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Home , Binary logarithm, Common logarithm, Exponential function, Function (mathematics), Inverse function, Logarithm

The as an inverse

In this section we concentrate on understanding the logarithm function. If the logarithm is understood as the inverse of the , then the properties of will naturally follow from our Elementary Functions understanding of exponents. Part 3, Exponential Functions & Logarithms The meaning of the logarithm. Lecture 3.3a, Logarithms: Basic Properties The logarithmic function g(x) = logb(x) is the inverse of the exponential function f(x) = bx.

Dr. Ken W. Smith y The meaning of y = logb(x) is b = x. Sam Houston State University The y 2013 b = x

is the “exponential form” for the logarithm y = logb(x).

The positive b is called the (of the logarithm.)

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Some worked exercises. Write each of the following logarithms in exponential form and then use that exponential form to solve for x. 1 4 log2( 8 ) = x x 1 −3 1 1 log2(8) = x Solution. The exponential form is 2 = 8 . Since 2 = 8 the answer Solution. The exponential form is 2x = 8. Since 23 = 8 the answer is is x = −3 . √ x = 3 . 3 5 log ( 2) = x 47 2 √ 2 log2(2 ) = x Solution. The exponential form is 2x = 3 2 = 21/3. So x = 1/3 . Solution. The exponential form is 2x = 247. So x = 47 . 1 3 log2( 2 ) = x 1 Notice how we use the exponential form in each problem! Solution. The exponential form is 2x = 1 . Since 2−1 = the answer 2 2 is x = −1 .

Smith (SHSU) Elementary Functions 2013 3 / 29 Smith (SHSU) Elementary Functions 2013 4 / 29 The graph of a logarithm function The graph of a logarithm function

The graph of y = 2x was drawn in an earlier lecture (see below.) The graph of y = log2 x :

The graph of the y = log2 x is obtained by reflecting the graph of y = 2x across the y = x.

Smith (SHSU) Elementary Functions 2013 5 / 29 Smith (SHSU) Elementary Functions 2013 6 / 29 The graph of a logarithm function The graph of a logarithm function

If we draw them together, we have the picture below. The graph of the exponential function y = 2x:

The graph of the logarithmic function y = log x: Smith (SHSU) Elementary Functions 2013 7 / 29 Smith (SHSU) Elementary Functions 2 2013 8 / 29 Logarithms Logarithms

We agreed earlier that the exponential function f(x) = bx has domain (−∞, ∞) and range (0, ∞).

Since g(x) = logb x is the inverse function of f(x) the domain of the log function will be the range of the exponential function, and vice versa. In summary, here are our abbreviations: So the domain of logb x is (0, ∞) and the range is (−∞, ∞). 1 ln x means the logarithm base e, The most useful base for logarithms is e. We will abbreviate loge(x) by 2 log x means the logarithm base 10 and ln(x) and speak of the “”. 3 lb x means the logarithm base 2. Sometimes, for historical reasons, we may use base 10. It is customary to speak then of the “” and abbreviate log10(x) by log(x), dropping the subscript. However (warning!), in higher mathematics and applications, log(x) usually means base e and is equivalent to ln(x).

In these notes we will use log(x) to mean log10(x). One more abbreviation – in science, because store Smith (SHSU) Elementary Functions 2013 9 / 29 Smith (SHSU) Elementary Functions 2013 10 / 29 data in binary (in of zeroes and ones), one uses base 2. Some Logarithmsabbreviate log2(x) as lb(x) and speak of the “binary” logarithm. Properties of exponential functions in terms of logarithms

The logarithm function plucks the exponent from an expression. For this A few more worked exercises. reason, the properties of exponents translate into properties of logarithms. Write each of the following logarithms in exponential form and then use For example, we know that when we multiply two terms with a common that exponential form to solve for x. base, we add the exponents:

1 log(1000) = x x y x+y Solution. The exponential form is 10x = 1000. Since 103 = 1000 the (b )(b ) = b (1) answer is x = 3 . Suppose we call the first term M := bx and the second term N := by. 1 Then one may ask the question, “What is the exponent on b in the 2 ln( ) = x e3 MN? Solution. The exponential form is ex = e−3 so the answer is −3 . The answer is “We add the exponents appearing in M and N.” In other 1 words (if we learn to translate “logb” as “the exponent on b that...”), we 3 lb(√ ) = x 2 can restate this exponent property as “when we multiply we add 1 √ their exponents”. This is the product property for logarithms: Solution. The exponential form is 2x = √ . Since 21/2 = 2 then 2 1 2−1/2 = √ and so the answer is x = −1/2 . logb(MN) = logb M + logb N (2) 2

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x What happens when we divide two terms with a common base? A third important property of exponents: when we raise a term like b to a , we multiply exponents. x b x−y = b (3) x xc by (b ) = b (5) When we do , we subtract exponents. So, in the language of In our “logarithm language” (thinking of M as bx) we have the exponent logarithms, we have the property, “the exponent in a quotient is property the difference of the two exponents”: c logb(M ) = c logb M (6) M log ( ) = log M − log N (4) b N b b Each of these three properties is merely a restatement, in the language of logarithms, of a property of exponents.

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We review the three logarithm rules we have developed so far.

Product Property of Logarithms:

logb(MN) = logb M + logb N

Quotient Property of Logarithms: In the next presentation, we develop several more properties of logarithms. M (END) log ( ) = log M − log N b N b b

Exponent Property of Logarithms: c logb(M ) = c logb M

Each of these properties is a restatement, in the language of logarithms, of a property of exponents. Smith (SHSU) Elementary Functions 2013 15 / 29 Smith (SHSU) Elementary Functions 2013 16 / 29 Logarithms

We review the three basic logarithm rules we have developed so far.

Product Property of Logarithms:

Elementary Functions logb(MN) = logb M + logb N Part 3, Exponential Functions & Logarithms Lecture 3.3b, Logarithms: Basic Properties, Continued Quotient Property of Logarithms: M log ( ) = log M − log N Dr. Ken W. Smith b N b b

Sam Houston State University Exponent Property of Logarithms: 2013 c logb(M ) = c logb M

Each of these three properties is merely a restatement of a property of exponents. Smith (SHSU) Elementary Functions 2013 17 / 29 Smith (SHSU) Elementary Functions 2013 18 / 29 Changing the base Changing the base

Suppose we want to change the base of our logarithm. This often occurs We began with when we want to use a “good” base like e on a problem which began with y = log x. a different base. b Suppose we want to work with base c but our problem began with base b: We rewrote this as log x y = c . logc b y = logb x. Rewrite this in exponential form: So y, which was originally equal to logb x is now y b = x. logc x logb x = Now take the log of both sides of the . If we want to work in base logc b c then let us log () to both sides of our equation. c Let’s call this the “change of base” equation or “change of base” y logc(b ) = logc(x). property. Now we use the exponent property pulling the exponent y outside the logarithm: One way to remember this is to note that on the left side of the equal y log (b) = log (x). (logb x), b is lower than x. c c Then on the right side of the equal sign ( log x ), b is still lower than x! Solve for y: log b log x Smith (SHSU) Elementaryy = Functionsc . 2013 19 / 29 Smith (SHSU) Elementary Functions 2013 20 / 29 logc b Logarithms More on the logarithm as an inverse function

We began this lecture by defining g(x) = logb(x) as the inverse function of f(x) = bx. Since these functions are inverses, we know then that

(f ◦ g)(x) = (g ◦ f)(x) = x. (7) Example. Suppose we want to compute log (17) but our only 2 Let us examine this in more detail. allows us to use the natural logarithm ln. Then, by the change of base equation we can write x x Note that (g ◦ f)(x) = g(f(x)) = g(b ) = logb(b ). Since the log function ln 17 and the exponential function are inverse functions, this must be equal to log (17) = ≈ 4.087463. just x and we have 2 ln 2 x logb(b ) = x.

This equation is really fairly easy to understand. If we translate “logb(x)” x as “the exponent on b that give x” then we should translate logb(b ) as “the exponent on b which gives bx.” Obviously this should be x since x is the exponent one places on b to get bx. (If that doesn’t make , read through it one more time slowly....) Smith (SHSU) Elementary Functions 2013 21 / 29 Smith (SHSU) Elementary Functions 2013 22 / 29 More on the logarithm as an inverse function Six properties of logarithms

In summary, we have developed the following six properties of logarithms. 1 The Product Property of Logarithms:

Since (f ◦ g)(x) = x we also have logb(MN) = logb M + logb N log x x = (f ◦ g)(x) = f(g(x)) = f(logb x) = b b . So

log x b b = x. 2 The Quotient Property of Logarithms:

This is almost as easy to understand as the previous equation. M log ( ) = log M − log N b N b b It says that if we place on b “the exponent you put on b to get x” (logb x) then we should just get x! 3 The Exponent Property of Logarithms:

c logb(M ) = c logb M

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If we understand the logarithm as the inverse of the exponential function then we are prepared to find the inverse of a variety of functions. Here are 4 The Change of Base Property some examples. Find the inverse function of: 2 logc x 1 f(x) = ex . logb x = 2 logc b 2 f(x) = ex −5. 3 f(x) = 5 + ex.

5 Inverse Property #1 4 f(x) = log2(x + 2) + 2. x logb b = x Solutions 2 2 1 To find the inverse of f(x) = ex y = ex and swap inputs and outputs 6 Inverse Property #2 2 x = ey . blogb x = x Take the natural logarithm of both sides ln x = y2 and solve for y by taking roots of both sides √ Smith (SHSU) Elementary Functions 2013 25 / 29 Smith (SHSU) Elementaryln Functionsx = y. 2013 26 / 29 √ More on the logarithm as an inverse function MoreSo on one the inverse logarithm function is asf an−1(x inverse) = ln x function.

4 To find the inverse of f(x) = log2(x + 2) + 2 we write x2−5 y2−5 2 To find the inverse of y = e we swap letters so that x = e , y = log (x + 2) + 2, take natural logs of both sides 2 change variables (to indicate that we are swapping inputs and ln x = y2 − 5, outputs) x = log2(y + 2) + 2, add 5 and take square roots so that and subtract 2 from both sides √ f −1(x) = ln x + 5 . x − 2 = log2(y + 2). At this it is best to write this logarithmic equation in exponential form. 3 To find the inverse of y = 5 + ex we swap variables, subtract 5 from 2x−2 = y + 2. both sides and then take the natural log to get ln(x − 5) = y. So Subtract 2 from both sides x−2 f −1(x) = ln(x − 5) . 2 − 2 = y and then write out our answer using inverse function notation. f −1(x) = 2x−2 − 2 Smith (SHSU) Elementary Functions 2013 27 / 29 Smith (SHSU) Elementary Functions 2013 28 / 29 Exponential Functions

In the next of lectures, we apply properties of logarithms.

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