8.5 Logarithmic Functions 821

Section 8.5 Logarithmic Functions 821

8.5 Logarithmic Functions

We can now apply the inverse function theory from the previous section to the exponential function. From Section 8.2, we know that the function f(x) = bx is either increasing (if b > 1) or decreasing (if 0 < b < 1), and therefore is one-to-one. Consequently, f has an inverse function f −1. As an example, let’s consider the exponential function f(x) = 2x. f is increasing, has domain Df = (−∞, ∞), and range Rf = (0, ∞). Its graph is shown in Figure 1(a). The graph of the inverse function f −1 is a reﬂection of the graph of f across the line y = x, and is shown in Figure 1(b). Since domains and ranges are interchanged, the domain of the inverse function is Df −1 = (0, ∞) and the range is Rf −1 = (−∞, ∞). y f y = x y y = x 5 5

f −1

x x 5 5

(a) (b) Figure 1. The graphs of f(x) = 2x and its inverse f −1(x) are reﬂec- tions across the line y = x.

Unfortunately, when we try to use the procedure given in Section 8.4 to ﬁnd a formula for f −1, we run into a problem. Starting with y = 2x, we then interchange x and y to obtain x = 2y. But now we have no algebraic method for solving this last equation for y. It follows that the inverse of f(x) = 2x has no formula involving the usual arithmetic operations and functions that we’re familiar with. Thus, the inverse function is a new function. The name of this new function is the logarithm of x to base −1 2, and it’s denoted by f (x) = log2(x). Recall that the deﬁning relationship between a function and its inverse (Property 14 in Section 8.4) simply states that the inputs and outputs of the two functions x are interchanged. Thus, the relationship between 2 and its inverse log2(x) takes the following form:

v v = log2(u) ⇐⇒ u = 2 More generally, for each exponential function f(x) = bx (b > 0, b 6= 1), the inverse −1 function f (x) is called the logarithm of x to base b, and is denoted by logb(x). The deﬁning relationship is given in the following deﬁnition.

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Version: Fall 2007 822 Chapter 8 Exponential and Logarithmic Functions

Deﬁnition 1. If b > 0 and b 6= 1, then the logarithm of u to base b is deﬁned by the relationship

v v = logb(u) ⇐⇒ u = b . (2)

In order to understand the logarithm function better, let’s work through a few simple examples.

I Example 3. Compute log2(8).

Label the required value by v, so v = log2(8). Then by (2), using b = 2 and u = 8, it follows that 2v = 8, and therefore v = 3 (solving by inspection). v In the last example, note that log2(8) = 3 is the exponent v such that 2 = 8. Thus, in general, one way to interpret the deﬁnition of the logarithm in (2) is that logb(u) is the exponent v such that bv = u. In other words, the value of the logarithm is the exponent!

I Example 4. Compute log10(10 000).

Again, label the required value by v, so v = log10(10 000). By (2), it follows that 10v = 10 000, and therefore v = 4. Note that here again we have found the exponent v=4 that is needed for base 10 in order to get 10v = 10 000.

1 I Example 5. Compute log3 9 .

1 v = log 3 9 1 =⇒ 3v = by (2) 9 1 =⇒ v = −2 since 3−2 = 9

I Example 6. Solve the equation log5(x) = 1.

log5(x) = 1 =⇒ 51 = x by (2) =⇒ x = 5

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I Example 7. Solve the equation logb(64) = 3 for b.

logb(64) = 3 =⇒ b3 = 64 by (2) √ =⇒ b = 3 64 = 4

I Example 8. Solve the equation log1/2(x) = −2.

log1/2(x) = −2 1−2 =⇒ = x by (2) 2 1 1 =⇒ x = = = 4 1 2 1 2 4

The composition relationships in Property 15 of Section 8.4, applied to bx and logb(x), become

Property 9.

x logb(b ) = x (10) and

blogb(x) = x. (11)

Both equations are important. Note that (11) again shows that the logb(x) is the exponent v such that bv = x. Equation (10) will be used frequently in this and later sections to help us solve exponential equations. Logarithmic functions are used in many areas of science and engineering. For example, they are used to deﬁne the Richter scale for the magnitudes of earthquakes, the decibel scale for the loudness of sound, and the astronomical scale for stellar brightness. They are also important tools for use in computation (as we will see in Section 8.8). Our main use of logarithms in this textbook will be to solve exponential equations, and thereby help us study physical phenomena that are described by exponential functions (as in Section 8.7). Computing Logarithms In Examples 3–8 above, we were able to compute the logarithms by converting to exponential equations that could be solved by inspection. But it’s easy to see that most of the time this won’t work. For example, how would we compute the value of log2(7)?

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Fortunately, mathematicians have found other methods for computing logarithms to high accuracy, and they can now be easily approximated using a calculator or computer.

Your calculator has built-in buttons for computing two diﬀerent logarithms, log10(x) and loge(x). log10(x) is called the common logarithm, and loge(x) is called the natural logarithm.

Common Logarithm: The common logarithm log10(x) is computed using the LOG button on your calculator. Notice also that its inverse function 10x, can be computed using the same button in conjunction with the 2ND button. The common logarithm is usually the most convenient one to use for computations involving scientiﬁc notation (because we use a base 10 number system), and therefore is the logarithm most often used in the physical sciences. Because of that, it’s often just abbreviated by log(x), and we’ll do that as well in the remainder of the text.

Common Logarithm. log(x) and log10(x) are equivalent notations. Thus, we have the deﬁning relationship

v = log(u) ⇐⇒ u = 10v.

The composition properties for the common logarithm are

log(10x) = x (12)

and

10log(x) = x.

Natural Logarithm: The natural logarithm loge(x) is computed using the LN button on your calculator. Its inverse function, ex, is computed using the same button in conjunction with the 2ND button. The natural logarithm turns out to be the most convenient one to use in mathematics, because a lot of formulas, especially in calcu- lus, are much simpler when the natural logarithm is used. The natural logarithm is abbreviated by ln(x).

Natural Logarithm. ln(x) and loge(x) are equivalent notations. Thus, we have the deﬁning relationship

v = ln(u) ⇐⇒ u = ev.

The composition properties for the common logarithm are

ln(ex) = x (13)

and

eln(x) = x.

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Note that when using your calculator to compute log(x) and ln(x), you will usually only obtain approximate values, as these values frequently are irrational numbers. What about other bases? You can also compute these on your calculator, but we’ll ﬁrst need to develop the Change of Base Formula in the next section. However, at this point, we can at least solve exponential equations involving bases 10 and e, as shown in the next two examples.

x I Example 14. Solve the equation 704 = 2(10) . The ﬁrst step is to isolate the exponential on the right side by dividing both sides by 2:

352 = 10x

Then simply apply the log10(x) function to both sides of the equation:

x log10(352) = log10(10 )

x But (10) implies that log10(10 ) = x. Therefore, x = log10(352) = log(352) is the exact solution. The approximate value, using a calculator, is 2.546542663 (see Figure 2). Alternatively, instead of taking the logarithm of both sides in the second step, you x can apply (2) to the equation 352 = 10 to get x = log10(352).

Figure 2. Approximation of log(352) = log10(352). This last example shows how logarithms can be used for solving exponential equations. The basic strategy is to ﬁrst isolate the exponential on one side of the equation, and then take appropriate logarithms of both sides. Here’s one more example for now, and then we’ll return to this process repeatedly in the remaining sections, especially when we work with application problems.

x I Example 15. Solve the equation 30 = 20e . First isolate the exponential on the right side by dividing both sides by 20:

1.5 = ex

This time, since the base of the exponential function is e, apply the natural logarithm function to both sides:

x loge(1.5) = loge(e )

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x Simplify the right side, since loge(e ) = x by (10):

loge(1.5) = x

Therefore, x = loge(1.5) = ln(1.5) is the exact solution. The approximate value, using a calculator, is 0.4054651081 (see Figure 3).

Figure 3. Approximation of ln 1.5 = loge(1.5). In the next section, we’ll learn how to solve exponential equations involving other bases. Graphs of Logarithmic Functions

At the beginning of this section, we looked at the graphs of f(x) = 2x and its inverse −1 function f (x) = log2(x). More generally, the graph of the exponential function f(x) = bx for b > 1 is shown in Figure 4(a), along with its inverse logarithmic function −1 f (x) = logb(x). According to Section 8.4, the two graphs are reﬂections across the line y = x. Similarly, the graph for 0 < b < 1 is shown in Figure 4(b).

y f y = x f y y = x

f −1 x x f −1

(a) b > 1 (b) 0 < b < 1 x −1 Figure 4. The graphs of f(x) = b and f (x) = logb(x) are reﬂec- tions across the line y = x.

Because domains and ranges of inverse functions are interchanged, it follows that

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Property 16.

Domain(logb(x)) = (0, ∞) and

Range(logb(x)) = (−∞, ∞). In particular, note that the logarithm of a negative number, as well as the logarithm of 0, are not deﬁned.

Two particular points on the graph of the logarithm are noteworthy. Since b0 = 1, it follows that logb(1) = 0, and therefore the x-intercept of the graph of logb(x) is (1, 0). 1 Similarly, since b = b, it follows that logb(b) = 1, and therefore (b, 1) is on the graph.

Property 17.

logb(1) = 0 and logb(b) = 1

x Finally, since the graph of b has a horizontal asymptote y = 0, the graph of logb(x) must have a vertical asymptote x = 0. This behavior is a consequence of the fact that inputs and outputs of inverse functions are interchanged, and can be observed in Figure 4. In the ﬁnal example below, we’ll apply a transformation to the logarithm and see how that aﬀects the graph.

I Example 18. Plot the graph of the function f(x) = log2(x + 1).

The graph of f(x) = log2(x + 1) will be the same as the graph of g(x) = log2(x) shifted one unit to the left. The graph of g is shown in Figure 1(b). The x-intercept (1, 0) on the graph of g will be shifted one unit to the left to (0, 0) on the graph of f. Likewise, the vertical asymptote x = 0 on the graph of g will be shifted one unit to the left to the line x = −1 on the graph of f. The ﬁnal graph of f is shown in Figure 5.

y 5

x 5

Figure 5. The graph of f(x) = log2(x + 1).

Version: Fall 2007