of a solution. Pure water has pH 7.0, substances3 4 with pH values greater than 7.0 are alkaline, and substances with pH values less than 7.0 are acidic. See Figure 34. It is customary to round pH values to the nearest tenth.

17 14

Acidic Neutral Alkaline

Figure 34

Classroom Example 2 Example 2 Finding pH

(a) Find the pH of a solution with + -4 + -8 (a) Find the pH of a solution with H3O = 2.5 * 10 . H3O = 6.8 * 10 . (b) Find the hydronium ion (b) Find the hydronium ion concentration of a solution with pH 7.1. 3 4 3 4 = concentration of a solution Solution with pH = 4.3. + Answers: (a) pH = -log H3O 5 (a) 7.2 (b) 5.0 * 10- -4 + -4 pH = -log32.5 *410 Substitute H3O = 2.5 * 10 . -4 pH = - log1 2.5 + log 102 Product property3 4 4 pH = -10.3979 - 4 2 log 10- = -4 pH = -10.3979 + 4 2 Distributive property pH ≈ 3.6 Add.

*A mole is the amount of a substance that contains the same number of molecules as the number of atoms in exactly 12 grams of carbon-12.

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+ (b) pH = -log H3O + 7.1 = -log3H3O 4 Substitute pH = 7.1. + -7.1 = log H33O 4 Multiply by -1. + -7.1 H3O = 10 3 4 Write in exponential form. + -8 -7.1 3H3O 4 ≈ 7.9 * 10 Evaluate 10 with a calculator. 3 4 ■✔ Now Try Exercises 29 and 33.

Note In the fourth line of the solution in Example 2(a), we use the equality symbol, =, rather than the approximate equality symbol, ≈, when replacing log 2.5 with 0.3979. This is often done for convenience, despite the fact that most logarithms used in applications are indeed approximations.

Example 3 Using pH in an Application Wetlands are classified as bogs, fens, marshes, and swamps based on pH values. A pH value between 6.0 and 7.5 indicates that the wetland is a “rich fen.” When the pH is between 3.0 and 6.0, it is a “poor fen,” and if the pH falls to 3.0 or less, the wetland is a “bog.” (Source: R. Mohlenbrock, “Summerby Swamp, Michigan,” Natural History.) Suppose that the hydronium ion concentration of a sample of water from a wetland is 6.3 * 10-5. How would this wetland be classified? + Solution pH = -log H3O Definition of pH pH log 6.3 10-5 Substitute for H O+ . Classroom Example 3 = - 3 *4 3 Refer to Example 3. The -5 pH = - log1 6.3 + log 102 Product property3 4 hydronium ion concentration of n a water sample from a wetland is pH = -1log 6.3 - -5 2 Distributive property; log 10 = n 4.5 * 10-3. Classify this wetland. pH = -log 6.3 + 15 2 Definition of subtraction Answer: bog pH ≈ 4.2 Use a calculator. The pH is between 3.0 and 6.0, so the wetland is a poor fen.

■✔ Now Try Exercise 37.

Classroom Example 4 Example 4 Measuring the Loudness of Sound Find the decibel rating of a sound The loudness of sounds is measured in decibels. We first assign an intensity of with intensity 10,000,000I0 . I0 to a very faint threshold sound. If a particular sound has intensity I, then the Answer: 70 decibel rating d of this louder sound is given by the following formula. I d 10 log I0

Find the decibel rating d of a sound with intensity 10,000I0 .

10,000I0 Solution d = 10 log Let I = 10,000I0. I0 I d 10 log 10,000 0 = 1 = I0 d = 10 4 log 10,000 = log 104 = 4 d = 401 2 Multiply. The sound has a decibel rating of 40. ■✔ Now Try Exercise 63.

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Natural Logarithms In most practical applications of logarithms, the irrational number e is used as the base. Logarithms with base e are natural logarithms because they occur in the life sciences and economics in natural situations that involve growth and decay. The base e logarithm of x is written ln x (read “el-en x”). The expression ln x represents the exponent to which e must be raised in order to obtain x.

y Natural Logarithm

f(x) = ln x For all positive numbers x, 2 x ln x log x. 0 e 2684 –2

Figure 35 A graph of the natural logarithmic function ƒ x = ln x is given in Figure 35. 1 2 Classroom Example 5 Example 5 Evaluating Natural Logarithms with a Calculator Use a calculator to find the values of ln e4, ln 341, and ln 0.06894. Use a calculator to find the values of Answers: 4; 5.831882477; ln e3, ln 142, and ln 0.005832. -2.674518718 Solution Figure 36 shows that the exact value of ln e3 is 3, and that

ln 142 ≈ 4.955827058 and ln 0.005832 ≈ -5.144395284.

Figure 36

■✔ Now Try Exercises 45, 51, and 53.

Figure 37 Figure 37 illustrates that ln x is the exponent to which e must be raised in order to obtain x.

Applications and Models with Natural Logarithms

Classroom Example 6 Example 6 Measuring the Age of Rocks Refer to the model given in Example 6. Geologists sometimes measure the age of rocks by using “atomic clocks.” By (a) How old is a rock in which measuring the amounts of argon-40 and potassium-40 in a rock, it is possible to A = K? find the age t of the specimen in years with the formula (b) How old is a rock in which A ln 1 8.33 A K = 0.325? + K t = 1.26 * 109 , Answers: ln 2 (a) 4.06 billion yr old 1 2 A A BB where A and K are the numbers of atoms of argon-40 and potassium-40, respec- (b) 2.38 billion yr old tively, in the specimen. (a) How old is a rock in which A = 0 and K 7 0? A (b) The ratio K for a sample of granite from New Hampshire is 0.212. How old is the sample?

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Solution A Looking Ahead To Calculus (a) If A = 0, then K = 0 and the equation is as follows. The natural logarithmic function A ƒ x = ln x and the reciprocal function ln 1 + 8.33 1 9 K g x = have an important relation- t = 1.26 * 10 Given formula 1 2 x ln 2 ship in calculus. The derivative of the 1 2 A A BB natural logarithmic function is the 1 2 9 ln 1 A t = 1.26 * 10 = 0, so ln 1 + 0 = ln 1 reciprocal function. Using Leibniz ln 2 K notation (named after one of the 1 92 1 2 co-inventors of calculus), we write t = 1.26 * 10 0 ln 1 = 0 d 1 this fact as ln x = . dx x t = 10 21 2 1 2 The rock is new (0 yr old). A (b) Because K = 0.212, we have the following. ln 1 + 8.33 0.212 t = 1.26 * 109 Substitute. ln 2 1 1 22 t ≈ 11.85 * 109 2 Use a calculator.

The granite is about 1.85 billion yr old. ■✔ Now Try Exercise 77.

Classroom Example 7 Example 7 Modeling Global Temperature Increase Refer to the equation given in Carbon dioxide in the atmosphere traps heat from the sun. The additional solar Example 7. Suppose that C = 2C0 and k = 14. radiation trapped by carbon dioxide is radiative forcing. It is measured in 2 (a) Find the radiative forcing watts per square meter w/m . In 1896 the Swedish scientist Svante Arrhenius given these conditions. modeled radiative forcing R caused by additional atmospheric carbon dioxide, (b) Find the average global using the logarithmic equation1 2 temperature increase to the C nearest degree Fahrenheit R = k ln , C under the same conditions. 0 Answers: where C0 is the preindustrial amount of carbon dioxide, C is the current carbon dioxide level, and k is a constant. Arrhenius determined that 10 k 16 when (a) 9.7 w/m2 (b) 10°F … … C = 2C0 . (Source: Clime, W., The Economics of Global Warming, Institute for International Economics, Washington, D.C.)

(a) Let C = 2C0. Is the relationship between R and k linear or logarithmic? (b) The average global temperature increase T (in °F) is given by T R = 1.03R. Write T as a function of k. 1 2 Solution C (a) If C = 2C0 , then C = 2, so R = k ln 2 is a linear relation, because ln 2 is a constant. 0

(b) T R = 1.03R C T1 k2 = 1.03k ln Use the given expression for R. C0 1 2 ■✔ Now Try Exercise 75.

Logarithms with Other Bases We can use a calculator to find the values of either natural logarithms (base e) or common logarithms (base 10). However, sometimes we must use logarithms with other bases. The change-of-base theorem can be used to convert logarithms from one base to another.

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Change-of-Base Theorem Looking Ahead To Calculus In calculus, natural logarithms are For any positive real numbers x, a, and b, where a 1 and b 1, the fol- more convenient to work with than lowing holds. logarithms with other bases. The log b x change-of-base theorem enables us to log a x convert any logarithmic function to a log b a natural logarithmic function.

Proof Let y = loga x. Then ay = x Write in exponential form. y logb a = logb x Take the base b logarithm on each side.

y logb a = logb x Power property

logb x y = Divide each side by logb a. logb a

logb x loga x = . Substitute loga x for y. logb a Any positive number other than 1 can be used for base b in the change-of-base theorem, but usually the only practical bases are e and 10 since most calculators Teaching Tip Show students give logarithms for these two bases only. that log2 10 can be evaluated using log 10 ln 10 Using the change-of-base theorem, we can graph an equation such as or . log 2 ln 2 log x ln x y = log2 x by directing the calculator to graph y = log 2 , or, equivalently, y = ln 2 . ■

Classroom Example 8 Example 8 Using the Change-of-Base Theorem Use the change-of-base theorem to find an approximation to four Use the change-of-base theorem to find an approximation to four decimal places decimal places for each logarithm. for each logarithm. (a) log 20 (b) log 0.7 4 2 (a) log5 17 (b) log2 0.1 Answers: Solution (a) 2.1610 (b) -0.5146 (a) We use natural logarithms to approximate this logarithm. Because log5 5 = 1 and log5 25 = 2, we can estimate log5 17 to be a number between 1 and 2. ln 17 1.7604 log5 17 = ≈ 1.7604 Check: 5 ≈ 17 ln 5

The first two entries in Figure 38(a) show that the results are the same whether natural or common logarithms are used.

(a) (b) Figure 38

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(b) We use common logarithms for this approximation.

log 0.1 3.3219 log2 0.1 = ≈ -3.3219 Check: 2- ≈ 0.1 log 2

The last two entries in Figure 38(a) show that the results are the same whether natural or common logarithms are used. Some calculators, such as the TI-84 Plus, evaluate these logarithms directly without using the change-of-base theorem. See Figure 38(b).

■✔ Now Try Exercises 79 and 81.

Example 9 Modeling Diversity of Species One measure of the diversity of the species in an ecological community is modeled by the formula

H = - P1 log2 P1 + P2 log2 P2 + g + Pn log2 Pn ,

where P1, P2, c , Pn 3are the proportions of a sample that belong4 to each of n species found in the sample. (Source: Ludwig, J., and J. Reynolds, Statistical Ecology: A Primer on Methods and Computing, © 1988, John Wiley & Sons, NY.) Find the measure of diversity in a community with two species where there are 90 of one species and 10 of the other. 90 Solution There are 100 members in the community, so P1 = 100 = 0.9 and 10 P2 = 100 = 0.1.

H = - 0.9 log2 0.9 + 0.1 log2 0.1 Substitute for P1 and P2. Classroom Example 9 Refer to the formula in Example 9. In Example 8(b),3 we found that log2 0.1 ≈ -43.32. Now we find log2 0.9. Find the measure of diversity in a community with two species where log 0.9 there are 60 of one species and 140 log2 0.9 = ≈ -0.152 Change-of-base theorem log 2 of the other. Answer: 0.881 Now evaluate H.

H = - 0.9 log2 0.9 + 0.1 log2 0.1 H ≈ -3 0.9 -0.152 + 0.1 -3.324 Substitute approximate values. H ≈ 0.4693 1 2 1 24 Simplify. Verify that H ≈ 0.971 if there are 60 of one species and 40 of the other. 1 As the proportions of n species get closer to n each, the measure of diversity increases to a maximum of log2 n. ■✔ Now Try Exercise 73.

We saw previously that graphing calculators are capable of fitting expo- y = 1.479 + 0.809 lnx nential curves to data that suggest such behavior. The same is true for logarithmic curves. For example, during the early 2000s on one particular day, interest rates 5 for various U.S. Treasury Securities were as shown in the table.

Time 3-mo 6-mo 2-yr 5-yr 10-yr 30-yr Yield 0.83% 0.91% 1.35% 2.46% 3.54% 4.58% −5 35 Source: U.S. Treasury. −2 Figure 39 shows how a calculator gives the best-fitting natural logarithmic curve Figure 39 for the data, as well as the data points and the graph of this curve. ■

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4.4 Exercises

1. increasing Concept Preview Answer each of the following. 2. increasing x -1 1. For the exponential function ƒ x = a , where a 7 1, is the function increasing or 3. ƒ x = log5 x decreasing over its entire domain? 4. log4 11 1 2 1 2 5. natural; common 2. For the logarithmic function g x log x, where a 1, is the function increasing ln 12 = a 7 6. ln 3 or decreasing over its entire domain? 7. There is no power of 2 that 1 2 3. If ƒ x 5x, what is the rule for ƒ -1 x ? yields a result of 0. = 8. 3 and 4 4. What1 2is the name given to the exponent1 2 to which 4 must be raised to obtain 11? 9. log 8 = 0.90308999 10. ln 2.75 = 1.0116009 5. A base e logarithm is called a(n) logarithm, and a base 10 logarithm is called a(n) logarithm. 11. 12 12. 7 6. How is log 12 written in terms of natural logarithms using the change-of-base 13. -1 14. -2 3 15. 1.7993 16. 1.9731 theorem? 17. 2.6576 18. 2.2596 - - 7. Why is log2 0 undefined? 19. 3.9494 20. 3.5505 21. 0.1803 22. 0.3503 8. Between what two consecutive integers must log2 12 lie? 23. 3.9494 24. 3.5505 9. The graph of y log x shows a point on the graph. Write the logarithmic equation 25. 0.1803 26. 0.3503 = associated with that point. 27. The logarithm of the product y of two numbers is equal to the (8, 0.90308999) sum of the logarithms of the 1 numbers. x 28. The logarithm of the quotient 0 48 of two numbers is equal to y = log x the difference between the –1 logarithm of the numerator and the logarithm of the 10. The graph of y = ln x shows a point on the graph. Write the logarithmic equation denominator. associated with that point. y

1 (2.75, 1.0116009) x 0 2468 –1 y = ln x

Find each value. If applicable, give an approximation to four decimal places. See Example 1. 11. log 1012 12. log 107 13. log 0.1 14. log 0.01 15. log 63 16. log 94 17. log 0.0022 18. log 0.0055 518 643 19. log 387 * 23 20. log 296 * 12 21. log 22. log 342 287 1 2 1 2 23. log 387 + log 23 24. log 296 + log 12 25. log 518 - log 342 26. log 643 - log 287

Answer each question. 27. Why is the result in Exercise 23 the same as that in Exercise 19? 28. Why is the result in Exercise 25 the same as that in Exercise 21?

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29. 3.2 30. 1.8 For each substance, find the pH from the given hydronium ion concentration. See 31. 8.4 32. 13.5 Example 2(a).

-4 -2 33. 2.0 * 10-3 34. 4.0 * 10-4 29. grapefruit, 6.3 * 10 30. limes, 1.6 * 10 -5 -7 35. 1.6 * 10 36. 3.2 * 10 9 14 31. crackers, 3.9 * 10- 32. sodium hydroxide (lye), 3.2 * 10- 37. poor fen 38. poor fen + 39. bog 40. bog Find the H3O for each substance with the given pH. See Example 2(b). 41. rich fen 42. rich fen 33. soda3 pop, 2.74 34. wine, 3.4 43. (a) 2.60031933 35. beer, 4.8 36. drinking water, 6.5 (b) 1.60031933 (c) 0.6003193298 Suppose that water from a wetland area is sampled and found to have the given hydro- (d) The whole number parts nium ion concentration. Determine whether the wetland is a rich fen, a poor fen, or a bog. will vary, but the decimal See Example 3. parts will be the same. 44. 4.3979; Each can be written 37. 2.49 * 10-5 38. 6.22 * 10-5 39. 2.49 * 10-2 as log 25 plus a logarithm of a -2 -7 -7 power of 10. 40. 3.14 * 10 41. 2.49 * 10 42. 5.86 * 10

45. 1.6 46. 5.8 Solve each problem. 47. -2 48. -4 1 1 43. Use a calculator to find an approximation for each logarithm. 49. 2 50. 3 51. 3.3322 52. 3.6636 (a) log 398.4 (b) log 39.84 (c) log 3.984 53. -8.9480 54. -4.8665 (d) From the answers to parts (a)–(c), make a conjecture concerning the decimal 55. 10.1449 56. 9.8386 values in the approximations of common logarithms of numbers greater than 1 57. 2.0200 58. 1.5976 that have the same digits. 59. 10.1449 60. 9.8386 61. 2.0200 62. 1.5976 44. Given that log 25 ≈ 1.3979, log 250 ≈ 2.3979, and log 2500 ≈ 3.3979, make a conjecture for an approximation of log 25,000. Why does this pattern continue? 63. (a) 20 (b) 30 (c) 50 (d) 60 (e) 3 decibels Find each value. If applicable, give an approximation to four decimal places. See 64. (a) 21 (b) 70 (c) 91 Example 5. (d) 120 (e) 140 1 45. ln e1.6 46. ln e5.8 47. ln e2 1 48. ln 49. ln e 50. ln 3 e e4 2 2 51. ln 28 52. ln 39 53. ln 0.00013

54. ln 0.0077 55. ln 27 * 943 56. ln 33 * 568 98 184 2 1 2 57. ln 58. ln 59. ln 27 + ln 943 13 17

60. ln 33 + ln 568 61. ln 98 - ln 13 62. ln 84 - ln 17

Solve each problem. See Examples 4, 6, 7, and 9. 63. Decibel Levels Find the decibel ratings of sounds having the following intensities.

(a) 100I0 (b) 1000I0 (c) 100,000I0 (d) 1,000,000I0 (e) If the intensity of a sound is doubled, by how much is the decibel rating increased? Round to the nearest whole number. 64. Decibel Levels Find the decibel ratings of the following sounds, having intensities as given. Round each answer to the nearest whole number.

(a) whisper, 115I0 (b) busy street, 9,500,000I0

(d) rock music, 895,000,000,000I0

(e) jetliner at takeoff, 109,000,000,000,000I0

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65. (a) 3 (b) 6 (c) 8 65. Earthquake Intensity The magnitude of an earthquake, measured on the Richter I 66. 1,258,900,000I0 scale, is log , where I is the amplitude registered on a seismograph 100 km from 10 I0 67. 631,000,000I0 the epicenter of the earthquake, and I0 is the amplitude of an earthquake of a certain 68. 1.995 times greater (small) size. Find the Richter scale ratings for earthquakes having the following 69. 106.6 thousand; We must amplitudes. assume that the model continues to be logarithmic. (a) 1000I0 (b) 1,000,000I0 (c) 100,000,000I0 70. 1590 million person-trips; 66. Earthquake Intensity On December 26, 2004, an earthquake struck in the Indian This is very close to the actual Ocean with a magnitude of 9.1 on the Richter scale. The resulting tsunami killed an number. estimated 229,900 people in several countries. Express this reading in terms of I0 to the nearest hundred thousand. 67. Earthquake Intensity On February 27, 2010, a massive earthquake struck Chile with a magnitude of 8.8 on the Richter scale. Express this reading in terms of I0 to the nearest hundred thousand. 68. Earthquake Intensity Comparison Compare the answers to Exercises 66 and 67. How many times greater was the force of the 2004 earthquake than that of the 2010 earthquake? 69. (Modeling) Bachelor’s Degrees in Psychology The table gives the number of bachelor’s degrees in psychology (in thousands) earned at U.S. colleges and univer- sities for selected years from 1980 through 2012. Suppose x represents the number of years since 1950. Thus, 1980 is represented by 30, 1990 is represented by 40, and so on.

Degrees Earned Year (in thousands) 1980 42.1 1990 54.0 2000 74.2 2010 97.2 2011 100.9 2012 109.0 Source: National Center for Education Statistics.

The following function is a logarithmic model for the data.

ƒ x = -273 + 90.6 ln x Use this function to estimate the1 number2 of bachelor’s degrees in psychology earned in the year 2016 to the nearest tenth thousand. What assumption must we make to estimate the number of degrees in years beyond 2012?

70. (Modeling) Domestic Leisure Travel U.S. Domestic Leisure Travel Volume The bar graph shows numbers of 2000 leisure trips within the United States (in millions of person-trips of 50 or 1500 more miles one-way) over the years 2009–2014. The function 1000

500 ƒ t = 1458 + 95.42 ln t, t Ú 1,

Person-Trips (in millions) Person-Trips 0 where1 2 t represents the number of years 2009 2010 2011 2012 2013 2014 since 2008 and ƒ t is the number Year of person-trips, in millions, approxi- Source: Statista 2014. mates the curve reasonably1 2 well. Use the function to approximate the number of person-trips in 2012 to the nearest million. How does this approximation compare to the actual number of 1588 million?

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71. (a) 2 (b) 2 (c) 2 (d) 1 71. (Modeling) Diversity of Species The number of species S n in a sample is 72. (a) 4 (b) 4 (c) 5 given by 73. 1 74. 1.589 1 2 75. between 7°F and 11°F n S n = a ln 1 + , 76. (a) T x = a 353 1.006 x - 1990 1 2 a b 6.4891 2 ln 280 1 2 where n is the number of individuals in the sample, and a is a constant that indicates (b) C(x) = 353(1.006)x−1990 the diversity of species in the community. If a 0.36, find S n for each value of n. C D = 2100 (Hint: S n must be a whole number.) 1 2 (a) 100 (b) 200 (c) 150 (d) 10 1 2 72. (Modeling) Diversity of Species In Exercise 71, find S n if a changes to 0.88. Use 1990 the following values of n. 2275 1 2 300 (a) 50 (b) 100 (c) 250 C T(x) 6.489 ln , where = 280 73. (Modeling) Diversity of Species Suppose a sample of a small community shows C(x) is deﬁned as given earlier. two species with 50 individuals each. Find the measure of diversity H. 15 74. (Modeling) Diversity of Species A virgin forest in northwestern Pennsylvania has 4 species of large trees with the following proportions of each:

hemlock, 0.521; beech, 0.324; birch, 0.081; maple, 0.074. 1990 2275 0 Find the measure of diversity H to the nearest thousandth. C is an exponential function, and T is a 75. (Modeling) Global Temperature Increase In Example 7, we expressed the aver- linear function over the age global temperature increase T (in °F) as same time period. While C the carbon dioxide levels T k = 1.03k ln , in the atmosphere C0 increase at an exponential 1 2 rate, the average global where C0 is the preindustrial amount of carbon dioxide, C is the current carbon temperature will rise at dioxide level, and k is a constant. Arrhenius determined that 10 … k … 16 when C

a linear rate. was double the value C0 . Use T k to find the range of the rise in global tempera- (c) The slope is 0.0388. ture T (rounded to the nearest degree) that Arrhenius predicted. (Source: Clime, This means that the W., The Economics of Global 1Warming,2 Institute for International Economics, temperature is expected Washington, D.C.) to rise at an average rate of 0.04°F per year 76. (Modeling) Global Temperature Increase (Refer to Exercise 75.) According to from 1990 to 2275. one study by the IPCC, future increases in average global temperatures (in °F) can (d) x ≈ 2208.9; be modeled by C ≈ 1307 ppm (about 4.67 times above C T C = 6.489 ln , preindustrial carbon 280 dioxide levels) 1 2 77. 1.13 billion yr where C is the concentration of atmospheric carbon dioxide (in ppm). C can be modeled by the function

C x = 353 1.006 x-1990,

where x is the year. (Source: International1 2 1 Panel2 on Climate Change (IPCC).) (a) Write T as a function of x. (b) Using a graphing calculator, graph C x and T x on the interval [1990, 2275] using different coordinate axes. Describe the graph of each function. How are C and T related? 1 2 1 2 (c) Approximate the slope of the graph of T. What does this slope represent? (d) Use graphing to estimate x and C x when T x = 10°F. 77. Age of Rocks Use the formula of Example1 2 6 to1 estimate2 the age of a rock sample A having K = 0.103. Give the answer in billions of years, rounded to the nearest hundredth.

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78. (a) 6 78. (Modeling) Planets’ Distances from the Sun Planet DP and Periods of Revolution The table contains the planets’ average distances D from the sun and Mercury 0.39 0.24 their periods P of revolution around the sun in Venus 0.72 0.62 −2 4 years. The distances have been normalized so that Earth 1 1 −2 Earth is one unit away from the sun. For example, Mars 1.52 1.89 Let x = ln D and y = ln P since Jupiter’s distance is 5.2, its distance from for each planet. From the the sun is 5.2 times farther than Earth’s. Jupiter 5.2 11.9 graph, the data appear to (a) Using a graphing calculator, make a scatter Saturn 9.54 29.5 be linear. diagram by plotting the point (ln D, ln P) for Uranus 19.2 84.0 (b) y 1.5x = each planet on the xy-coordinate axes. Do Neptune 30.1 164.8 6 the data points appear to be linear? Source: Ronan, C., The Natural (b) Determine a linear equation that models the History of the Universe, MacMillan data points. Graph the line and the data on Publishing Co., New York. −2 4 the same coordinate axes. −2 (c) Use this linear model to predict the period of Pluto if its distance is 39.5. The points 0, 0 and Compare the answer to the actual value of 248.5 yr. 3.40, 5.10 determine the line y =11.5x2 or Use the change-of-base theorem to find an approximation to four decimal places for ln1 P = 1.5 ln2 D. each logarithm. See Example 8. (Answers will vary.) 79. log 5 80. log 9 81. log 0.59 82. log 0.71 (c) P ≈ 248.3 yr 2 2 8 8

79. 2.3219 80. 3.1699 83. log1/2 3 84. log1/3 2 85. logp e 86. logp 2 81. -0.2537 82. -0.1647 2 87. log 12 88. log 5 89. log0.32 5 90. log0.91 8 83. 1.5850 84. 0.6309 13 19 - - 2 2 85. 0.8736 86. 0.3028 87. 1.9376 88. 1.0932 Let u = ln a and v = ln b. Write each expression in terms of u and v without using the 89. -1.4125 90. -22.0488 ln function. 3 3 a a 3 91. 4v 1 u 92. 3u 2v 91. ln b4 a 92. ln 93. ln 94. ln a # b4 + 2 - b2 b5 3 5 1 2 2 93. 2 u - 2 v 94. 3 u + 4v B ConceptA Check B Use the various properties of exponential and logarithmicA functionsB to 1 95. (a) 4 (b) 25 (c) e evaluate the expressions in parts (a) – (c). 1 96. (a) 2 (b) ln 3 (c) ln 9 95. Given g x = ex, find (a) g ln 4 (b) g ln 52 (c) g ln . 97. (a) 6 (b) ln 3 (c) ln 9 e 98. (a) 7 (b) 1 (c) 2 96. Given ƒ x 3x, find (a) ƒ log 2 (b) ƒ log ln 3 (c) ƒ log 2 ln 3 . 1 2 = 1 32 1 3 2 A 3B 6 ln 3 2 ln 3 99. D 100. D 97. Given ƒ1x2 = ln x, find (a) ƒ1e 2 (b) ƒ1e 1 22 (c) ƒ1e 1 . 22 7 log 2 2 log 2 98. Given ƒ x = log2 x, find (a) ƒ 2 (b) ƒ 2 2 (c) ƒ 2 2 . 101. domain: -∞, 0 ´ 0, ∞ ; 1 2 1 2 1 2 1 2 range: -∞, ∞ ; symmetric 1 2 1 2 1 2 1 2 with respect1 to the2 y-axis1 2 Work each problem. 1 2 102. (a) -∞, 0 ´ 0, ∞ 99. Concept Check Which of the following is equivalent to 2 ln 3x for x 7 0? (b) 2 1 2 f(1x) = log2 3 ͉x͉ A. ln 9 + ln x B. ln 6x C. ln 6 + ln x D. ln 9x 4 1 2 100. Concept Check Which of the following is equivalent to ln 4x - ln 2x for x 7 0? ln 4x −4 4 A. 2 ln x B. ln 2x C. 1 2D. ln 12 2 ln 2x

−4 101. The function ƒ x = ln x plays a prominent role in calculus. Find its domain, its range, and the symmetries of its graph. (c) Because of limited reso- 1 2 lution, the graph appears 0 0 102. Consider the function ƒ x = log3 x . to show a point with (a) What is the domain of this function? x-value of 0, which does 1 2 0 0 not exist on this graph. (b) Use a graphing calculator to graph ƒ x = log3 x in the window -4, 4 by -4, 4 . 103. ƒ x 2 ln x, so it is 1 2 3 4 = + (c) How might one easily misinterpret the domain0 0of the function by merely the graph of g x = ln x 3 4 translated1 2 2 units up. observing the calculator graph? 1 2 Copyright Pearson. All Rights Reserved.

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104. ƒ x = ln x - 1, so it is the Use properties of logarithms to rewrite each function, and describe how the graph of the graph of g x = ln x trans- given function compares to the graph of g x = ln x. lated1 2 1 unit down. 1 2 1 2x x 105. ƒ x = ln x - 2, so it is the 103. ƒ x = ln e2x 104. ƒ x = ln 105. ƒ x = ln e e2 graph of g x = ln x trans- lated1 2 2 units down. 1 2 1 2 1 2 1 2 1 2

Chapter 4 Quiz (Sections 4.1—4.4)

3 [4.1] 1. For the one-to-one function ƒ x = 3x - 6, find ƒ-1 x . x3 6 1. ƒ -1 x = + 2 3 2. Solve 42x+1 = 83x-6. 1 2 1 2 1 2 [4.2] 3. Graph ƒ x = -3x. Give the domain and range. 2. 4 3. y domain: 4. Graph ƒ1x2 = log4 x + 2 . Give the domain and range. 5 6 2 -∞, ∞ ; 0 5. Future Value Suppose that $15,000 is deposited in a bank certificate of deposit at x range: 1 2 1 2 –3 3 an annual rate of 2.7% for 8 yr. Find the future value if interest is compounded as –3 1-∞, 0 2 f(x) = –3x follows. –9 1 2 (a) annually (b) quarterly (c) monthly (d) daily (365 days)

[4.3] 6. Use a calculator to evaluate each logarithm to four decimal places. 4. y domain: (a) log 34.56 (b) ln 34.56 2, ∞ ; 4 - 7. What is the meaning of the expression log6 25? –2 range: x 1 , 2 0 2 -∞ ∞ 8. Solve each equation.

1 2 (a) x 3log3 4 (b) log 25 2 (c) log x 2 f(x) = log (x + 2) = x = 4 = - 4 9. Assuming all variables represent positive real numbers, use properties of logarithms [4.2] to rewrite 5. (a) $18,563.28 (b) $18,603.03 (c) $18,612.02 (d) $18,616.39 x # y log . 3 2pq4 [4.4] 6. (a) 1.5386 (b) 3.5427 10. Given logb 9 = 3.1699 and logb 5 = 2.3219, find the value of logb 225.

[4.3] 11. Find the value of log3 40 to four decimal places. 7. The expression log 25 repre- 6 12. If ƒ x 4x, what is the value of ƒ log 12 ? sents the exponent to which 6 = 4 must be raised to obtain 25. [4.4] 1 2 1 2 1 1 8. (a) 4 (b) 5 (c) 16 9. 2 log3 x + log3 y - log3 p - 4 log3 q 10. 7.8137 11. 3.3578 12. 12 5 6 5 6 E F 4.5 Exponential and Logarithmic Equations

■ Exponential Equations Exponential Equations We solved exponential equations in earlier sec- ■ Logarithmic Equations tions. General methods for solving these equations depend on the property below, ■ Applications and which follows from the fact that logarithmic functions are one-to-one. Models

Property of Logarithms

If x 7 0, y 7 0, a 7 0, and a 1, then the following holds.

x y is equivalent to log a x log a y.

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Classroom Example 1 Example 1 Solving an Exponential Equation Solve 8x 21. Give the solution to = x the nearest thousandth. Solve 7 = 12. Give the solution to the nearest thousandth. Answer: 1.464 Solution The properties of exponents cannot be used to solve this equation, 5 6 so we apply the preceding property of logarithms. While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.

15 7x = 12 ln 7x = ln 12 Property of logarithms −2 5 x ln 7 = ln 12 Power property This is exact. ln 12 −20 x = Divide by ln 7. ln 7 As seen in the display at the bottom Use a calculator. of the screen, when rounded to three x ≈ 1.277 decimal places, the solution of This is approximate. x The solution set is 1.277 . 7 - 12 = 0 agrees with that found in Example 1. ■✔ Now Try Exercise 11. 5 6

ln 12 12 Caution Do not confuse a quotient like ln 7 in Example 1 with ln 7 , which can be written as ln 12 - ln 7. We cannot change the quotient of two logarithms to a difference of logarithms.

ln 12 12 ln ln 7 7

Classroom Example 2 Example 2 Solving an Exponential Equation Solve 52x-3 8x+1. Give the = 2x-1 x+2 solution to the nearest thousandth. Solve 3 = 0.4 . Give the solution to the nearest thousandth. 2x 1 x 2 Answer: 6.062 Solution 3 - = 0.4 + 2x-1 x+2 5 6 ln 3 = ln 0.4 Take the natural logarithm on each side. Teaching Tip Caution students 2x - 1 ln 3 = x + 2 ln 0.4 Power property against applying the quotient 2x ln 3 ln 3 x ln 0.4 2 ln 0.4 Distributive property property of logarithms incorrectly 1 -2 = 1 2 + in Example 2. Write so that the terms with x are 2x ln 3 - x ln 0.4 = 2 ln 0.4 + ln 3 on one side. x 2 ln 3 - ln 0.4 = 2 ln 0.4 + ln 3 Factor out x. 1 2 2 ln 0.4 + ln 3 x = Divide by 2 ln 3 - ln 0.4. 2 ln 3 - ln 0.4 ln 0.42 + ln 3 x = Power property ln 32 - ln 0.4 ln 0.16 + ln 3 3 x = Apply the exponents. ln 9 - ln 0.4 This is exact. ln 0.48 −4 4 x = Product and quotient properties ln 22.5 x ≈ -0.236 Use a calculator. −3 This is approximate. This screen supports the solution The solution set is -0.236 . found in Example 2. ■✔ Now Try Exercise 19. 5 6 Copyright Pearson. All Rights Reserved.

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Classroom Example 3 Example 3 Solving Base e Exponential Equations Solve each equation. Give solutions to the nearest Solve each equation. Give solutions to the nearest thousandth. 2 thousandth. (a) e x = 200 (b) e2x+1 # e-4x = 3e (a) e x = 50 Solution (b) e40 x0 # e x-1 = 5e x2 Answers: (a) e = 200 2 (a) {3.912 (b) 0.722 ln e x = ln 200 Take the natural logarithm on each side.

2 2 5 6 5 6 x = ln 200 ln ex = x2 Remember both roots. x = { ln 200 Square root property 2 x ≈ {2.302 Use a calculator. The solution set is {2.302 .

(b) e2x+1 # e-4x5= 3e 6 2x 1 e- + = 3e am # an = am+n -2x + 1 -2x e -2x+1-1 -2x e = 3 Divide by e; e1 = e = e . ln e-2x = ln 3 Take the natural logarithm on each side. -2x ln e = ln 3 Power property -2x = ln 3 ln e = 1

1 1 x = - ln 3 Multiply by - . 2 2 x ≈ -0.549 Use a calculator. The solution set is -0.549 . ■✔ Now Try Exercises 21 and 23. 5 6

Classroom Example 4 Example 4 Solving an Exponential Equation (Quadratic in Form) Solve e2x 6e x 5 0. Give - + = 2x x exact value(s) for x. Solve e - 4e + 3 = 0. Give exact value(s) for x. x Answer: 0, ln 5 Solution If we substitute u = e , we notice that the equation is quadratic in form. 5 6 e2x - 4e x + 3 = 0

x 2 x e - 4e + 3 = 0 amn = an m

2 x 1 2 u - 4u + 3 = 0 Let u =1 e 2. u - 1 u - 3 = 0 Factor. u - 1 = 01 or21 u - 32 = 0 Zero-factor property u = 1 or u = 3 Solve for u. e x = 1 or e x = 3 Substitute e x for u. Take the natural logarithm ln e x ln 1 or ln e x ln 3 = = on each side. x = 0 or x = ln 3 ln e x = x; ln 1 = 0 Both values check, so the solution set is 0, ln 3 . ■✔ Now Try Exercise 35. 5 6 Copyright Pearson. All Rights Reserved.

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Logarithmic Equations The following equations involve logarithms of variable expressions.

Classroom Example 5 Example 5 Solving Logarithmic Equations Solve each equation. Give exact values. Solve each equation. Give exact values. 3 (a) 4 ln x = 36 (a) 7 ln x = 28 (b) log2 x - 19 = 3 3 (b) log3 x - 5 = 1 Solution 1 2 Answers:1 2 (a) e9 (b) 2 (a) 7 ln x = 28

5 6 5 6 loge x = 4 ln x = loge x; Divide by 7. x = e4 Write in exponential form. The solution set is e4 .

3 (b) log2 5x -6 19 = 3 3 3 1 x - 192 = 2 Write in exponential form. x3 - 19 = 8 Apply the exponent. x3 = 27 Add 19. 3 x = 27 Take cube roots. 2 3 x = 3 27 = 3 2 The solution set is 3 . ■✔ Now Try Exercises 41 and 49. 5 6

Classroom Example 6 Example 6 Solving a Logarithmic Equation Solve Solve log x + 6 - log x + 2 = log x. Give exact value(s). log 2x + 1 + log x = log x + 8 . Solutio1n Recall2 that1 logarithms2 are defined only for nonnegative numbers. Give1 exact value(s).2 1 2 Answer: 2 log x + 6 - log x + 2 = log x x + 6 5 6 1 2 log1 2 = log x Quotient property x 2 Teaching Tip Tell students that + a logarithmic equation cannot be x + 6 = x Property of logarithms solved with two logarithms on the x 2 same side of the equation. The + equation must be transformed x + 6 = x x + 2 Multiply by x + 2. so that each side contains an 2 expression with at most one x + 6 = x1 + 2x2 Distributive property logarithm. x2 + x - 6 = 0 Standard form x + 3 x - 2 = 0 Factor. x + 3 = 01 or21 x - 2 = 0 Zero-factor property x = -3 or x = 2 Solve for x. The proposed negative solution -3 is not in the domain of log x in the original equation, so the only valid solution is the positive number 2. The solution set is 2 . 1 2 ■✔ Now Try Exercise 69. 5 6 Copyright Pearson. All Rights Reserved.

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Caution Recall that the domain of y = loga x is 0, ∞ . For this reason, it is always necessary to check that proposed solutions of a logarithmic equation result in logarithms of positive numbers in1 the original2 equation.

Classroom Example 7 Example 7 Solving a Logarithmic Equation Solve log3 4x + 1 x + 1 = 3. Give exact value(s). Solve log2 3x - 7 x - 4 = 3. Give exact value(s). 31 21 24 13 Solution log 3x 7 x 4 3 Answer: - 4 , 2 31 221 -24 - = 3x - 7 x - 4 = 23 Write in exponential form. E F 31 21 24 2 1 3x - 1921x + 282 = 8 Multiply. Apply the exponent. 3x2 - 19x + 20 = 0 Standard form 3x - 4 x - 5 = 0 Factor. 3x - 4 1= 0 or21 x - 52 = 0 Zero-factor property 4 x = or x = 5 Solve for x. 3 A check is necessary to be sure that the argument of the logarithm in the given equation is positive. In both cases, the product 3x - 7 x - 4 leads to 8, and log 8 3 is true. The solution set is 4 , 5 . 2 = 3 1 21 2 ■✔ Now Try Exercise 53. E F Classroom Example 8 Example 8 Solving a Logarithmic Equation Solve Solve log 3x + 2 + log x - 1 = 1. Give exact value(s). log2 2x - 5 + log2 x - 3 = 3. Solution1 log 23x + 21 + log2 x - 1 = 1 Give 1exact value(s).2 1 2 log10 3x + 2 x - 1 = 1 log x = log10 x; product property Answer: 11 + 65 1 2 1 2 42 1 31 3x + 212 x - 241 = 10 Write in exponential form. U V 2 1 1 3x 21- x - 2 = 10 Multiply; 10 = 10. 3x2 - x - 12 = 0 Subtract 10. -b { b2 - 4ac x = 22a Quadratic formula

- -1 { -1 2 - 4 3 -12 x = 2 2 3 1 2 1 2 1 21 2 Substitute a = 3, b = -1, c = -12. The two proposed solutions are 1 2

1 - 145 1 + 145 and . 26 26

The first proposed solution, 1 - 145 , is negative. Substituting for x in 62 log x - 1 results in a negative argument, which is not allowed. Therefore, this solution must be rejected. 1The second2 proposed solution, 1 + 145 , is positive. Substituting it for x in 62 log 3x + 2 results in a positive argument. Substituting it for x in log x + 1 also results in a positive argument. Both are necessary conditions. Therefore, the 1 2 1 2 solution set is 1 + 145 . 62 ■✔ Now Try Exercise 77. Copyright Pearson.E All Rights FReserved.

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Note We could have replaced 1 with log10 10 in Example 8 by first writing

log 3x + 2 + log x - 1 = 1 Equation from Example 8

log1 10 3x2 + 2 x1 - 1 2 = log10 10 Substitute. 31 3x + 212 x - 241 = 10, Property of logarithms and then continuing1 as shown21 on2 the preceding page.

Classroom Example 9 Example 9 Solving a Base e Logarithmic Equation Solve ln eln x ln x 4 ln 5. - - = ln x Give exact value(s). Solve ln e - ln x - 3 = ln 2. Give exact value(s). 1 2 Answer: 5 Solution This1 logarithmic2 equation differs from those in Examples 7 and 8 because the expression on the right side involves a logarithm. 5 6 ln eln x - ln x - 3 = ln 2

ln x ln x - ln 1x - 32 = ln 2 e = x x ln1 2 = ln 2 Quotient property x - 3 x = 2 Property of logarithms x - 3 x = 2 x - 3 Multiply by x - 3. x = 21x - 6 2 Distributive property x = 6 Solve for x. Check that the solution set is 6 . ■✔ Now Try Exercise 79. 5 6

Solving an Exponential or Logarithmic Equation

To solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a 7 0 and a 1, and follow the guidelines. ƒ x 1. a b Solve1 2 by taking logarithms on each side.

2. log a ƒ x b Solve by changing to exponential form ab ƒ x . 1 2 = 3. log a ƒ x log a g x 1 2 The given equation is equivalent to the equation ƒ x g x . Solve 1 2 1 2 = algebraically. 1 2 1 2 4. In a more complicated equation, such as e2x+1 # e-4x = 3e, See Example 3(b). ƒ x it may be necessary to first solve for a or loga ƒ x and then solve the resulting equation using one of the methods1 2 given above. 1 2 5. Check that each proposed solution is in the domain.

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Applications and Models

Classroom Example 10 Example 10 Applying an Exponential Equation to the Strength -kt The equation T = T0 + Ce can of a Habit be used to describe Newton’s law of cooling. Solve this equation The strength of a habit is a function of the number of times the habit is repeated. for k. If N is the number of repetitions and H is the strength of the habit, then, accord-

1 T - T0 ing to psychologist C.L. Hull, Answer: k = - t ln C H = 1000 1 - e-kN , A B where k is a constant. Solve this equation1 for k. 2 Solution H 1000 1 e-kN First solve for = - e-kN. H = 1 - e1-kN 2 Divide by 1000. 1000 H - 1 = -e-kN Subtract 1. 1000

Classroom Example 11 kN H e- = 1 - Multiply by -1 and rewrite. Refer to the model in Example 11. 1000 (a) Use the function ƒ t to Now solve estimate the number of tablets H Take the natural logarithm for k. ln e-kN = ln 1 - sold in 2011. How 1does2 this 1000 on each side. figure compare to the actual a b figure of 24.1 million? H -kN = ln 1 - ln ex = x (b) If this trend continues, 1000 approximately when will a b annual sales reach 58 million 1 H 1 k = - ln 1 - Multiply by - . tablets? N 1000 N Answers: a b With the final equation, if one pair of values for H and N is known, k can be found, (a) 24.8 million; This is fairly close to the actual figure. and the equation can then be used to find either H or N for given values of the other variable. (b) Annual sales will reach ■✔ Now Try Exercise 91. 58 million in 2019.

Example 11 Modeling PC Tablet Sales in the U.S.

Sales The table gives U.S. tablet sales (in millions) for several years. The data can be Year (in millions) modeled by the function 2010 10.3 ƒ t = 20.57 ln t + 10.58, t Ú 1, 2011 24.1 2012 35.1 where t is the number of1 years2 after 2009. 2013 39.8 (a) Use the function to estimate the number of tablets sold in the United States 2014 42.1 in 2015. Source: Forrester Research. (b) If this trend continues, approximately when will annual sales reach 60 million?

Solution (a) The year 2015 is represented by t = 2015 - 2009 = 6. ƒ t = 20.57 ln t + 10.58 Given function ƒ 162 = 20.57 ln 6 + 10.58 Let t = 6. ƒ162 ≈ 47.4 Use a calculator. Based on this model,1 2 47.4 million tablets were sold in 2015.

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(b) Replace ƒ t with 60 and solve for t.

1 2 ƒ t = 20.57 ln t + 10.58 Given function 1 602 = 20.57 ln t + 10.58 Let ƒ t = 60. 49.42 = 20.57 ln t Subtract1 2 10.58. 49.42 ln t = Divide by 20.57 and rewrite. 20.57 t = e49.42/20.57 Write in exponential form. t ≈ 11.05 Use a calculator. Adding 11 to 2009 gives the year 2020. Based on this model, annual sales will reach 60 million in 2020. ■✔ Now Try Exercise 111.

4.5 Exercises

1. B 2. F Concept Preview Match each equation in Column I with the best first step for 3. E 4. A solving it in Column II. 5. D 6. C I II log 19 ln 19 1. 10x 150 A. Use the product rule for exponents. 7. log7 19; log 7 ; ln 7 = log 10 1 ln 10 2. e2x-1 24 B. Take the common logarithm on each side. 8. log3 10; log 3 , or log 3 ; ln 3 = C. Write the sum of logarithms as the log 12 3. log x2 10 2 ln 12 4 - = logarithm of a product. 9. log1/2 12; 1 ; 1 log ln 2x x 2 2 4. e #1e = 2e 2 D. Let u = ex and write the equation in log 4 ln 4 quadratic form. 5. 2e2x - 5ex - 3 = 0 10. log1/3 4; 1 ; 1 log 3 ln 3 E. Change to exponential form. 6. log 2x - 1 + log x + 4 = 1 F. Take the natural logarithm on each side. 11. 1.771 12. 1.594 1 2 1 2 13. 2.322 14. 1.631 Concept Preview An exponential equation such as 5 - 6 5 - 6 15. 6.213 16. 2.151 x 5 - 6 5 - 6 5 = 9 17. 1.710 18. 0.823 5 - 6 5 6 can be solved for its exact solution using the meaning of logarithm and the change-of- 19. 3.240 5 6 5 6 base theorem. Because x is the exponent to which 5 must be raised in order to obtain 9, the exact solution is 5 6 log 9 ln 9 log 9, or , or . 5 log 5 ln 5 For each equation, give the exact solution in three forms similar to the forms above. 1 x 1 x 7. 7x = 19 8. 3x = 10 9. = 12 10. = 4 2 3 a b a b Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35–40, give solutions in exact form. See Examples 1–4. 1 x 11. 3x = 7 12. 5x = 13 13. = 5 2 1 x a b 14. = 6 15. 0.8x = 4 16. 0.6x = 3 3 a b 17. 4x-1 = 32x 18. 2x+3 = 52x 19. 6x+1 = 42x-1

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2 4 20. -5.057 20. 3x-4 = 72x+5 21. ex = 100 22. ex = 1000 21. 2.146 22. 1.621 5{ 6 { 1 x 23. 9.386 24. 0.347 23. e3x-7 # e-2x = 4e 24. e1-3x # e5x = 2e 25. = -3 5 6 5 6 3 25. ∅ 26. ∅ 5 6 5 6 1 x a b 27. 32.950 28. 6.579 26. = -9 27. 0.05 1.15 x = 5 28. 1.2 0.9 x = 0.6 9 29. 7.044 30. 1 5 6 5 6 a b 1 2 1 2 31. 25.677 32. 9.095 29. 3 2 x-2 + 1 = 100 30. 5 1.2 3x-2 + 1 = 7 31. 2 1.05 x + 3 = 10 5 6 5 6 33. 2011.568 34. 1917.096 x x-1980 x-1900 5 6 5 6 32. 311.42 - 4 = 60 33. 511.0152 = 8 34. 611.0242 = 9 35. ln 2, ln 4 36. ln 3, ln 5 5 6 5 6 35. e21x - 26ex + 8 = 0 36. e21x - 8e2x + 15 = 0 37. 2e12x + e2x = 6 37. 5ln 3 6 38. 5ln 1 6 2 3 38. 3e2x + 2ex = 1 39. 52x + 3 5x = 28 40. 32x - 12 3x = -35 39. log5 4 40. Elog F 5, log 7 E F 1 2 1 2 5 3 6 3 Solve each equation. Give solutions in exact form. See Examples 5–9. 41. 5e2 42.6 e3 e1.5 e5 41. 5 ln x = 10 42. 3 ln x = 9 43. 5 4 6 44. 5 2 6 43. ln 4x = 1.5 44. ln 2x = 5 45. 2 - 10 E F E F 45. log 2 - x = 0.5 46. log 3 - x = 0.75 24 46. 3 - 1000 E F 47. log 2x 4 2 48. log 8 3x 3 2 61 +2 = 51 - 2 = 47. 16 48. -39 49. log x3 37 3 50. log x3 65 0 49. E3 50.F 4 4 1 + 2 = 7 1 + 2 = 5 6 5 - 6 51. e 52. 10 51. ln x + ln x2 = 3 52. log x + log x2 = 3 5 6 5 6 1 2 1 2 14 53. log x 5 x 3 2 54. log 3x 8 x 6 3 53. 5 -66, 4 54. 5 - 63 , 8 3 + - = 4 + - = 8 55. 5 8, 06 56. , 0 55. log2 312x + 218 x + 244 = 5 56. log5 313x + 521x + 124 = 1 - E - 3 F 57. log x log x 15 2 58. log x log 2x 1 1 57. 55 6 58. 2 31+ 21+ 24= 31+ 21+ 24= 59. 5 60. E 1 F 5 -6 5 -6 59. log x + 251 = log 2x + 10 + log 4 60. log 3x + 51 - log 22x + 4 = 0 61. ∅ 62. ∅ 5 6 5 6 61. log 1x - 102 - log 1x - 6 2= log 2 62. log 1x2 - 92 - log 1x - 3 2= log 5 63. -2 64. -5 9 63. ln 71 - x +2 ln 1 -1 x =2ln 25 - x 64. ln 31 - x +2 ln 5 -1 x =2ln 50 - 6x 65. 50 6 66. 52 6 65. log18 x +22 + log1 8 x2+ 4 =1 log8 8 2 66. log12 5x -2 6 -1log2 2x + 1 1= log2 32 67. 5126 68. 3 E F 2 69. 25 70. 5 67. log2 x - 100 - log2 x + 10 = 1 68. log2 x - 2 + log2 x - 1 = 1 5 6 5 6 1 2 1 2 1 2 1 2 71. ∅ 72. ∅ 5 6 5 6 69. log x1+ log x 2- 21 =1log 1002 70. log x1+ log2 3x - 131 = log2 10 73. 5 74. 4 2 71. log 9x + 51 = 3 +2log x + 2 72. log 11x + 91 = 3 +2log x + 3 75. 3 76. 8 E F 5 6 73. ln 41x - 2 2- ln 4 = -ln1 x -22 74. ln 51 + 4x -2 ln 3 + x 1= ln 32 77. 5 16+ 41 78. 5 16+ 85 42 62 75. log15 x + 22 + log5 x - 21 = 1 2 76. log12 x - 72 + log1 2 x =23 79. U6 V 80. U6 V 77. log2 12x - 23 + log12 x + 12 = 1 78. log5 13x + 2 + log5 x - 1 = 1 81. 4 82. 1, 10 5 6 5 6 79. ln ex 2 ln e ln e4 80. ln ex ln e3 ln e3 83. 1, 100 84. 2, 2 1- 2 = 1 2 1- 2= 1 2 5 6 5 - 6 81. log log x 1 82. log x log x 85. Proposed5 6 solutions5 that cause6 2 2 = = any argument of a logarithm 2 12 2 2 2 3 83. log x = log x 84. log2 2x = to be negative or zero must be 2 rejected. The statement is not 2 1 2 correct. For example, the 85. Concept Check Consider the following statement: “We must reject any negative solution set of proposed solution when we solve an equation involving logarithms.” Is this correct? Why or why not? log -x + 99 = 2 86. Concept Check What values of x could not possibly be solutions of the following is -1 . 1 2 equation? 86. any real numbers less than or 5 67 equal to 2 4 loga 4x - 7 + loga x + 4 = 0

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87. x = ek/ p-a Solve each equation for the indicated variable. Use logarithms with the appropriate 88. t = e p1-r /k2 bases. See Example 10. 1 1 2 T - T0 89. t = - k log T T k 1 - 0 87. p = a + , for x 88. r = p - k ln t, for t ln x log A - Pr A A B 90. n = - Pr log 1 r -kt + 89. T = T0 + T1 - T0 10 , for t 90. A = , for n A B 1 - 1 + r -n 91. t 2 ln 1 RI = - R 1 - E2 1 2 E K1 2 K - y -Rt/2 ln 91. I = 1 - e , for t 92. y = bx , for b ayA B R 1 + ae- 92. b = x - 1 2 M AA BB y 93. y A B 1 e-Cx , for x 94. m 6 2.5 log , for M ln + - = + - = - B M0 93. x = -C 1 2 I A B 95. log A log B C log x, for A 96. d 10 log , for I 94. M M 10 6-m /2.5 = - = = 0 # I0 1 2 95. A B = xC r t n d/10 97. A = P 1 + , for t 98. D = 160 + 10 log x, for x 96. I = I0 # 10 n A a b log P 97. t = To solve each problem, refer to the formulas for compound interest. n log 1 r + n r tn A P 1 and A Pert 98. x 10D/10-16 = A B n a b 99. $11,611.84 100. $6885.64 99. Compound Amount If $10,000 is invested in an account at 3% annual interest 101. 2.6 yr 102. 5.55 yr compounded quarterly, how much will be in the account in 5 yr if no money is 103. 2.64% 104. 3.09% withdrawn?

105. (a) 10.9% (b) 35.8% 100. Compound Amount If $5000 is invested in an account at 4% annual interest com- (c) 84.1% pounded continuously, how much will be in the account in 8 yr if no money is 106. (a) $3891 (b) $6990 withdrawn? (c) $8495 101. Investment Time Kurt wants to buy a $30,000 truck. He has saved $27,000. Find the number of years (to the nearest tenth) it will take for his $27,000 to grow to $30,000 at 4% interest compounded quarterly. 102. Investment Time Find t to the nearest hundredth of a year if $1786 becomes $2063 at 2.6%, with interest compounded monthly. 103. Interest Rate Find the interest rate to the nearest hundredth of a percent that will produce $2500, if $2000 is left at interest compounded semiannually for 8.5 yr. 104. Interest Rate At what interest rate, to the nearest hundredth of a percent, will $16,000 grow to $20,000 if invested for 7.25 yr and interest is compounded quarterly?

(Modeling) Solve each application. See Example 11. 105. In the central Sierra Nevada (a mountain range in California), the percent of moisture that falls as snow rather than rain is approximated reasonably well by ƒ x = 86.3 ln x - 680, where x is the altitude in feet and1 2 ƒ x is the percent of moisture that falls as snow. Find the percent of moisture, to the nearest tenth, that falls as snow at each altitude. 1 2 (a) 3000 ft (b) 4000 ft (c) 7000 ft 106. Northwest Creations finds that its total sales in dollars, T x , from the distribution of x thousand catalogues is approximated by 1 2 T x = 5000 log x + 1 . Find the total sales, to the nearest1 2 dollar, resulting1 2 from the distribution of each number of catalogues. (a) 5000 (b) 24,000 (c) 49,000

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107. 2019 107. Average Annual Public University Costs The Year Average Annual Cost 108. (a) 11.65 m per sec table shows the cost of a year’s tuition, room 2006 $12,837 (b) 2.48 sec and board, and fees at 4-year public colleges 109. (a) 62% (b) 1989 for the years 2006–2014. Letting y represent 2007 $13,558 the cost in dollars and x the number of years 110. (a) The graph of ƒ -x is 2008 $14,372 symmetric to the graph since 2006, the function 2009 $15,235 of ƒ x with respect1 2 to ƒ x = 13,017 1.05 x 2010 $16,178 the y-axis. 1 2 (b) 984 ft (c) 39 ft models the data1 2 quite well.1 According2 to this 2011 $17,156 function, in what year will the 2006 cost be 2012 $17,817 doubled? 2013 $18,383 2014 $18,943 Source: The College Board, Annual Survey of Colleges.

108. Race Speed At the World Championship races held at Rome’s Olympic Stadium in 1987, American sprinter Carl Lewis ran the 100-m race in 9.86 sec. His speed in meters per second after t seconds is closely modeled by the function

ƒ t = 11.65 1 - e-t/1.27 . (Source: Banks, Robert B., Towing1 2 Icebergs,1 Falling2 Dominoes, and Other Adven- tures in Applied Mathematics, Princeton University Press.) (a) How fast, to the nearest hundredth, was he running as he crossed the finish line? (b) After how many seconds, to the nearest hundredth, was he running at the rate of 10 m per sec? 109. Women in Labor Force The percent of women in the U.S. civilian labor force can be modeled fairly well by the function

67.21 ƒ x = , 1 + 1.081e-x/24.71 1 2 where x represents the number of years since 1950. (Source: Monthly Labor Review, U.S. Bureau of Labor Statistics.) (a) What percent, to the nearest whole number, of U.S. women were in the civilian labor force in 2014? (b) In what year were 55% of U.S. women in the civilian labor force?

110. Height of the Eiffel Tower One side of y the Eiffel Tower in Paris has a shape that can be approximated by the graph of the function x f(x) = –301 ln , x > 0 x 207 ƒ x = -301 ln , x 7 0, 207 1 2 where x and ƒ x are both measured in 100 feet. (Source: Banks, Robert B., Towing x Icebergs, Falling1 2 Dominoes, and Other 100 Adventures in Applied Mathematics, Princeton University Press.) (a) Why does the shape of the left side of the graph of the Eiffel Tower have the formula given by ƒ -x ? (b) The short horizontal segment at the top of the figure has length 7.8744 ft. How 1 2 tall, to the nearest foot, is the Eiffel Tower? (c) How far from the center of the tower is the point on the right side that is 500 ft above the ground? Round to the nearest foot.

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0.0034 0.0053x 111. (a) P x = 1 - e- - 111. CO2 Emissions Tax One action that government could take to reduce carbon (b) P(x) = 1 − e−0.0034 − 0.0053x emissions into the atmosphere is to levy a tax on fossil fuel. This tax would be 1 2 1 based on the amount of carbon dioxide emitted into the air when the fuel is burned. The cost-benefit equation

ln 1 - P = -0.0034 - 0.0053x

0 1000 models the approximate relationship1 2 between a tax of x dollars per ton of carbon 0 and the corresponding percent reduction P (in decimal form) of emissions of carbon dioxide. (Source: Nordhause, W., “To Slow or Not to Slow: The Economics of (c) P 60 ≈ 0.275, or 27.5%; The reduction in carbon the Greenhouse Effect,” Yale University, New Haven, Connecticut.) emissions1 2 from a tax of (a) Write P as a function of x. $60 per ton of carbon is (b) Graph P for 0 … x … 1000. Discuss the benefit of continuing to raise taxes on 27.5%. carbon. (d) x = $130.14 (c) Determine P, to the nearest tenth, when x $60. Interpret this result. 112. (a) R ≈ 4.4 w/m2 = (b) T ≈ 4.5°F; (d) What value of x will give a 50% reduction in carbon emissions?

1 112. Radiative Forcing Radiative forcing, R, measures the influence of carbon dioxide 113. ƒ - x = ln x + 5; in altering the additional solar radiation trapped in Earth’s atmosphere. The Inter- domain: 0, ∞ ; 1 2 national Panel on Climate Change (IPCC) in 1990 estimated k to be 6.3 in the range: -∞, ∞ 1 2 radiative forcing equation 114. ƒ -1 x = ln x - 10 ; 1 2 domain: 10, ∞ ; C 1 2 1 2 R = k ln , range: -∞, ∞ C0 1 2 115. ƒ -1 x = ln x + 4 - 1; 1 2 where C0 is the preindustrial amount of carbon dioxide and C is the current level. domain: -4, ∞ ; 1 2 1 2 (Source: Clime, W., The Economics of Global Warming, Institute for International range: -∞, ∞ Economics, Washington, D.C.) 1 2 116. ƒ -1 x = ex - 2; (a) Use the equation R 6.3 ln C to determine the radiative forcing R (in watts 1 2 = C0 domain: -∞, ∞ ; 1 2 per square meter to the nearest tenth) expected by the IPCC if the carbon diox- range: -2, ∞ 1 2 ide level in the atmosphere doubles from its preindustrial level. -1 1 x/2 117. ƒ x 1= 3 e 2; (b) Determine the global temperature increase T, to the nearest tenth, that the IPCC domain: -∞, ∞ ; 1 2 predicted would occur if atmospheric carbon dioxide levels were to double, range: 0, ∞ given T R 1.03R. 1 2 = 118. ƒ -1 x = ex-6 + 1; 1 2 1 2 domain: -∞, ∞ ; Find ƒ-1 x , and give the domain and range. 1 2 range: 1, ∞ 1 2 113. ƒ x1 2= ex-5 114. ƒ x = ex + 10 115. ƒ x = ex+1 - 4 1 2 119. 1.52 116. ƒ1x2 = ln x + 2 117. ƒ1x2 = 2 ln 3x 118. ƒ1x2 = ln x - 1 + 6 120. -0.93, 1.35 5 6 1 2 1 2 1 2 1 2 1 2 121. 0 Use a graphing calculator to solve each equation. Give irrational solutions correct to 5 6 122. 0.69, 1.10 the nearest hundredth. 5 6 123. 2.45, 5.66 x x x -x 5 6 119. e + ln x = 5 120. e - ln x + 1 = 3 121. 2e + 1 = 3e 124. 0.23 3 5 6 122. ex + 6e-x = 5 123. log x =1x2 - 82x + 14 124. ln x = - x + 3 125. When5 6dividing each side by 125. Find the error in the following “proof” that 2 1. 2 1 6 log 3 , the direction of the inequality symbol should be 1 1 6 True statement 1 9 3 reversed because log 3 is negative. 1 2 1 6 Rewrite the left side. 3 3 a 1 b 2 1 log 6 log Take the logarithm on each side. 3 3 a b1 1 2 log 6 1 log Property of logarithms; identity property 3 3 1 2 6 1 Divide each side by log 3 . Copyright Pearson. All Rights Reserved.

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4.6 Applications and Models of Exponential Growth and Decay

■ The Exponential The Exponential Growth or Decay Function In many situations in Growth or Decay Function ecology, biology, economics, and the social sciences, a quantity changes at a rate proportional to the amount present. The amount present at time t is a special ■ Growth Function Models function of t called an exponential growth or decay function. ■ Decay Function Models Exponential Growth or Decay Function

Let y0 be the amount or number present at time t = 0. Then, under certain conditions, the amount y present at any time t is modeled by

Looking Ahead To Calculus kt y y e , where k is a constant. The exponential growth and decay 0 function formulas are studied in calculus in conjunction with the topic known as differential equations. The constant k determines the type of function. • When k 7 0, the function describes growth. Examples of exponential growth include compound interest and atmospheric carbon dioxide. • When k 6 0, the function describes decay. One example of exponential decay is radioactive decay.

Growth Function Models The amount of time it takes for a quantity that grows exponentially to become twice its initial amount is its doubling time.

Example 1 Determining a Function to Model Exponential Growth

Carbon Dioxide Earlier in this chapter, we discussed the growth of atmospheric carbon dioxide Year (ppm) over time using a function based on the data from the table. Now we determine 1990 353 such a function from the data. 2000 375 (a) Find an exponential function that gives the amount of carbon dioxide y in 2075 590 year x. 2175 1090 (b) Estimate the year when future levels of carbon dioxide will be double the 2275 2000 preindustrial level of 280 ppm. Source: International Panel on Solution Climate Change (IPCC). (a) The data points exhibit exponential growth, so the equation will take the form y y ekx. = 0

We must find the values of y0 and k. The data begin with the year 1990, so to simplify our work we let 1990 correspond to x = 0, 1991 correspond to x = 1, and so on. Here y0 is the initial amount and y0 = 353 in 1990 when x = 0. Thus the equation is

kx y = 353e . Let y0 = 353. From the last pair of values in the table, we know that in 2275 the carbon dioxide level is expected to be 2000 ppm. The year 2275 corresponds to 2275 - 1990 = 285. Substitute 2000 for y and 285 for x, and solve for k.

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Classroom Example 1 y = 353ekx Solve for k. Refer to the table given with Example 1. 2000 = 353ek 285 Substitute 2000 for y and 285 for x. (a) Find a different exponential 2000 1 2 model using the data for 2000 = e285k Divide by 353. and 2175. Let the year 2000 353 correspond to x = 0. 2000 (b) Use the model from part (a) ln = ln e285k Take the natural logarithm on each side. 353 to estimate when future levels of carbon dioxide will triple 2000 from the preindustrial level of ln = 285k ln ex = x, for all x. 353 280 ppm. Answers: 1 2000 1 k = # ln Multiply by 285 and rewrite. (a) y = 375e0.00610x 285 353 (b) It will triple from the k ≈ 0.00609 Use a calculator. preindustrial level during 2132. A function that models the data is

y = 353e0.00609x. (b) y = 353e0.00609x Solve the model from part (a) for the year x. 0.00609x 560 = 353e To double the level 280, let y = 2 280 = 560. 560 1 2 = e0.00609x Divide by 353. 353 560 ln = ln e0.00609x Take the natural logarithm on each side. 353 Teaching Tip In Example 1(b), 560 stress that 75.8 does not represent ln = 0.00609x ln ex = x, for all x. 353 the time for carbon dioxide to double. It represents the time since 1 560 1 1990 that the preindustrial level of x = # ln Multiply by and rewrite. 0.00609 353 0.00609 280 ppm took to double. Mention that the solution to x ≈ 75.8 Use a calculator. 2 = e0.00609x Since x = 0 corresponds to 1990, the preindustrial carbon dioxide level will yields a doubling time of 114 yr. double in the 75th year after 1990, or during 2065, according to this model.

■✔ Now Try Exercise 43.

Classroom Example 2 Example 2 Finding Doubling Time for Money How long will it take for money in an account that accrues interest How long will it take for money in an account that accrues interest at a rate of at a rate of 2.25%, compounded 3%, compounded continuously, to double? continuously, to double? Solution A = Per t Continuous compounding formula Answer: about 31 yr 0.03t 2P = Pe Let A = 2P and r = 0.03. 2 = e0.03t Divide by P. Teaching Tip Have students solve Example 2 using interest ln 2 = ln e0.03t Take the natural logarithm on each side. rates of 2%, 4%, and 10%. x Explain that the Rule of 72, ln 2 = 0.03t ln e = x 72 ln 2 t = , = t Divide by 0.03. r 0.03 is often used to approximate 23.10 ≈ t Use a calculator. doubling times. Have students check their answers against this rule. It will take about 23 yr for the amount to double. ■✔ Now Try Exercise 31.

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Example 3 Using an Exponential Function to Model Population Growth According to the U.S. Census Bureau, the world population reached 6 billion people during 1999 and was growing exponentially. By the end of 2010, the population had grown to 6.947 billion. The projected world population (in billions of people) t years after 2010 is given by the function

ƒ t = 6.947e0.00745t. (a) Based on this model, what 1will2 the world population be in 2025? (b) If this trend continues, approximately when will the world population reach Classroom Example 3 9 billion? Refer to the model in Example 3. Solution (a) Based on the model, what will the world population be in (a) Since t = 0 represents the year 2010, in 2025, t would be 2025 - 2010 = 2020? 15 yr. We must find ƒ t when t is 15. (b) If this trend continues, 0.00745t approximately when will ƒ t1 2= 6.947e Given function the world population reach 0.00745 15 ƒ 15 6.947e Let t = 15. 8 billion? 1 2 = 1 2 ƒ 15 ≈ 7.768 Use a calculator. Answers: 1 2 (a) 7.484 billion The population will1 be2 7.768 billion at the end of 2025. (b) It will reach 8 billion during the year 2028. (b) ƒ t = 6.947e0.00745t Given function 0.00745t 1 92 = 6.947e Let ƒ t = 9. 9 1 2 = e0.00745t Divide by 6.947. 6.947 9 ln = ln e0.00745t Take the natural logarithm on each side. 6.947 9 ln = 0.00745t ln ex = x, for all x. 6.947 9 ln 6.947 t = Divide by 0.00745 and rewrite. 0.00745 t ≈ 34.8 Use a calculator. Thus, 34.8 yr after 2010, during the year 2044, world population will reach 9 billion. ■✔ Now Try Exercise 39.

Decay Function Models Half-life is the amount of time it takes for a quantity that decays exponentially to become half its initial amount.

Note In Example 4 on the next page, the initial amount of substance is given as 600 g. Because half-life is constant over the lifetime of a decaying 1 quantity, starting with any initial amount, y0 , and substituting 2 y0 for y in k t y = y0 e would allow the common factor y0 to be divided out. The rest of the work would be the same.

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Classroom Example 4 Example 4 Determining an Exponential Function to Model Suppose 800 g of a radioactive Radioactive Decay substance are present initially and 2.5 yr later only 400 g remain. Suppose 600 g of a radioactive substance are present initially and 3 yr later only (a) Determine an exponential 300 g remain. function that models this (a) Determine an exponential function that models this decay. decay. (b) How much of the substance (b) How much of the substance will be present after 6 yr? will be present after 4 yr? Solution Answers: (a) We use the given values to find k in the exponential equation (a) y = 800e-0.277t (b) 264 g y y ekt. = 0 kt Because the initial amount is 600 g, y0 = 600, which gives y = 600e . The initial amount (600 g) decays to half that amount (300 g) in 3 yr, so its half- life is 3 yr. Now we solve this exponential equation for k.

kt y = 600e Let y0 = 600. 3k 300 = 600e Let y = 300 and t = 3. 0.5 = e3k Divide by 600. ln 0.5 = ln e3k Take the natural logarithm on each side. ln 0.5 = 3k ln e x = x, for all x. ln 0.5 = k Divide by 3. 3 k ≈ -0.231 Use a calculator. A function that models the situation is y = 600e-0.231t. (b) To find the amount present after 6 yr, let t = 6. y = 600e-0.231t Model from part (a) 0.231 6 y = 600e- Let t = 6. 1 2 y = 600e-1.386 Multiply. y ≈ 150 Use a calculator.

After 6 yr, 150 g of the substance will remain. ■✔ Now Try Exercise 19.

Classroom Example 5 Example 5 Solving a Carbon Dating Problem Suppose that the skeleton of a woman who lived in the Classical Carbon-14, also known as radiocarbon, is a radioactive form of carbon that is Greek period was discovered in found in all living plants and animals. After a plant or animal dies, the radiocar- 2005. Carbon-14 testing at that bon disintegrates. Scientists can determine the age of the remains by comparing time determined that the skeleton the amount of radiocarbon with the amount present in living plants and animals. 3 contained 4 of the carbon-14 of This technique is called carbon dating. The amount of radiocarbon present a living woman of the same size. after t years is given by Estimate the year in which the Greek woman died. y y e-0.0001216t, = 0 Answer: 361 bce where y0 is the amount present in living plants and animals. (a) Find the half-life of carbon-14. 1 (b) Charcoal from an ancient fire pit on Java contained 4 the carbon-14 of a living sample of the same size. Estimate the age of the charcoal.

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Solution 1 (a) If y0 is the amount of radiocarbon present in a living thing, then 2 y0 is half this initial amount. We substitute and solve the given equation for t.

y y e-0.0001216t Given equation = 0 1 -0.0001216t Let y 1 y . y y e 0 = 0 = 0 2 2 1 -0.0001216t = e Divide by y0 . 2 1 ln = ln e-0.0001216t Take the natural logarithm on each side. 2 1 ln = -0.0001216t ln e x = x, for all x. 2 1 ln 2 = t Divide by -0.0001216. -0.0001216 5700 ≈ t Use a calculator. The half-life is 5700 yr. 1 (b) Solve again for t, this time letting the amount y = 4 y0 . y y e-0.0001216t Given equation = 0 1 -0.0001216t 1 y y e Let y y . 0 = 0 = 0 4 4 1 -0.0001216t = e Divide by y0 . 4 1 ln = ln e-0.0001216t Take the natural logarithm on each side. 4 1 ln 4 = t ln e x = x; Divide by -0.0001216. -0.0001216 t ≈ 11,400 Use a calculator.

The charcoal is 11,400 yr old. ■✔ Now Try Exercise 23.

Example 6 Modeling Newton’s Law of Cooling Newton’s law of cooling says that the rate at which a body cools is proportional to the difference in temperature between the body and the environment around it. The temperature ƒ t of the body at time t in appropriate units after being introduced into an environment having constant temperature T0 is 1 2 -kt ƒ t = T0 + Ce , where C and k are constants. A pot of coffee with1 2 a temperature of 100°C is set down in a room with a temperature of 20°C. The coffee cools to 60°C after 1 hr. (a) Write an equation to model the data. (b) Find the temperature after half an hour. (c) How long will it take for the coffee to cool to 50°C?

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Classroom Example 6 Solution Newton’s law of cooling also applies to warming. Mary took a (a) We must find values for C and k in the given formula. As given, when t = 0, leg of lamb out of her refrigerator, T0 = 20, and the temperature of the coffee is ƒ 0 = 100. which is set at 34°F, and placed -kt it in her oven, which she had ƒ t = T0 + Ce Given function 1 2 preheated to 350°F. After 1 hr, her 100 20 Ce-0k Let t 0, ƒ 0 100, and T 20. meat thermometer registered 70°F. 1 2 = + = = 0 = (a) Write an equation to model 100 = 20 + C e0 = 1 1 2 the data. 80 = C Subtract 20. (b) Find the temperature 90 min after the leg of lamb was The following function models the data. placed in the oven. -kt (c) Mary wants to serve the ƒ t = 20 + 80e Let T0 = 20 and C = 80. leg of lamb medium rare, The coffee cools to 60°C after 1 hr, so when t 1, ƒ 1 60. which requires an internal 1 2 = = temperature of 145°F. What is -kt ƒ t = 20 + 80e Above function with T0 =1 202 and C = 80 the total amount of time it will -1k take to cook the leg of lamb? 1 602 = 20 + 80e Let t = 1 and ƒ 1 = 60. Answers: 40 = 80e-k Subtract 20. 1 2 (a) ƒ t = 350 - 316e-0.121t 1 k (b) 86.5°F = e- Divide by 80. 1 2 2 (c) 3.576 hr, or about 3 hr, 35 min 1 ln = ln e-k Take the natural logarithm on each side. 2 1 ln = -k ln e x = x, for all x. 2

k ≈ 0.693 Multiply by -1, rewrite, and use a calculator.

Thus, the model is ƒ t = 20 + 80e-0.693t. 1 1 (b) To find the temperature1 2 after 2 hr, let t = 2 in the model from part (a). ƒ t = 20 + 80e-0.693t Model from part (a) 1 1 2 -0.693 1/2 1 ƒ = 20 + 80e Let t = 2 . 2 1 21 2 a1 b ƒ ≈ 76.6°C Use a calculator. 2 a b (c) To find how long it will take for the coffee to cool to 50°C, let ƒ t = 50. 0.693t ƒ t = 20 + 80e- Model from part (a) 1 2 -0.693t 1 502 = 20 + 80e Let ƒ t = 50. 30 = 80e-0.693t Subtract1 2 20. 3 = e-0.693t Divide by 80. 8 3 ln = ln e-0.693t Take the natural logarithm on each side. 8 3 ln = -0.693t ln e x = x, for all x. 8 3 ln 8 t = Divide by -0.693 and rewrite. -0.693 t ≈ 1.415 hr, or about 1 hr, 25 min ■✔ Now Try Exercise 27.

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4.6 Exercises

1. B 2. D Concept Preview Population Growth A population is increasing according to the 3. C 4. A exponential function

0.02x 5. B 6. D y = 2e , 7. C 8. A where y is in millions and x is the number of years. Match each question in Column I 1 1 1 1 with the correct procedure in Column II to answer the question. 9. 3 ln 3 10. 6 ln 3 I II 11. 1 ln 1 12. 1 ln 1 100 2 200 2 1. How long will it take for the population to A. Evaluate y = 2e0.02 1/3 . 1 1 1 1 triple? 1 2 13. 2 ln 4 14. 4 ln 9 2. When will the population reach 3 million? B. Solve 2e0.02x 6. 15. (a) 440 g (b) 387 g = (c) 264 g (d) 22 yr 3. How large will the population be in 3 yr? C. Evaluate y = 2e0.02 3 . 1 2 4. How large will the population be in 4 months? D. Solve 2e0.02x = 3.

Concept Preview Radioactive Decay Strontium-90 decays according to the exponential function -0.0241t y = y0 e , where t is time in years. Match each question in Column I with the correct procedure in Column II to answer the question. I II -0.0241t 5. If the initial amount of Strontium-90 is 200 g, A. Solve 0.75y0 = y0 e . how much will remain after 10 yr?

6. If the initial amount of Strontium-90 is 200 g, B. Evaluate y = 200e-0.0241 10 . how much will remain after 20 yr? 1 2

1 -0.0241t 7. What is the half-life of Strontium-90? C. Solve 2 y0 = y0 e . 8. How long will it take for any amount of D. Evaluate y = 200e-0.0241 20 . Strontium-90 to decay to 75% of its initial 1 2 amount?

(Modeling) The exercises in this set are grouped according to discipline. They involve exponential or logarithmic models. See Examples 1–6. Physical Sciences (Exercises 9–28)

An initial amount of a radioactive substance y0 is given, along with information about the amount remaining after a given time t in appropriate units. For an equation of the kt form y = y0 e that models the situation, give the exact value of k in terms of natural logarithms.

9. y0 = 60 g; After 3 hr, 20 g remain. 10. y0 = 30 g; After 6 hr, 10 g remain.

11. y0 = 10 mg; The half-life is 100 days. 12. y0 = 20 mg; The half-life is 200 days.

13. y0 = 2.4 lb; After 2 yr, 0.6 lb remains. 14. y0 = 8.1 kg; After 4 yr, 0.9 kg remains.

Solve each problem. 15. Decay of Lead A sample of 500 g of radioactive lead-210 decays to polonium-210 according to the function A t = 500e-0.032t, where t is time in years. Find the amount1 2 of radioactive lead remaining after (a) 4 yr, (b) 8 yr, (c) 20 yr. (d) Find the half-life.

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16. (a) 404 g (b) 327 g 16. Decay of Plutonium Repeat Exercise 15 for 500 g of plutonium-241, which decays -0.053t (c) 173 g (d) 13 yr according to the function A t = A0 e , where t is time in years. 17. 1600 yr 18. 12 yr 17. Decay of Radium Find the1 half-life2 of radium-226, which decays according to the 19. 3.6 g 20. 0.88 g -0.00043t function A t = A0 e , where t is time in years. 21. 16 days 22. Magnitude 1 is about 6.3 18. Decay of Tritium1 2 Find the half-life of tritium, a radioactive isotope of hydrogen, -0.056t times as great as magnitude 3. which decays according to the function A t = A0 e , where t is time in years. 23. 9000 yr 24. 4200 yr 19. Radioactive Decay If 12 g of a radioactive1 2 substance are present initially and 4 yr 25. 15,600 yr 26. 43°C later only 6.0 g remain, how much of the substance will be present after 7 yr? 27. 6.25°C 28. 81.25°C 20. Radioactive Decay If 1 g of strontium-90 is present initially, and 2 yr later 0.95 g 29. (a) 3% compounded quarterly remains, how much strontium-90 will be present after 5 yr? (b) $826.95 21. Decay of Iodine How long will it take any quantity of iodine-131 to decay to 25% of its initial amount, knowing that it decays according to the exponential function -0.087t A t = A0 e , where t is time in days? 22. Magnitude1 2 of a Star The magnitude M of a star is modeled by

5 I M = 6 - log , 2 I0

where I0 is the intensity of a just-visible star and I is the actual intensity of the star being measured. The dimmest stars are of magnitude 6, and the brightest are of magnitude 1. Determine the ratio of light intensities between a star of magnitude 1 and a star of magnitude 3. 23. Carbon-14 Dating Suppose an Egyptian mummy is discovered in which the amount of carbon-14 present is only about one-third the amount found in living human beings. How long ago did the Egyptian die? 24. Carbon-14 Dating A sample from a refuse deposit near the Strait of Magellan had 60% of the carbon-14 of a contemporary sample. How old was the sample? 25. Carbon-14 Dating Paint from the Lascaux caves of France contains 15% of the normal amount of carbon-14. Estimate the age of the paintings. 26. Dissolving a Chemical The amount of a chemical that will dissolve in a solution increases exponentially as the (Celsius) temperature t is increased according to the model A t = 10e0.0095t. At what temperature will 15 g dissolve?1 2 27. Newton’s Law of Cooling Boiling water, at 100°C, is placed in a freezer at 0°C. The temperature of the water is 50°C after 24 min. Find the temperature of the water to the nearest hundredth after 96 min. (Hint: Change minutes to hours.) 28. Newton’s Law of Cooling A piece of metal is heated to 300°C and then placed in a cooling liquid at 50°C. After 4 min, the metal has cooled to 175°C. Find its temperature to the nearest hundredth after 12 min. (Hint: Change minutes to hours.)

Finance (Exercises 29–34) 29. Comparing Investments Russ, who is self-employed, wants to invest $60,000 in a pension plan. One investment offers 3% compounded quarterly. Another offers 2.75% compounded continuously. (a) Which investment will earn more interest in 5 yr? (b) How much more will the better plan earn?

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30. 5.61 yr 30. Growth of an Account If Russ (see Exercise 29) chooses the plan with continuous 31. 27.73 yr compounding, how long will it take for his $60,000 to grow to $70,000? 32. The time will be divided by 3. 31. Doubling Time Find the doubling time of an investment earning 2.5% interest if 33. 36.62 yr interest is compounded continuously. 34. 39 yr, 347 days 32. Doubling Time If interest is compounded continuously and the interest rate is 35. (a) 315 (b) 229 (c) 142 tripled, what effect will this have on the time required for an investment to double? 36. 1968 33. Growth of an Account How long will it take an investment to triple if interest is 37. (a) P0 = 1; a ≈ 1.01355 compounded continuously at 3%? (b) 1.3 billion (c) 2030 34. Growth of an Account Use the Table feature of a graphing calculator to find how 38. (a) 961,000 long it will take $1500 invested at 2.75% compounded daily to triple in value. (b) 7.2 yr Zoom in on the solution by systematically decreasing the increment for x. Find the (c) 17.3 yr answer to the nearest day. (Find the answer to the nearest day by eventually letting 1 the increment of x equal 365 . The decimal part of the solution can be multiplied by 365 to determine the number of days greater than the nearest year. For example, if the solution is determined to be 16.2027 yr, then multiply 0.2027 by 365 to get 73.9855. The solution is then, to the nearest day, 16 yr, 74 days.) Confirm the answer algebraically.

Social Sciences (Exercises 35–44) 35. Legislative Turnover The turnover of leg- islators is a problem of interest to political scientists. It was found that one model of legislative turnover in a particular body was

M t = 434e-0.08t, where M t represents1 2 the number of continuously serving members at time t. Here, t = 0 represents1 2 1965, t = 1 represents 1966, and so on. Use this model to approximate the number of continuously serving members in each year. (a) 1969 (b) 1973 (c) 1979

36. Legislative Turnover Use the model in Exercise 35 to determine the year in which the number of continuously serving members was 338. 37. Population Growth In 2000 India’s population reached 1 billion, and it is projected to be 1.4 billion in 2025. (Source: U.S. Census Bureau.) x-2000 (a) Find values for P0 and a so that P x = P0 a models the population of India in year x. Round a to five decimal places. 1 2 (b) Predict India’s population in 2020 to the nearest tenth of a billion. (c) In what year is India’s population expected to reach 1.5 billion? 38. Population Decline A midwestern city finds its residents moving to the suburbs. Its population is declining according to the function

-0.04t P t = P0 e ,

where t is time measured in years and1 2 P0 is the population at time t = 0. Assume that P0 = 1,000,000. (a) Find the population at time t = 1 to the nearest thousand. (b) How long, to the nearest tenth of a year, will it take for the population to decline to 750,000? (c) How long, to the nearest tenth of a year, will it take for the population to decline to half the initial number?

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39. (a) $8412 39. Health Care Spending Out-of-pocket spending in the United States for health care (b) 2010 increased between 2008 and 2012. The function 40. $998 billion ƒ x = 7446e0.0305x 41. (a) $14,542 (b) $16,162 models average annual expenditures1 2 per household, in dollars. In this model, x rep- (c) $17,494 resents the year, where x = 0 corresponds to 2008. (Source: U.S. Bureau of Labor 42. (a) 348 yr Statistics.) (b) 3974 yr (a) Estimate out-of-pocket household spending on health care in 2012 to the nearest (c) 2288 yr dollar. (b) In what year did spending reach $7915 per household? 40. Recreational Expenditures Personal consumption expenditures for recre- ation in billions of dollars in the United States during the years 2000–2013 can be approximated by the function A t = 632.37e0.0351t, where t =102 corresponds to the year 2000. Based on this model, how much were personal consumption expenditures in 2013 to the nearest billion? (Source: U.S. Bureau of Economic Analysis.) 41. Housing Costs Average annual per-household spending on housing over the years 2000–2012 is approximated by H = 12,744e0.0264t, where t is the number of years since 2000. Find H to the nearest dollar for each year. (Source: U.S. Bureau of Labor Statistics.) (a) 2005 (b) 2009 (c) 2012 42. Evolution of Language The number of years, n, since two independently evolving languages split off from a common ancestral language is approximated by n ≈ -7600 log r, where r is the proportion of words from the ancestral language common to both languages. Find each of the following to the nearest year. (a) Find n if r = 0.9. (b) Find n if r = 0.3. (c) How many years have elapsed since the split if half of the words of the ancestral language are common to both languages? 43. School District Growth Student enrollment in the Wentzville School District, one of the fastest-growing school districts in the state of Missouri, has projected growth as shown in the graph.

y Enrollment Projections

20,000 19,915

19,000

18,000

17,000

16,000

15,000 Projected Enrollment 14,000 14,225 0 x 13–1414–15 15–16 16–17 17–18 18–19 19–20 20–21 21–22 22–23 23–24 School Year Source: Wentzville School District.

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0.034x kx 43. (a) y = 14,225e (a) Use the model y = y0 e to find an exponential function that gives the projected (b) 2025–26 enrollment y in school year x. Let the school year 2013–14 correspond to x = 0, 44. (a) y = 100e0.309x 2014–15 correspond to x = 1, and so on, and use the two points indicated on the (b) 1,061,475 graph. (b) Estimate the school year for which projected enrollment will be 21,500 students. 45. (a) 15,000 (b) 9098 44. YouTube Views The number of views of a YouTube video increases after the num- (c) 5249 ber of hours posted as shown in the table. 46. 14 days 47. (a) 611 million Hour Number of Views (b) 746 million 20 100 (c) 1007 million 48. 6.9 days 25 517 49. 13.2 hr 30 2015 35 10,248

kx (a) Use the model y = y0 e to find an exponential function that gives projected number of views y after number of hours x. Let hour 20 correspond to x = 0, hour 25 correspond to x = 5, and so on, and use the first and last data values given in the table. (b) Estimate the number of views after 50 hr.

Life Sciences (Exercises 45–50) 45. Spread of Disease During an epidemic, the number of people who have never had the disease and who are not immune (they are susceptible) decreases exponentially according to the function ƒ t = 15,000e-0.05t, where t is time in days. Find the number1 2 of susceptible people at each time. (a) at the beginning of the epidemic (b) after 10 days (c) after 3 weeks 46. Spread of Disease Refer to Exercise 45 and determine how long it will take, to the nearest day, for the initial number of people susceptible to decrease to half its amount. 47. Growth of Bacteria The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product the number of bacteria present is given by ƒ t = 500e0.1t, where t is time in days and the value1 2of ƒ t is in millions. Find the number of bacteria, in millions, present at each time. 1 2 (a) 2 days (b) 4 days (c) 1 week 48. Growth of Bacteria How long will it take the bacteria population in Exercise 47 to double? Round the answer to the nearest tenth. 49. Medication Effectiveness Drug effectiveness decreases over time. If, each hour, a drug is only 90% as effective as the previous hour, at some point the patient will not be receiving enough medication and must receive another dose. If the initial dose was 200 mg and the drug was administered 3 hr ago, the expression 200 0.90 3, which equals 145.8, represents the amount of effective medication still in the system. (The exponent is equal to the number of hours since the drug was administered.)1 2 The amount of medication still available in the system is given by the function ƒ t = 200 0.90 t. In this model, t is in hours and ƒ 1t 2 is in milligrams.1 2 How long will it take for this initial dose to reach the dangerously low level of 50 mg? Round the answer to the nearest tenth. 1 2

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50. (a) 250,000 50. Population Size Many environmental situations place effective limits on the growth G(x) = 100+2400e−x of the number of an organism in an area. Many such limited-growth situations are 2500 described by the logistic function MG G x 0 , = -kMx G0 + M - G0 e 0 8 0 1 2 where G0 is the initial number present, 1M is the maximum2 possible size of the popu- (b) 590; 589 lation, and k is a positive constant. The screens illustrate a typical logistic function (c) 2.8; 2.7726 calculation and graph.

51. 2020 52. (a) S 1 ≈ 45,000; 70 S 3 ≈ 37,000 1 2 (b) S 2 ≈ 72,000; 1 2 S 10 ≈ 49,000 1 2 53. 6.9 yr 54. 27.5 yr 1 2 −1 65 −5

Assume that G0 = 100, M = 2500, k = 0.0004, and x = time in decades (10-yr periods). (a) Use a calculator to graph the function, using 0 … x … 8 and 0 … y … 2500. (b) Estimate the value of G 2 from the graph. Then evaluate G 2 algebraically to find the population after 20 yr. 1 2 1 2 (c) Find the x-coordinate of the intersection of the curve with the horizontal line y = 1000 to estimate the number of decades required for the population to reach 1000. Then solve G x = 1000 algebraically to obtain the exact value of x.

Economics (Exercises 51–56)1 2 51. Consumer Price Index The U.S. Consumer Price Index for the years 1990–2013 is approximated by A t = 100e0.0264t, where t represents the number of 1years2 after 1990. (Since A 16 is about 153, the amount of goods that could be purchased for $100 in 1990 cost about $153 in 2006.) Use the function to determine the year in which costs will be1 125%2 higher than in 1990. (Source: U.S. Bureau of Labor Statistics.) 52. Product Sales Sales of a product, under relatively stable market conditions but in the absence of promotional activities such as advertising, tend to decline at a constant yearly rate. This rate of sales decline varies considerably from product to product, but it seems to remain the same for any particular product. The sales decline can be expressed by the function

-at S t = S0 e ,

where S t is the rate of sales at time1 2 t measured in years, S0 is the rate of sales at time t = 0, and a is the sales decay constant. 1 2 (a) Suppose the sales decay constant for a particular product is a = 0.10. Let S0 = 50,000 and find S 1 and S 3 to the nearest thousand. (b) Find S 2 and S 10 to the nearest thousand if S 80,000 and a 0.05. 1 2 1 2 0 = = 53. Product Sales1 2 Use 1the2 sales decline function given in Exercise 52. If a = 0.10, S0 = 50,000, and t is time measured in years, find the number of years it will take for sales to fall to half the initial sales. Round the answer to the nearest tenth. 54. Cost of Bread Assume the cost of a loaf of bread is $4. With continuous compounding, find the number of years, to the nearest tenth, it would take for the cost to triple at an annual inflation rate of 4%.

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55. 11.6 yr 56. 34.7 yr 55. Electricity Consumption Suppose that in a certain area the consumption of electricity has increased at a continuous rate of 6% per year. If it continued to increase 57. (a) 0.065; 0.82; Among at this rate, find the number of years, to the nearest tenth, before twice as much elec- people age 25, 6.5% have tricity would be needed. some CHD, while among people age 65, 82% have 56. Electricity Consumption Suppose a conservation campaign, together with higher some CHD. rates, caused demand for electricity to increase at only 2% per year. (See Exercise 55.) (b) 48 yr Find the number of years, to the nearest tenth, before twice as much electricity 58. (a) would be needed.

(Modeling) Solve each problem that uses a logistic function. 57. Heart Disease As age increases, so does the likelihood of coronary heart disease (CHD). The fraction of people x years old with some CHD is modeled by The maximum height 0.9 appears to be 50 ft. ƒ x = . 1 + 271e-0.122x (b) 50 y = 1 2 1+47.5e−0.22x (Source: Hosmer, D., and S. Lemeshow, Applied Logistic Regression, John Wiley 60 and Sons.) (a) Evaluate ƒ 25 and ƒ 65 to the nearest hundredth. Interpret the results. (b) At what age, to the nearest year, does this likelihood equal 50%? 1 2 1 2 10 70 0 58. Tree Growth The height of a certain tree in feet The horizontal asymptote after x years is modeled by is y 50. It indicates = 50 that this tree cannot grow ƒ x = 0.22x . taller than 50 ft. 1 + 47.5e- (c) 19.4 yr 1 2 (a) Make a table for ƒ starting at x = 10, and incrementing by 10. What appears to be the maximum height of the tree? (b) Graph ƒ and identify the horizontal asymptote. Explain its significance. (c) After how many years was the tree 30 ft tall? Round to the nearest tenth.

Summary Exercises on Functions: Domains and Defining Equations

Finding the Domain of a Function: A Summary To find the domain of a function, given the equation that defines the function, remember that the value of x input into the equation must yield a real number for y when the function is evaluated. For the functions studied so far in this book, there are three cases to consider when determining domains.

Guidelines for Domain Restrictions

1. No input value can lead to 0 in a denominator, because division by 0 is undefined. 2. No input value can lead to an even root of a negative number, because this situation does not yield a real number. 3. No input value can lead to the logarithm of a negative number or 0, because this situation does not yield a real number.

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Unless otherwise specified, we determine domains as follows. • The domain of a polynomial function is the set of all real numbers. • The domain of an absolute value function is the set of all real numbers for which the expression inside the absolute value bars (the argument) is defined. • If a function is defined by a rational expression, the domain is the set of all real numbers for which the denominator is not zero. • The domain of a function defined by a radical with even root index is the set of all real numbers that make the radicand greater than or equal to zero. If the root index is odd, the domain is the set of all real numbers for which the radicand is itself a real number. • For an exponential function with constant base, the domain is the set of all real numbers for which the exponent is a real number. • For a logarithmic function, the domain is the set of all real numbers that make the argument of the logarithm greater than zero.

Determining Whether an Equation Defines y as a Function of x For y to be a function of x, it is necessary that every input value of x in the domain leads to one and only one value of y. To determine whether an equation such as

x - y3 = 0 or x - y2 = 0 represents a function, solve the equation for y. In the first equation above, doing so leads to

3 y = x. 2 Notice that every value of x in the domain (that is, all real numbers) leads to one and only one value of y. So in the first equation, we can write y as a function of x. However, in the second equation above, solving for y leads to

y = { x. 2 If we let x 4, for example, we get two values of y: 2 and 2. Thus, in the 1. ∞, ∞ = - - second equation, we cannot write y as a function of x. 7 2. 12 , ∞ 2 3. -∞, ∞ C 4. ∞, B6 6, ∞ Exercises 1- 2´ 5. ∞, ∞ Find the domain of each function. Write answers using interval notation. 1- 2 1 2 6. ∞, 3 3, ∞ 1- -2 ´ 1. ƒ x = 3x - 6 2. ƒ x = 2x - 7 3. ƒ x = x + 4 7. ∞, 3 3, 3 3, ∞ 1- - 4 ´ 3- 2 ´ 2 1 2 x + 2 1 2 -2 1 2 8. -∞, ∞ 4. ƒ x = 5. ƒ x = 6. ƒ x = 0 x2 -0 9 1 2 1 2 1 2 x - 6 x2 + 7 9. 4, 4 2 1- 2 2 1 2 x + 7 1 2 3 3 1 2 2 10. -∞, -7 ´ 3, ∞ 7. ƒ x = 8. ƒ x = x + 7x - 4 9. ƒ x = log5 16 - x 1 2 x2 - 9 11. -∞, -1 ´ 8, ∞ 2 1 2 1 2 1 2 1 2 1 2 1 2 12. ∞, 0 0, ∞ x + 7 2 1/x 1- ´4 3 2 10. ƒ x = log 11. ƒ x = x - 7x - 8 12. ƒ x = 2 13. ∞, ∞ x - 3 1- 2 1 2 2 1 2 1 2 1 2 14. -∞, -5 ´ -5, ∞ 1 x2 - 25 1 2 13. ƒ x = 14. ƒ x = 15. ƒ x = x3 - 1 15. 1, ∞ 2x2 - x + 7 x + 5 1 2 1 2 2 1 2 1 2 1 2 3 2 Copyright Pearson. All Rights Reserved.

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3 16. ∞, 5 5, 5 2 x2 x 4 x - 1 - - ´ - 16. ƒ x = ln x - 5 17. ƒ x = e + + 18. ƒ x = 2 2 2 x2 - 1 ´ 5, ∞ 1 2 1 2 1 2 A 2 B A B 0 0 17. -∞, ∞ -1 3 1 19. ƒ x = 3 20. ƒ x = 3 18. A∞, 1 B 1, 1 1, ∞ x - 1 x - 8 1- -2 ´ - ´ 19. -∞, 1 1 2 B 1 2 B 1 2 1 2 1 2 21. ƒ x = ln x2 + 1 22. ƒ x = x - 3 x + 2 x - 4 20. ∞, 2 2, ∞ 1- 2 ´ 2 x 2 2 21. ∞, ∞ 1 2 1 + 2 1 2 121 212 21 2 1- 2 1 2 23. ƒ x = log 24. ƒ x = 4 - x x + 3 x - 3 22. -2, 3 ´ 4, ∞ 2 1 2 1 2 a b 1 2 1 2 1 2 23. -∞, -2 ´ -2, 3 ´ 3, ∞ 1 3 4 3 2 25. ƒ x = e 1/x 26. ƒ x = 24. 3, ∞ x2 7 1 - 2 1 2 1 2 0 0 - 25. -∞, 0 ´ 0, ∞ 1 2 1 2 3 2 27. ƒ x = x100 - x50 + x2 + 5 28. ƒ x = 0 -x2 -0 9 26. 1-∞, -2 71 ´ 2 - 7, 7 4 23 2 2 2 29. ƒ1x2 = 16 - x4 30. ƒ1x2 = 16 - x4 ´ 7, ∞ A B A B 2 2 27. 2, 28. 1 2 2 1 2 -∞ ∞ ∅ x - 2x - 63 5 A B 31. ƒ x = 32. ƒ x = 5 - x 29. -2, 2 30. -∞, ∞ x2 + x - 12 1 2 2 31. -∞, -7 ´ -4, 3 ´ 9, ∞ 1 2 B 1 2 3 4 1 2 -1 32. -∞, ∞ 33. ƒ x = 5 - x 34. ƒ x = 1 4 1 2 3 2 x - 3 33. -∞, 5 34. -∞, 3 2 1 2 1 2 P P 1 2 B 1 2 35. -∞, 4 ´ 4, ∞ 35. ƒ x = log 36. ƒ x = 6x -9 1 4 1 2 4 x 36. ∞, ∞ - 1- 2 1 2 1 2 `2 ` 1 2 3 2 37. -∞, -5 ´ 5, ∞ 37. ƒ x = 6 x -25 38. ƒ x = 6 x -25 1 2 2 2 38. -∞, ∞ 1 4 3 2 1 2 -3 1 2 -2 39. -2, 6 39. ƒ x = ln 40. ƒ x = 1 2 x + 2 x - 6 log x 40. 0, 1 1, ∞ 1 ´2 1 2 a b 1 2 1 21 2 41. 1A 2 1 42.2 B 43. C 44. D Determine which one of the choices (A, B, C, or D) is an equation in which y can be 45. A 46. B written as a function of x. 47. D 48. C 41. A. 3x + 2y = 6 B. x = y C. x = y + 3 D. x2 + y2 = 9 49. C 50. B 2 2 2 2 2 42. A. 3x + 2y = 36 B. x + y -0 20 = 0 C. x - 0 y = 00 D. x = y - 4 1 43. A. x = y2 B. x = log y2 C. x3 +0 y30 = 5 D. x = y2 + 3 2 x2 y2 x2 y2 44. A. + = 1 B. x = 5y2 - 3 C. - = 1 D. x = 10y 4 4 4 9

2 - y 4 45. A. x = B. x = ln y + 1 2 C. x = y + 1 D. x = y2 y + 3 2 2 2 1 2 46. A. ey = x B. ey+2 = x C. e y = x 0 0 D. 10 y+2 = x 0 0 0 0 1 1 1 1 47. A. x2 = B. x + 2 = C. 3x = D. 2x = y2 y2 y4 y3 1 48. A. x y B. x y2 C. x D. x4 y4 81 = = = y + = x0 2 0 y02 0 y2 0 x2 0 x y x2 y2 49. A. - = 1 B. - = 1 C. - = 0 D. - = 0 4 9 4 9 4 9 4 9

M04_LHSD7659_12_AIE_C04_pp405-496.indd 487 06/10/15 6:48 PM NOT FOR SALE 488 Chapter 4 Inverse, Exponential, and Logarithmic Functions

Chapter 4 Test Prep

Key Terms

4.1 one-to-one function future value 4.3 logarithm 4.4 common logarithm inverse function present value base pH 4.2 exponential function compound amount argument natural logarithm exponential equation continuous logarithmic equation 4.6 doubling time compound interest compounding logarithmic function half-life

New Symbols

ƒ1 x inverse of ƒ x log x common (base 10) logarithm of x e a constant, approximately 2.718281828459045 ln x natural (base e) logarithm of x 1 2 1 2 log a x logarithm of x with the base a

Quick Review

Concepts Examples

4.1 Inverse Functions One-to-One Function In a one-to-one function, each x-value corresponds to only The function y = ƒ x = x2 is not one-to-one, because one y-value, and each y-value corresponds to only one y = 16, for example, corresponds to both x = 4 and x-value. x = -4. 1 2 A function ƒ is one-to-one if, for elements a and b in the domain of ƒ,

a 3 b implies ƒ a 3 ƒ b .

Horizontal Line Test 1 2 1 2 A function is one-to-one if every horizontal line intersects The graph of ƒ x = 2x - 1 is a straight line with slope 2. the graph of the function at most once. ƒ is a one-to-one function by the horizontal line test. 1 2 Inverse Functions Let ƒ be a one-to-one function. Then g is the inverse func- Find the inverse of ƒ. tion of ƒ if ƒ x = 2x - 1 Given function ƒ g x x for every x in the domain of g y 2x 1 Let y ƒ x . 1 2 = - = and x 2y 1 Interchange x and y. 1 2 1 2 = - 1 2 g ƒ x x for every x in the domain of ƒ. x + 1 y = Solve for y. 2 To find1 g x2 1, interchange2 x and y in y = ƒ x , solve for y, -1 x + 1 and replace y with g x , which is ƒ x . ƒ -1 x = Replace y with ƒ -1 x . 1 2 1 2 2

M04_LHSD7659_12_AIE_C04_pp405-496.indd 488 06/10/15 6:48 PM NOT FOR SALE chapter 4 Test Prep 489

Concepts Examples

4.2 Exponential Functions Additional Properties of Exponents For any real number a 7 0, a 1, the following hold true. (a) ax is a unique real number for all real numbers x. (a) 2x is a unique real number for all real numbers x.

(b) ab = ac if and only if b = c. (b) 2x = 23 if and only if x = 3. (c) If a 7 1 and m 6 n, then am 6 an. (c) 25 6 210, because 2 7 1 and 5 6 10. 5 10 m n 1 1 1 (d) If 0 6 a 6 1 and m 6 n, then a 7 a . (d) 2 7 2 because 0 6 2 6 1 and 5 6 10. Exponential Function If a 7 0 and a 1, then the exponential function with base ƒ x A =B 3x isA theB exponential function with base 3. a is ƒ x ax. x y y 1 2 x Graph of ƒ x ax f(x) = 3 1 2 1 1 -1 3 1. The points -1, , 0, 1 , and 1, a are on the graph. 4 1 2 a 0 1 (1, 3) 1 3 2 2. If a 7 1, then ƒ is an1 increasing2 1function.2 –1, 1 A B Q 3R (0, 1) If 0 6 a 6 1, then ƒ is a decreasing function. x 0 1 3. The x-axis is a horizontal asymptote.

4. The domain is -∞, ∞ , and the range is 0, ∞ . 1 2 1 2 4.3 Logarithmic Functions Logarithm 4 For all real numbers y and all positive numbers a and x, log3 81 = 4 is equivalent to 3 = 81. y where a 1, y log a x is equivalent to x a . Logarithmic Function If a 7 0, a 1, and x 7 0, then the logarithmic function ƒ x = log3 x is the logarithmic function with base 3. with base a is ƒ x log a x. 1 2 x y y Graph of ƒ x log a x f(x) = log x 1 2 1 3 -1 1 3 (3, 1) 1. The points1 2 a , -1 , 1, 0 , and a, 1 are on the graph. (1, 0) 1 0 1 2. If a 1, then ƒ is an increasing function. 3 1 x 7 1 2 1 2 0 13 If 0 a 1,A then Bƒ is a decreasing function. 1 6 6 –2 , –1 Q 3 R 3. The y-axis is a vertical asymptote.

4. The domain is 0, ∞ , and the range is -∞, ∞ . Properties of Logarithms1 2 1 2 For x 7 0, y 7 0, a 7 0, a 1, and any real number r, the following properties hold.

log a xy log a x log a y Product property log2 3 # 5 = log2 3 + log2 5 x 3 1 2 log a log a x log a y Quotient property log2 = log2 3 - log2 5 y 5 r 5 log a x r log a x Power property log6 3 = 5 log6 3

log a 1 0 Logarithm of 1 log10 1 = 0

log a a 1 Base a logarithm of a log10 10 = 1 Theorem on Inverses For a 7 0 and a 1, the following properties hold.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 489 06/10/15 6:48 PM NOT FOR SALE 490 Chapter 4 Inverse, Exponential, and Logarithmic Functions

Concepts Examples

4.4 Evaluating Logarithms and the Change-of-Base Theorem Common and Natural Logarithms For all positive numbers x, base 10 logarithms and base e Approximate log 0.045 and ln 247.1. logarithms are written as follows.

log x log10 x Common logarithm log 0.045 ≈ -1.3468 Use a calculator. ln x log e x Natural logarithm ln 247.1 ≈ 5.5098

Change-of-Base Theorem For any positive real numbers x, a, and b, where a 1 and Approximate log8 7. b 1, the following holds.

log b x log 7 ln 7 log a x log8 7 = = ≈ 0.9358 Use a calculator. log b a log 8 ln 8

4.5 Exponential and Logarithmic Equations Property of Logarithms If x 7 0, y 7 0, a 7 0, and a 1, then the following holds. Solve. 5x x y is equivalent to log a x log a y. e = 10 ln e5x = ln 10 Take natural logarithms. 5x = ln 10 ln e x = x, for all x. ln 10 x = Divide by 5. 5 x ≈ 0.461 Use a calculator. The solution set can be written with the exact value, ln 10 5 , or with the approximate value, 0.461 .

2 5 6 E logF 2 x - 3 = log2 6 x2 3 6 Property of logarithms 1 - 2 = x2 = 9 Add 3. x = {3 Take square roots. Both values check, so the solution set is {3 . 5 6 4.6 Applications and Models of Exponential Growth and Decay Exponential Growth or Decay Function Let y0 be the amount or number present at time t = 0. The formula for continuous compounding, Then, under certain conditions, the amount present at any A Pe r t, time t is modeled by is an example of exponential growth. Here, A is the y y ekt, where k is a constant. 0 compound amount if P dollars are invested at an annual interest rate r for t years.

If P = $200, r = 3%, and t = 5 yr, find A. A = Pert A = 200e0.03 5 Substitute. 1 2 A ≈ $232.37 Use a calculator.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 490 20/10/15 11:05 AM NOT FOR SALE chapter 4 Review Exercises 491

Chapter 4 Review Exercises

Determine whether each function as graphed or defined is one-to-one. 1. not one-to-one 2. one-to-one 1. y 2. y 3. y 3. one-to-one 4. one-to-one 5. not one-to-one x x x 6. not one-to-one 0 0 0

3 7. ƒ -1 x = x + 3 8. not possible2 1 2 4. y = x3 + 1 5. y = x + 3 2 6. y = 3x2 + 2 9. It represents the number 2 of years after 2004 for the 1 2 investment to reach $50,000. Find the inverse of each function that is one-to-one. 10. yes 7. ƒ x x3 3 8. ƒ x 25 x2 11. one-to-one 12. true = - = - 2 1 2 1 2 13. B 14. A Concept Check Work each problem. 15. C 16. D 9. Suppose ƒ t is the amount an investment will grow to t years after 2004. What does -1 17. log2 32 = 5 ƒ $50,000 represent? 1 2 18. log 10 1 100 = 2 10. The1 graphs of2 two functions are shown. Based on their graphs, are these functions 4 19. log3/4 3 = -1 inverses? 20. y 10 6 1 x + 2 f(x) = ؊ 1 Q 5R 4 2 −16.1 16.1 ؊3 3 0 x y = ؊1 –2 −10 -∞, ∞ ; -1, ∞ 11. To have an inverse, a function must be a(n) function. 21. 101 3 = 10002 1 22.2 93/2 = 27 23. e1/2 = e 12. True or false? The x-coordinate of the x-intercept of the graph of y = ƒ x is the 1 2 y-coordinate of the y-intercept of the graph of y = ƒ - x . 1 2 1 2 Match each equation with the figure that most closely resembles its graph.

x x 13. y = log0.3 x 14. y = e 15. y = ln x 16. y = 0.3 A. y B. y C. y D. y

x x x x 0 0 0 0

Write each equation in logarithmic form. 3 -1 4 17. 25 = 32 18. 1001/2 = 10 19. = 4 3 1 x + 2 a b 20. Graph ƒ x = 5 - 1. Give the domain and range. 1 2 Write each equationA inB exponential form. 3 1 21. log 1000 = 3 22. log9 27 = 23. ln e = 2 2 2 Copyright Pearson. All Rights Reserved.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 491 06/10/15 6:48 PM NOT FOR SALE 492 Chapter 4 Inverse, Exponential, and Logarithmic Functions

24. 3 24. Concept Check What is the base of the logarithmic function whose graph contains 25. 2 the point 81, 4 ? 25. Concept Check What is the base of the exponential function whose graph contains 26. 2 log5 x + 4 log5 y + 1 2 1 the point 4, 1 ? 5 3 log5 m + log5 p - 16 27. log m log n 31 + 3 - 2 log3 5 - log3 r A B 28. This cannot be simplified. Use properties of logarithms to rewrite each expression. Simplify the result if possible. Assume all variables represent positive real numbers.

29. -1.3862 30. 1.6590 2 4 5 3 mn 2 26. log5 x y m p 27. log3 28. log7 7k + 5r 31. 11.8776 32. 6.1527 5r 33. 1.1592 34. 6.0486 2 A B 1 2 22 35. 5 36. 1.792 Use a calculator to find an approximation to four decimal places for each logarithm. 37. 3.667 38. 52.2696 29. log 0.0411 30. log 45.6 31. ln 144,000 39. E -13.257F 40. 2.386 5 6 5 6 5 41. -0.485 42. -0.123 32. ln 470 33. log 34. log 769 5 6 5 6 2/3 8 3 43. 2.102 44. 17.531 5 6 5 - 6 45. 2.487 46. 2 5 - 6 5 6 47. 3 48. 140.011 Solve each equation. Unless otherwise specified, give irrational solutions as decimals 5 6 5 6 correct to the nearest thousandth. 49. ln 3 50. 5 6 5∅ 6 51. 6.959 35. 16x+4 83x-2 36. 4x 12 37. 32x-5 13 5 6 = = = x+3 x x+3 x x-1 52. 5A, C, D6 38. 2 = 5 39. 6 = 4 40. e = 4 2 x 5x 2 3x 7 16 41. e - 12 42. 2e + 8 43. 10e - 5 13/3 e = = = 53. e 54. 5 x 2 2x 1 x 3 4x 1 8x 2x 20 4 44. 5 + = 2 - 45. 6 - = 3 + 46. e # e = e 55. 5 10 6- 7 56. e3 2 2 E F 47. e6x # e x = e21 48. 100 1.02 x/4 = 200 49. 2e2x - 5e x - 3 = 0 57. 3 58. 6 U V 5{6 (Give exact form.) 4 x 1 2 59. , 5 60. 6 1 x 5 -63 5 6 50. + 2 = 0 51. 4 1.06 + 2 = 8 2 10 61. 1, 3 62. 546 E F 52. Concepta b Check Which one or more of the following1 2 choices is the solution set of 63. 2 64. 3 5 6 5x = 9? 65. E -3 F 5 6 5 6 log 9 ln 9 66. n = a eS/a - 1 A. log 9 B. log 5 C. D. 5 6 5 9 log 5 ln 5 67. I I 0 = 101d/10 2 5 6 5 6 e f e f 68. x = 10D/100-2 Solve each equation. Give solutions in exact form. 69. 1.315 53. 3 ln x = 13 54. ln 5x = 16 5 6 55. log 2x + 7 = 0.25 56. ln x + ln x3 = 12 3 2 57. log21 x + 52 = 5 58. log3 x - 9 = 3 ln x 59. log4 1 3x + 21 x - 4 = 2 60. ln e 1 - ln 2x - 4 = ln 3

61. log x31+ log 1321- 3x 24= 1 62. log7 3x + 21 - log2 7 x - 2 = 1 1 2 1 2 1 1 2 63. ln 6x - ln x + 1 = ln 4 64. log16 x + 1 = 4 2 1 2 1 2 n 65. ln ln e-x ln 3 66. S a ln 1 , for n = = + a 3 4 I a b 67. d = 10 log , for I0 68. D = 200 + 100 log x, for x I0

69. Use a graphing calculator to solve the equation e x = 4 - ln x. Give solution(s) to the nearest thousandth.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 492 06/10/15 6:48 PM NOT FOR SALE chapter 4 Review Exercises 493

70. 89 decibels is about twice as Solve each problem. loud as 86 decibels. This is a 100% increase. 70. (Modeling) Decibel Levels Decibel rating of the loudness of a sound is modeled by

71. (a) 1,000,000I0 I d 10 log , (b) 158,500,000I0 = I0 (c) 158.5 times greater

72. (a) 199,500,000I0 where I is the intensity of a particular sound, and I0 is the intensity of a very faint (b) 12,600,000I0 threshold sound. A few years ago, there was a controversy about a proposed gov- (c) The 1906 earthquake ernment limit on factory noise. One group wanted a maximum of 89 decibels, had a magnitude almost while another group wanted 86. Find the percent by which the 89-decibel intensity 16 times greater than the exceeds that for 86 decibels. 1989 earthquake. 71. Earthquake Intensity The magnitude of an earthquake, measured on the Richter 73. 3.0% 74. 4.0 yr scale, is log I , where I is the amplitude registered on a seismograph 100 km from 75. $20,363.38 I0 76. $17,458.04 the epicenter of the earthquake, and I0 is the amplitude of an earthquake of a certain (small) size. On August 24, 2014, the Napa Valley in California was shaken by an 77. 17.3 yr earthquake that measured 6.0 on the Richter scale. 78. (a) For t = x, A(x) = x2 − x + 350 (a) Express this reading in terms of I0. 500 (b) On April 1, 2014, a quake measuring 8.2 on the Richter scale struck off the coast of Chile. It was the largest earthquake in 2014. Express the magnitude of an 8.2 reading in terms of I0 to the nearest hundred thousand. (c) How much greater than the force of the 6.0 earthquake was the force of the earth- 0 10 quake that measured 8.2? 0 72. Earthquake Intensity The San Francisco earthquake of 1906 had a Richter scale (b) For t = x, A(x) = 350 log (x + 1) rating of 8.3. 500 (a) Express the magnitude of this earthquake in terms of I0 to the nearest hundred thousand. (b) In 1989, the San Francisco region experienced an earthquake with a Richter scale rating of 7.1. Express the magnitude of this earthquake in terms of I0 to the 0 10 nearest hundred thousand. 0 (c) Compare the magnitudes of the two San Francisco earthquakes discussed in (c) For t = x, parts (a) and (b). A(x) = 350 (0.75)x 500 73. Interest Rate What annual interest rate, to the nearest tenth, will produce $4700 if $3500 is left at interest compounded annually for 10 yr? 74. Growth of an Account Find the number of years (to the nearest tenth) needed for $48,000 to become $53,647 at 2.8% interest compounded semiannually. 0 10 0 75. Growth of an Account Manuel deposits $10,000 for 12 yr in an account paying 3% (d) For t = x, interest compounded annually. He then puts this total amount on deposit in another A(x) = 100 (0.95)x account paying 4% interest compounded semiannually for another 9 yr. Find the 500 total amount on deposit after the entire 21-yr period. 76. Growth of an Account Anne deposits $12,000 for 8 yr in an account paying 2.5% interest compounded annually. She then leaves the money alone with no further deposits at 3% interest compounded annually for an additional 6 yr. Find the total 0 10 0 amount on deposit after the entire 14-yr period. Function (c) best describes A t . 77. Cost from Inflation Suppose the inflation rate is 4%. Use the formula for continu- 1 2 ous compounding to find the number of years, to the nearest tenth, for a $1 item to cost $2. 78. (Modeling) Drug Level in the Bloodstream After a medical drug is injected directly into the bloodstream, it is gradually eliminated from the body. Graph the following functions on the interval 0, 10 . Use 0, 500 for the range of A t . Determine the function that best models the amount A t (in milligrams) of a drug remaining in the body after t hours if 3503 mg4 were initially3 4injected. 1 2 1 2 (a) A t = t2 - t + 350 (b) A t = 350 log t + 1 (c) A t 350 0.75 t (d) A t 100 0.95 t 1 2 = 1 2 = 1 2 Copyright Pearson.1 2 All1 Rights2 Reserved. 1 2 1 2

M04_LHSD7659_12_AIE_C04_pp405-496.indd 493 06/10/15 6:48 PM NOT FOR SALE 494 Chapter 4 Inverse, Exponential, and Logarithmic Functions

79. 2016 79. (Modeling) Chicago Cubs’ Payroll The table shows the total payroll (in millions 80. (a) See the graph in part (c). of dollars) of the Chicago Cubs baseball team for the years 2010–2014. (b) An exponential function best describes the data. (c) Answers will vary. Total Payroll One answer is Year (millions of dollars) ƒ x = 973,716(1.395)x. 2010 145.4 9 1.51 ×2 10 2011 134.3 2012 111.0 2013 107.4

0 25 2014 92.7 −108 Source: www.baseballprospectus.com/ (d) 7,797,000,000 compensation 81. (a) $15,207 (b) $10,716 Letting ƒ x represent the total payroll and x represent the number of years since (c) $4491 2010, we find that the function (d) They are the same. 1 2 ƒ x = 146.02e-0.112x models the data quite well. According1 2 to this function, when will the total payroll halve its 2010 value? 80. (Modeling) Transistors on Computer Chips Computing power has increased dra- matically as a result of the ability to place an increasing number of transistors on a single processor chip. The table lists the number of transistors on some popular computer chips made by Intel.

Year Chip Transistors 1989 486DX 1,200,000 1994 Pentium 3,300,000 2000 Pentium 4 42,000,000 2006 Core 2 Duo 291,000,000 2008 Core 2 Quad 820,000,000 2010 Core (2nd gen.) 1,160,000,000 2012 Core (3rd gen.) 1,400,000,000 Source: Intel.

(a) Make a scatter diagram of the data. Let the x-axis represent the year, where x = 0 corresponds to 1989, and let the y-axis represent the number of transistors. (b) Decide whether a linear, a logarithmic, or an exponential function best describes the data. (c) Determine a function ƒ that approximates these data. Plot ƒ and the data on the same coordinate axes. (d) Assuming that this trend continues, use ƒ to estimate the number of transistors on a chip, to the nearest million, in the year 2016. 81. Financial Planning The traditional IRA (individual retirement account) is a common tax-deferred saving plan in the United States. Earned income deposited into an IRA is not taxed in the current year, and no taxes are incurred on the interest paid in subsequent years. However, when the money is withdrawn from the account after 1 age 59 2 , taxes must be paid on the entire amount withdrawn. Suppose we deposited $5000 of earned income into an IRA, we can earn an annual interest rate of 4%, and we are in a 25% tax bracket. (Note: Interest rates and tax brackets are subject to change over time, but some assumptions must be made to evaluate the investment.) Also, suppose that we deposit the $5000 at age 25 and withdraw it at age 60, and that interest is compounded continuously.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 494 06/10/15 6:48 PM NOT FOR SALE chapter 4 Test 495

2 82. (a) log4 2x - x = (a) How much money will remain after we pay the taxes at age 60? ln 2x2 - x (b) Suppose that instead of depositing the money into an IRA, we pay taxes on ln1 4 2 1 2 the money and the annual interest. How much money will we have at age 60? (b) ln(2x2 − x) y (Note: We effectively start with $3750 (75% of $5000), and the money earns 3% = ln4 2.5 (75% of 4%) interest after taxes.) (c) To the nearest dollar, how much additional money will we earn with the IRA? −2.5 2.5 (d) Suppose we pay taxes on the original $5000 but are then able to earn 4% in a tax-free investment. Compare the balance at age 60 with the IRA balance.

2 −5 82. Consider ƒ x = log4 2x - x . (c) 1 , 0 , 1, 0 (a) Use the change-of-base theorem with base e to write log 2x2 x in a suitable - 2 1 2 1 2 4 - 1 form to graph with a calculator. (d) x = 0, x =1 2 2 1 2 (e) ThereA isB no y-intercept (b) Graph the function using a graphing calculator. Use the window -2.5, 2.5 by because the domain of -5, 2.5 . 3 4 a logarithm must be a (c) What are the x-intercepts? 3 4 positive number, and the (d) Give the equations of the vertical asymptotes. argument 2x2 - x is 1 (e) Why is there no y-intercept? positive for -∞, 0 ´ 2 , ∞ only. 1 2 A B

Chapter 4 Test

3 [4.1] 1. Consider the function ƒ x = 2x - 7. 1. (a) -∞, ∞ ; -∞, ∞ (a) What are the domain and range2 of ƒ? (b) The graph is a stretched 1 2 1 1 2 1 2 3 (b) Explain why ƒ - exists. translation of y = x, 1 which passes the horizon-2 (c) Write an equation for ƒ - x . tal line test and is thus a (d) What are the domain and range of ƒ -1? one-to-one function. 1 2 -1 x3 (e) Graph both ƒ and ƒ . How are the two graphs related with respect to the line (c) ƒ -1 x + 7 = 2 y = x? (d) ∞, ∞ ; ∞, ∞ - 1 2 - 3 2. Match each equation with its graph. (e) y –1 f (x) = x + 7 1 2 1 22 1 x 10 x 3 (a) y = log1/3 x (b) y = e (c) y = ln x (d) y = f(x) = Ί2x – 7 3 –10 0 x a b 10 A. y B. y C. y D. y –10 The graphs are reflections of each other across the x x x x 0 0 0 0 line y = x. [4.2, 4.3] 2. (a) B (b) A (c) C (d) D 1 2x-3 3. Solve = 16x+1. 1 8 3. 2 3 a b 3/2 4. (a) log4 8 = 2 4. (a) Write 4 = 8 in logarithmic form. E F 2/3 (b) 8 = 4 2 (b) Write log8 4 = 3 in exponential form. [4.1–4.3] 1 x 5. y 5. Graph ƒ x = 2 and g x = log1/2 x on the same axes. What is their relationship? 4 6. Use properties of logarithms to rewrite the expression. Assume all variables repre- 1 x 1 2 1 2 f(x) = ( 2) 1 sent positive realA Bnumbers. x 0 41 x2 4 y g(x) = log1/2 x log 7 z23 They are inverses. Copyright Pearson. All Rights Reserved.

M04_LHSD7659_12_AIE_C04_pp405-496.indd 495 06/10/15 6:48 PM NOT FOR SALE 496 Chapter 4 Inverse, Exponential, and Logarithmic Functions

[4.3] Use a calculator to find an approximation to four decimal places for each logarithm. 1 6. 2 log7 x + 4 log7 y - 3 log7 z 7. log 2388 8. ln 2388 9. log9 13 [4.4] 10. Solve x2/3 = 25. 7. 3.3780 8. 7.7782 9. 1.1674 Solve each equation. Give irrational solutions as decimals correct to the nearest thou- [4.2] sandth. 10. {125 11. 12x = 1 12. 9x = 4 13. 162x+1 = 83x 11. 0 5 6 14. 2x+1 = 3x-4 15. e0.4x = 4x-2 [4.5]5 6 16. 2e2x 5ex 3 0 (Give both exact and approximate values.) 12. 0.631 - + =

[4.2]5 6 Solve each equation. Give solutions in exact form. 13. 4 9 5 6 17. logx = 2 18. log2 x - 4 x - 2 = 3 [4.5] 16 14. 12.548 1 19. log x log x 2 3 20. ln x 314 ln 321 ln 24 x 15. 2.811 2 + 2 + = - = 5 6 5 3 1 2 16. 50, ln 26 ; 0, 0.405 21. log3 x + 1 - log3 x - 3 = 2 5 6 22. A friend is taking another mathematics course and says, “I have no idea what an [4.3] 1 2 1 2 E F expression like log 27 really means.” Write an explanation of what it means, and 3 5 17. 4 tell how we can find an approximation for it with a calculator.

[4.5]E F 18. 0, 6 Solve each problem. 19. 2 5 6 23. (Modeling) Skydiver Fall Speed A skydiver in free fall travels at a speed modeled by 20. ∅ 5 6 -0.18t 7 v t = 176 1 - e 21. 2 feet per second after t seconds. 1How2 long,1 to the nearest2 second, will it take for the [4.4]E F skydiver to attain a speed of 147 ft per sec (100 mph)? 22. The expression log5 27 represents the exponent to which 5 24. Growth of an Account How many years, to the nearest tenth, will be needed for must be raised in order to $5000 to increase to $18,000 at 3.0% annual interest compounded (a) monthly obtain 27. To approximate (b) continuously? it with a calculator, use the 25. Tripling Time For any amount of money invested at 2.8% annual interest com- change-of-base theorem. pounded continuously, how long, to the nearest tenth of a year, will it take to triple? log 27 log5 27 = ≈ 2.0478 26. (Modeling) Radioactive Decay The amount of a certain radioactive material, in log 5 grams, present after t days is modeled by

[4.6] A t = 600e-0.05t. 23. 10 sec (a) Find the amount present after 12 days, to the nearest tenth of a gram. 24. (a) 42.8 yr (b) 42.7 yr 1 2 25. 39.2 yr (b) Find the half-life of the material, to the nearest tenth of a day. 26. (a) 329.3 g (b) 13.9 days

M04_LHSD7659_12_AIE_C04_pp405-496.indd 496 06/10/15 6:48 PM