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Section 4.2 Logarithmic Functions

Before Starting this Section, Review

1 Finding the inverse of a one-to-one function (Section 2.9 , page 298)
2 Polynomial and rational inequalities (Section 1.6 , page 151)
3 Graphing and transformations (Section 2.7 , page 263)

Objectives

1 Define logarithmic functions.
2 Evaluate logarithms.
3 Derive basic properties of logarithmic functions.
4 Find the domains of logarithmic functions.
5 Graph logarithmic functions.
6 Use natural logarithms in applications.

Detecting Possible Fraud

Listed below in two columns is the size (in square kilometers) of the first 50 countries of the world by alphabetical order. One column contains the real data, and the other column contains fictitious data. Can you guess which column contains the real data? In Example 9, we show how to detect the real data.

Country	Size I	Size II	Country	Size I	Size II
Afghanistan	608,239	645,807	Bermuda	54	53
Albania	29,562	28,748	Bhutan	47,560	46,650
Algeria	2,481,743	2,381,741	Bolivia	998,582	1,098,581
American Samoa	201	197	Bosnia and Herzegovina	49,857	51,129
Andorra	465	464	Botswana	601,780	581,730
Angola	997,852	1,246,700	Bouvet Island	48	49
Anguilla	98	96	Brazil	8,645,521	8,544,418
Antarctica	15,207,012	13,209,000	British Virgin Islands	201	151
Antigua & Barbuda	445	442	Brunei Darussalam	6,123	5,765
Antilles, Netherlands	801	800	Bulgaria	99,949	110,994
Arabia, Saudi	2,250,790	2,149,690	Burkina Faso	269,013	267,950
Argentina	2,875,510	2,777,409	Burundi	27,735	27,834
Armenia	29,851	29,743	Cambodia	201,137	181,035
Aruba	201	193	Cameroon	485,447	475,442
Australia	7,801,663	7,682,557	Canada	9,976,231	9,976,137
Austria	83,760	83,858	Cape Verde	4,121	4,033
Azerbaijan	86,640	86,530	Cayman Islands	260	259
Bahamas	20,112	13,962	Central African Republic	623,541	622,436
Bahrain	695	694	Chad	999,494	1,284,000
Bangladesh	99,763	142,615	Chile	756,589	755,482
Barbados	423	431	China	9,806,441	9,806,391
Belarus	207,599	207,600	Christmas Island	241	135
Belgium	31,618	30,518	Cocos (Keeling) Islands	21	14
Belize	23,105	22,966	Colombia	986,779	1,141,748
Benin	99,879	112,622	Comoros	2,001	1,862

Logarithmic Functions

1 Define logarithmic functions.

Recall from Section 4.1 that every exponential function

f (x) = a x, a > 0, a \neq 1,

$f (x) = a^{x}, a > 0, a \neq 1,$

is a one-to-one function and therefore has an inverse function. Let’s find the inverse of the function $f (x) = 3^{x} .$ If we think of an exponential function f as “putting an exponent on” a particular base, then the inverse function $f^{- 1}$ must “lift the exponent off” to undo the effect of f, as illustrated below:

Let’s try to find the inverse of $f (x) = 3^{x}$ by the procedure outlined in Section 2.9.

Step 1 Replace f(x) in the equation $f (x) = 3^{x}$ with y: $y = 3^{x} .$
Step 2 Interchange x and y in the equation in Step 1: $x = 3^{y} .$
Step 3 Solve the equation for y in Step 2.

In Step 3, we need to solve the equation $x = 3^{y}$ for y. Finding y is easy when $3 = 3^{y}$ or $27 = 3^{y},$ but what if $5 = 3^{y}$ ? There actually is a real number y that solves the equation, $5 = 3^{y},$ but we have not introduced the name for this number yet. (We were in a similar position concerning the solution of $y^{3} = 7$ before we introduced the name for the solution, namely, $\sqrt[3]{7}$ .) We can say that y is “the exponent on 3 that gives 5.” Instead, we use the word logarithm (or log for short) to describe this exponent. We use the notation $\log_{3} 5$ (read as “log of 5 with base 3”) to express the solution of the equation $5 = 3^{y} .$ This log notation is a new way to write information about exponents: $y = \log_{3} x$ means $x = 3^{y} .$ In other words, $\log_{3} x$ is the exponent to which 3 must be raised to obtain x. The inverse function of the exponential function $f (x) = 3^{x}$ is the logarithmic function with base 3. So

$f^{- 1} (x) = \log_{3} x or y = \log_{3} x, where y = f^{- 1} (x) .$

The previous discussion could be repeated for any base a instead of 3, provided that $a > 0$ and $a \neq 1$ (see Figure 4.9).

Figure 4.9

Logarithmic and exponential functions are inverses of each other

The definition of the logarithmic function says that the two equations

$\begin{array}{rcl} y & = & \log_{a} x (logarithmic form) and \\ x & = & a^{y} (exponential form) \end{array}$

are equivalent. For instance, $\log_{3} 81 = 4$ because 3 must be raised to the fourth power to obtain 81.

Example 1 Converting from Exponential to Logarithmic Form

Write each exponential equation in logarithmic form.

$4^{3} = 64$
${(\frac{1}{2})}^{4} = \frac{1}{16}$
$a^{- 2} = 7$

Solution

$4^{3} = 64$ is equivalent to $\log_{4} 64 = 3 .$
${(\frac{1}{2})}^{4} = \frac{1}{16}$ is equivalent to $\log_{1 / 2} \frac{1}{16} = 4 .$
$a^{- 2} = 7$ is equivalent to $\log_{a} 7 = - 2 .$

Practice Problem 1

Write each exponential equation in logarithmic form.
1. $2^{10} = 1024$
2. $9^{- 1 / 2} = \frac{1}{3}$
3. $p = a^{q}$

Side Note

To change from exponential form to logarithmic form, use the pattern

$a^{u} = v means \log_{a} v = u .$

The exponent u is the logarithm.

Example 2 Converting from Logarithmic Form to Exponential Form

Write each logarithmic equation in exponential form.

$\log_{3} 243 = 5$
$\log_{2} 5 = x$
$\log_{a} N = x$

Solution

$\log_{3} 243 = 5$ is equivalent to $243 = 3^{5} .$
$\log_{2} 5 = x$ is equivalent to $5 = 2^{x} .$
$\log_{a} N = x$ is equivalent to $N = a^{x} .$

Practice Problem 2

Write each logarithmic equation in exponential form.
1. $\log_{2} 64 = 6$
2. $\log_{v} u = w$

Side Note

To change from logarithmic form to exponential form, use

Evaluating Logarithms

2 Evaluate logarithms.

The technique of converting from logarithmic form to exponential form can be used to evaluate some logarithms by inspection.

Example 3 Evaluating Logarithms

Find the value of each of the following logarithms.

$\log_{5} 25$
$\log_{2} 16$
$\log_{1 / 3} 9$
$\log_{7} 7$
$\log_{6} 1$
$\log_{4} \frac{1}{2}$

Solution

Logarithmic Form	Exponential Form	Value
$\log_{5} 25 = y$	$25 = 5^{y} or 5^{2} = 5^{y}$	$y = 2$
$\log_{2} 16 = y$	$16 = 2^{y} or 2^{4} = 2^{y}$	$y = 4$
$\log_{1 / 3} 9 = y$	$9 = {(\frac{1}{3})}^{y} or 3^{2} = 3^{- y}$	$y = - 2$
$\log_{7} 7 = y$	$7 = 7^{y} or 7^{1} = 7^{y}$	$y = 1$
$\log_{6} 1 = y$	$1 = 6^{y} or 6^{0} = 6^{y}$	$y = 0$
$\log_{4} \frac{1}{2} = y$	$\frac{1}{2} = 4^{y} or 2^{- 1} = 2^{2 y}$	$y = - \frac{1}{2}$

Practice Problem 3

Evaluate.
1. $\log_{3} 9$
2. $\log_{9} \frac{1}{3}$
3. $\log_{1 / 2} 32$

Basic Properties of Logarithms

3 Derive basic properties of logarithmic functions.

We use the definition of the logarithm as an exponent to derive two basic properties for $a > 0$ and $a \neq 1,$

Because $a^{1} = a,$ we have $\log_{a} a = 1 .$
Because $a^{0} = 1,$ we have $\log_{a} 1 = 0 .$

The next two properties follow from the fact that a logarithmic function is the inverse of the appropriate exponential function.

Let $f (x) = a^{x}$ and $f^{- 1} (x) = \log_{a} x;$ then

$x = f^{- 1} (f (x)) = f^{- 1} (a^{x}) = \log_{a} a^{x}$

and
$x = f (f^{- 1} (x)) = f (\log_{a} x) = a^{\log_{a} x} .$

We summarize these basic properties of logarithms.

Example 4 Using Basic Properties of Logarithms

Evaluate.

$\log_{3} 3$
$5^{\log_{5} 7}$

Solution

Because $\log_{a} a = 1$ (property (1)), we have $\log_{3} 3 = 1 .$
Because $a^{\log_{a} x} = x$ (property (4)), we have $5^{\log_{5} 7} = 7 .$

Practice Problem 4

Evaluate.
1. $\log_{5} 1$
2. $\log_{3} 3^{5}$
3. $7^{\log_{7} 5}$

Domains of Logarithmic Functions

4 Find the domains of logarithmic functions.

Because the exponential function $f (x) = a^{x}$ has domain $(- \infty, \infty)$ and range $(0, \infty),$ its inverse function $y = \log_{a} x$ has domain $(0, \infty)$ and range $(- \infty, \infty) .$ Therefore, the logarithms of 0 and of negative numbers are not defined; so expressions such as $\log_{a} (- 2)$ and $\log_{a} (0)$ are meaningless.

Side Note

The $□$ represents any expression.

$\log_{a} □ \overset{}{\to} □ > 0$

Example 5 Finding the Domain

Find the domain.

$f (x) = \log_{3} (2 - x) .$
$g (x) = \log_{2} (\frac{x + 1}{x - 3})$

Solution

Because the domain of a logarithmic function is $(0, \infty),$ the expression $(2 - x)$ must be positive. The domain of f is the set of all real numbers x, where

$\begin{array}{rcl} 2 - x & > & 0 \\ 2 & > & x Solve for x . \\ x & < & 2. \end{array}$

So, the domain of f is $(- \infty, 2) .$
The domain of g consists of all real numbers x for which $\frac{x + 1}{x - 3} > 0 .$ We can solve this inequality using the test-point method (page 151).

Figure 4.10

Solving $\frac{x + 1}{x - 3} > 0$ by the test-point method

From Figure 4.10, we see that $\frac{x + 1}{x - 3} > 0$ on $(- \infty, - 1) \cup (3, \infty) .$ So the domain of g is $(- \infty, - 1) \cup (3, \infty) .$

Practice Problem 5

Find the domain of $f (x) = \log_{10} \sqrt{1 - x} .$

Graphs of Logarithmic Functions

5 Graph logarithmic functions.

We now look at other properties of logarithmic functions and begin with an example showing how to graph a logarithmic function.

Example 6 Sketching a Graph

Sketch the graph of $y = \log_{3} x .$

Plotting points (Method 1) To find selected ordered pairs on the graph of $y = \log_{3} x,$ we choose the x-values to be powers of 3. We can easily compute the logarithms of these values by using Property 3, namely, $\log_{3} 3^{x} = x .$ We compute the y-values in Table 4.6.

Table 4.6

`x`	$y = \log_{3} x$	(`x`, `y`)
$\frac{1}{27} = 3^{- 3}$	$y = \log_{3} \frac{1}{27} = \log_{3} 3^{- 3} = - 3$	$(\frac{1}{27}, - 3)$
$\frac{1}{9} = 3^{- 2}$	$y = \log_{3} \frac{1}{9} = \log_{3} 3^{- 2} = - 2$	$(\frac{1}{9}, - 2)$
$\frac{1}{3} = 3^{- 1}$	$y = \log_{3} \frac{1}{3} = \log_{3} 3^{- 1} = - 1$	$(\frac{1}{3}, - 1)$
$1 = 3^{0}$	$y = \log_{3} 1 = 0$	(1, 0)	Property (2)
$3 = 3^{1}$	$y = \log_{3} 3 = 1$	(3, 1)	Property (3)
$9 = 3^{2}$	$y = \log_{3} 9 = \log_{3} 3^{2} = 2$	(9, 2)

Plotting these ordered pairs and connecting them with a smooth curve gives us the graph of $y = \log_{3} x,$ shown in Figure 4.11.

Figure 4.11

Graph $y = \log_{3} x$ by plotting points

Using the inverse function (Method 2) Because $y = \log_{3} x$ is the inverse of the function $y = 3^{x},$ we first graph the exponential function $y = 3^{x},$ then reflect the graph of $y = 3^{x}$ about the line $y = x .$ Both graphs are shown in Figure 4.12. Note that the horizontal asymptote $y = 0$ of the graph of $y = 3^{x}$ is reflected as the vertical asymptote $x = 0$ for the graph of $y = \log_{3} x$ .

Figure 4.12

Graph $y = \log_{3} x$ by using the inverse function $y = 3^{x}$

Practice Problem 6

Sketch the graph of

$y = \log_{2} x .$

We can graph $y = \log_{1 / 3} x$ either by plotting points or by using the graph of its inverse. See Figure 4.13. The graph of $y = \log_{3} x$ shown in Figure 4.12 is rising. The graph of $y = \log_{1 / 3} x$ shown in Figure 4.13, on the other hand, is falling.

Figure 4.13

Graph of $y = \log_{1 / 3} x$ by using its inverse $y = {(\frac{1}{3})}^{x}$

In general, graphs of logarithmic functions have two basic shapes, determined by the base a. See Figure 4.14.

Figure 4.14

Basic shapes of $y = \log_{a} x$

Next, we summarize important properties of the exponential and logarithmic functions.

Properties of Exponential and Logarithmic Functions

Exponential Function $f (x) = a^{x}$	Logarithmic Function $f (x) = \log_{a} x$
1. The domain is $(- \infty, \infty),$ and the range is $(0, \infty) .$	The domain is $(0, \infty),$ and the range is $(- \infty, \infty) .$
2. The `y`-intercept is 1, and there is no `x`-intercept.	The `x`-intercept is 1, and there is no `y`-intercept.
3. The `x`-axis $(y = 0)$ is the horizontal asymptote.	The `y`-axis $(x = 0)$ is the vertical asymptote.
4. The function is one-to-one; that is, $a^{u} = a^{v}$ if and only if $u = v .$	The function is one-to-one; that is, $\log_{a} u = \log_{a} v$ if and only if $u = v .$
5. If $a > 1$ , $f (x) = a^{x}$ is an increasing function; $f (x) \to \infty$ as $x \to \infty$ and $f (x) \to 0$ as $x \to - \infty .$ If $0 < a < 1$ , $f (x) = a^{x}$ is a decreasing function; $f (x) \to 0$ as $x \to \infty$ and $f (x) \to \infty$ as $x \to - \infty .$	If $a > 1$ , $f (x) = \log_{a} x$ is an increasing function; $f (x) \to \infty$ as $x \to \infty$ and $f (x) \to - \infty$ as $x \to 0^{+} .$ If $0 < a < 1$ , $f (x) = \log_{a} x$ is a decreasing function; $f (x) \to - \infty$ as $x \to \infty$ and $f (x) \to \infty$ as $x \to 0^{+} .$

As $x \to \infty,$ it is important to compare the end behavior of the exponential, logarithmic, and linear functions (the symbol $<<$ indicates the different magnitude of growth):

$\begin{array}{c} Logarithmic function & << Linear function & << Exponential function . \\ (base a > 1) & (slope m > 0) & (base a > 1) \end{array}$

A linear function (with slope $m > 0$ ) represents steady growth. A logarithmic function (with base $a > 1$ ) represents diminishing growth and grows extremely slowly for very large values of x, with growth much slower than that of any linear function. In contrast, an exponential function (with base $a > 1$ ) represents extended growth and grows extremely fast for very large values of x, with growth much faster than that of any linear function.

Side Note

Slow growth of a logarithmic function:

$\log (1, 000, 000) = 6$

Fast growth of an exponential function:

$10^{6} = 1, 000, 000$

Once we know the shape of a logarithmic graph, we can shift it vertically or horizontally, stretch it, compress it, check answers with it, and interpret the graph.

To expedite the process of sketching the transformations on the graph of the logarithmic function $f (x) = \log_{a} x$ , we focus on tacking the movement of the vertical asymptote and the movement of the three points $(\frac{1}{a}, - 1), (1, 0)$ , and (a, 1) (see Figure 4.15, where points A, B, and C represent the final position of these three points, respectively, and the vertical asymptote is shifted to the right). The final position of the vertical asymptote will determine the domain of the function produced by the sequence of transformations. If the final graph is on the right of the vertical asymptote $x = k$ the domain is $(k, \infty),$ and if the final graph is on the left of the vertical asymptote $x = k$ the domain is $(- \infty, k)$ . Note that vertical shifts do not affect the location of the vertical asymptote.

Figure 4.15

Characteristic features of the logarithmic functions

The transformations on the graph of $y = \log_{a} x$ are summarized in Table 4.7.

Table 4.7 Transformations on the graph of $y = \log_{a} x$

Transformation (in each case $c > 0$ )	Change the graph point (`x`, `y`) on $y = \log_{a} x$ to	Description
Vertical shift $y = \log_{a} x + c$ $y = \log_{a} x - c$	$(x, y + c)$ $(x, y - c)$	Shift the graph of $y = \log_{a} x$ up `c` units. Shift the graph of $y = \log_{a} x$ down `c` units.
Horizontal shift $y = \log_{a} (x - c)$ $y = \log_{a} (x + c)$	$(x + c, y)$ $(x - c, y)$	Shift the graph of $y = \log_{a} x$ to the right `c` units. Vertical asymptote: $x = c .$ Shift the graph of $y = \log_{a} x$ to the left `c` units. Vertical asymptote: $x = - c .$
Reflection $y = - \log_{a} x$ $y = \log_{a} (- x)$	$(x, - y)$ $(- x, y)$	Reflect the graph of $y = \log_{a} x$ about the `x`-axis. Reflect the graph of $y = \log_{a} x$ about the `y`-axis.
Vertical stretching or compressing $y = c \log_{a} x$	(`x`, `cy`)	Vertically stretch the graph of $y = \log_{a} x$ if $c > 1 .$ Vertically compress the graph of $y = \log_{a} x$ if $0 < c < 1 .$
Horizontal stretching or compressing $y = \log_{a} (b x)$	$(\frac{x}{b}, y)$	Horizontally compress the graph of $y = \log_{a} x$ if $b > 1 .$ Horizontally stretch the graph of $y = \log_{a} x$ if $0 < b < 1$ .

Transformation (in each case

$c > 0$ )

Change the graph point (x, y) on

$y = \log_{a} x$ to

Description

Vertical shift

$y = \log_{a} x + c$

$y = \log_{a} x - c$

$(x, y + c)$

$(x, y - c)$

Shift the graph of $y = \log_{a} x$ up c units.

Shift the graph of $y = \log_{a} x$ down c units.

Horizontal shift

$y = \log_{a} (x - c)$

$y = \log_{a} (x + c)$

$(x + c, y)$

$(x - c, y)$

Shift the graph of $y = \log_{a} x$ to the right c units.

Vertical asymptote: $x = c .$

Shift the graph of $y = \log_{a} x$ to the left c units.

Vertical asymptote: $x = - c .$

Reflection

$y = - \log_{a} x$

$y = \log_{a} (- x)$

$(x, - y)$

$(- x, y)$

Reflect the graph of $y = \log_{a} x$ about the x-axis.

Reflect the graph of $y = \log_{a} x$ about the y-axis.

Vertical stretching or compressing

$y = c \log_{a} x$

(x, cy)

Vertically stretch the graph of $y = \log_{a} x$ if $c > 1 .$

Vertically compress the graph of $y = \log_{a} x$ if $0 < c < 1 .$

Horizontal stretching or compressing

$y = \log_{a} (b x)$

$(\frac{x}{b}, y)$

Horizontally compress the graph of $y = \log_{a} x$ if $b > 1 .$

Horizontally stretch the graph of $y = \log_{a} x$ if $0 < b < 1$ .

Example 7 Using Transformations on $y = \log_{3} x$

Start with the graph of $y = \log_{3} x$ and use transformations to sketch the graph of each function.

$f (x) = \log_{3} x + 2$
$f (x) = \log_{3} (x - 1)$
$f (x) = - \log_{3} x$
$f (x) = \log_{3} (- x)$

State the domain and range and the vertical asymptote for the graph of each function.

Solution

We start with the graph of $y = \log_{3} x$ and use the transformations shown in Figure 4.16.

Figure 4.16

Transformations on $y = \log_{3} x$

Practice Problem 7

Use transformations on the graph $f (x) = \log_{2} x$ to sketch the graph of

$y = - \log_{2} (x - 3) .$

Example 8 Using Transformations on Logarithmic Functions

Sketch the graph of $f (x) = 2 + \log_{2} (x - 1) .$ State the domain, range, and the vertical asymptote of f.

Solution

Figure 4.17 shows a sequence of transformations on the graph of $y = \log_{2} x$ to produce the graph of f.

Figure 4.17

Graphing $f (x) = 2 + \log_{2} (x - 1)$

From the graph of f in Figure 4.17(c), we see that the domain of f is $(1, \infty),$ its range is $(- \infty, \infty),$ and the vertical asymptote is the line $x = 1 .$

Practice Problem 8

Sketch the graph of $f (x) = 3 - \log_{2} x .$

Common Logarithm

The logarithm with base 10 is called the common logarithm and is denoted by omitting the base, so

$\log x = \log_{10} x .$

Applying the basic properties of logarithms (see page 447) to common logarithms, we have the following:

$\log 10 = 1$
$\log 1 = 0$
$\log 10^{x} = x$ , x any real number
$10^{\log x} = x, x > 0$

We use property (3) to evaluate the common logarithms of numbers that are powers of 10. For example,

$\begin{array}{rcl} \log 1000 & = & \log 10^{3} = 3 \\ and \log 0.01 & = & \log 10^{- 2} = - 2. \end{array}$

We use a calculator to find the common logarithms of numbers that are not powers of 10 by pressing the LOG key.

The graph of $y = \log x$ is similar to the graph in Figure 4.14(a) because the base for log x is $10 > 1$ .

Example 9 Benford’s Law

Benford’s Law states that the chances of the digit m occurring as the first decimal digit (from 1 to 9) from many real-life data sources is

$P_{m} = \log (m + 1) - \log m .$

This means that about $100 P_{m} %$ of the data can be expected to have m as the first digit.

Find $P_{1} .$ Interpret your result.
Use part a to decide which column (size I or size II) represents the actual data for the size of countries listed on page 444.

Solution

$\begin{array}{rcll} P_{m} & = & \log (m + 1) - \log m & Benford' s Law \\ P_{1} & = & \log 2 - \log 1 & m = 1 \\ = & \log 2 & \log 1 = \log_{10} 1 = 0 \\ \approx & 0.301 & Use a calculator . \end{array}$

This means that about 30.1% of the data is expected to have 1 as the first digit.
We count the number of countries having 1 as the first digit in the columns with headings Size I and Size II. We find the following:

Number of countries with 1 as the first digit:

$\begin{array}{rcl} Size I column & = & 1 \\ Size II column & = & 16 \end{array}$

30% of 50 $countries = (0.30) (50) = 15 .$ So according to Benford’s Law, size II column represents the actual data.

Practice Problem 9

Find $P_{2}$ and interpret your result.

Natural Logarithm

In most applications in calculus and the sciences, the convenient base for logarithms is the number e. The logarithm with base e is called the natural logarithm and is denoted by ln x (read “ell en x”) so that

$\ln x = \log_{e} x .$

Applying the basic properties of logarithms (see page 447) to natural logarithms, we have the following:

$\ln e = 1$
$\ln 1 = 0$
${\ln e}^{x} = x$ , x any real number
$e^{\ln x} = x, x > 0$

We can use property (3) to evaluate the natural logarithms of powers of e.

Example 10 Evaluating the Natural Logarithm Function

Evaluate each expression.

${\ln e}^{4}$
$\ln \frac{1}{e^{2.5}}$
ln 3

Solution

$\ln e^{4} = 4 Property (3)$
$\ln \frac{1}{e^{2.5}} = \ln e^{- 2.5} = - 2.5 Property (3)$
$\ln 3 \approx 1.0986123 Use a calculator .$

Practice Problem 10

Evaluate each expression.
1. $\ln \frac{1}{e}$
2. ln 2

Summary of Basic Properties of Logarithms

Base $a > 0, a \neq 1$	Base 10	Base `e`
1. $\log_{a} a = 1$	$\log 10 = 1$	$\ln e = 1$
2. $\log_{a} 1 = 0$	$\log 1 = 0$	$\ln 1 = 0$
3. $\log_{a} a^{x} = x$	$\log 10^{x} = x$	${\ln e}^{x} = x$
4. $a^{\log_{a} x} = x$	$10^{\log x} = x$	$e^{\ln x} = x$

Investments

6 Use natural logarithms in applications.

We recall (page 435), the continuous compound interest formula:

$A = P e^{r t}$ (5)

We express equation (5) in logarithmic form:

$\begin{array}{rcll} \frac{A}{P} & = & e^{r t} & Divide both sides of (5) by P . \\ \ln \frac{A}{P} & = & r t & Logarithmic form \end{array}$

Continuous Compound Interest

Exponential Form	Logarithmic Form
$A = P e^{r t}$	$\ln \frac{A}{P} = r t$

The exponential form is used when we need to find A or P; logarithmic form is useful in calculating r or t.

Example 11 Doubling Your Money

How long will it take to double your money if it earns 6.5% compounded continuously?
At what rate of return, compounded continuously, would your money double in 5 years?

Solution

If P dollars is invested and you want to double it, then the final amount $A = 2 P .$

$\begin{array}{rcll} \ln (\frac{A}{P}) & = & r t & Continuous compounding formula, logarithmic form \\ \ln (\frac{2 P}{P}) & = & 0.065 t & A = 2 P, r = 0.065 \\ \ln 2 & = & 0.065 t & \frac{2 P}{P} = 2 \\ \frac{\ln 2}{0.065} & = & t & Divide both sides by 0.065. \\ t & \approx & 10.66 & Use a calculator . \end{array}$

It will take approximately 11 years to double your money.
$\begin{array}{rcll} \ln (\frac{A}{P}) & = & r t & Continuous compounding formula, logarithmic form \\ \ln (\frac{2 P}{P}) & = & r (5) & A = 2 P, t = 5 \\ \ln 2 & = & 5 r & \frac{2 P}{P} = 2 \\ \frac{\ln 2}{5} & = & r & Divide both sides by 5. \\ r & \approx & 0.1386 & Use a calculator . \end{array}$

Your investment will double in 5 years at the approximate rate of 13.86%.

Practice Problem 11

Repeat Example 11 for tripling $(A = 3 P)$ your money.

Newton’s Law of Cooling

When a cool drink is removed from a refrigerator, the drink warms to the temperature of the room. When removed from the oven, a pizza baked at a high temperature cools to the temperature of the room.

In situations such as these, the rate at which an object’s temperature changes at any given time is proportional to the difference between its temperature and the temperature of the surrounding medium. This observation is called Newton’s Law of Cooling, although, as in the case of a cool drink, it applies to warming as well.

Newton’s Law of Cooling states that

$T (t) = T_{s} + (T_{0} - T_{s}) e^{- k t}$ (6)

where T(t) is the temperature of the object at time t, $T_{s}$ is the surrounding temperature, $T_{0}$ is the value of T(t) at $t = 0$ , and k is a positive constant that depends on the object.

We can express equation (6) in logarithmic form:

$\begin{array}{rcll} T (t) - T_{s} & = & (T_{0} - T_{s}) e^{- k t} & Subtract T_{s} from both sides of equation (6) . \\ \frac{T (t) - T_{s}}{T_{0} - T_{s}} & = & e^{- k t} & Divide both sides by T_{0} - T_{s} . \\ \ln (\frac{T (t) - T_{s}}{T_{0} - T_{s}}) & = & - k t & Logarithmic form \end{array}$

Newton’s Law of Cooling

Exponential Form	Logarithmic Form
$T (t) = T_{s} + (T_{0} - T_{s}) e^{- k t}$	$\ln (\frac{T (t) - T_{s}}{T_{0} - T_{s}}) = - k t$

Example 12 McDonald’s Hot Coffee

The local McDonald’s franchise has discovered that when coffee is poured from a coffeemaker whose contents are $180 ° F$ into a noninsulated pot, after 1 minute the coffee cools to $165 ° F$ if the room temperature is $72 ° F .$ How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at $125 ° F ?$

Solution

From the given data, we have $T = 165$ when $t = 1, T_{0} = 180,$ and $T_{s} = 72 .$

$\begin{array}{rcl} \ln (\frac{165 - 72}{180 - 72}) = \ln (\frac{93}{108}) & = & - k \cdot 1 & \begin{array}{l} Substitute given data in the logarithmic \\ form . \end{array} \\ k & = & - \ln (\frac{93}{108}) & Solve for k . \end{array}$

With this value of k, we find t when $T = 125 .$

$\begin{array}{rcll} \ln (\frac{125 - 72}{180 - 72}) = \ln (\frac{53}{108}) & = & - (- \ln (\frac{93}{108})) \cdot t & \begin{array}{l} Substitute data in logarithmic \\ form . \end{array} \\ [\frac{1}{\ln (\frac{93}{108})}] \ln (\frac{53}{108}) & = & t & Solve for t . \\ t & \approx & 4.76 & Use a calculator . \end{array}$

The employees should wait approximately 4.76 minutes (realistically, about 5 minutes) to deliver the coffee at $125 ° F$ to the customers.

Practice Problem 12

Repeat Example 12 , assuming that the coffee is to be delivered to the customers at $120 ° F .$

Section 4.2 Exercises

Concepts and Vocabulary

The domain of the function $y = \log_{a} x$ is , and its range is .
The logarithmic form $y = \log_{a} x$ is equivalent to the exponential form .
The logarithm with base 10 is called the logarithm, and the logarithm with base e is called the natural logarithm.
$a^{\log_{a} x} = \underline{}$ , and $\log_{a} a^{x} = \underline{}$ .
True or False. The graph of $y = \log_{a} x, a > 0, a \neq 1,$ is the graph of an increasing function.
True or False. The graph of $y = \log_{a} x, a > 0,$ and $a \neq 1,$ has no horizontal asymptote.
True or False. The domain of $f (x) = \log_{a} x, a > 0,$ and $a \neq 1$ is $(- \infty, \infty)$ .
True or False. If $y = \log_{a} x,$ then $x = y$ .

Building Skills

In Exercises 9–20, write each exponential equation in logarithmic form.

$5^{2} = 25$
${(49)}^{- 1 / 2} = \frac{1}{7}$
${(\frac{1}{16})}^{- 1 / 2} = 4$
${(a^{2})}^{2} = a^{4}$
$10^{0} = 1$
$10^{4} = 10, 000$
${(10)}^{- 1} = 0.1$
$3^{x} = 5$
$a^{2} + 2 = 7$
$a^{e} = π$
$2 a^{3} - 3 = 10$
$5 \cdot 2^{c t} = 11$

In Exercises 21–32, write each logarithmic equation in exponential form.

$\log_{2} 32 = 5$
$\log_{7} 49 = 2$
$\log_{10} 100 = 2$
$\log_{10} 10 = 1$
$\log_{10} 1 = 0$
$\log_{a} 1 = 0$
$\log_{10} 0.01 = - 2$
$\log_{1 / 5} 5 = - 1$
$3 \log_{8} 2 = 1$
$1 + \log 1000 = 4$
$\ln 2 = x$
$\ln π = a$

In Exercises 33–42, evaluate each expression without using a calculator.

$\log_{5} 125$
$\log_{9} 81$
log 10,000
$\log_{3} \frac{1}{3}$
$\log_{2} \frac{1}{8}$
$\log_{4} \frac{1}{64}$
$\log_{3} \sqrt{27}$
$\log_{27} 3$
$\log_{16} 2$
$\log_{5} \sqrt{125}$

In Exercises 43–54, evaluate each expression.

$\log_{3} 1$
$\log_{1 / 2} 1$
$\log_{7} 7$
$\log_{1 / 9} \frac{1}{9}$
$\log_{6} 6^{7}$
$\log_{1 / 2} {(\frac{1}{2})}^{5}$
$3^{\log_{3}}^{5}$
$7^{\log_{7} \frac{1}{2}}$
$2^{\log_{2} 7} + \log_{5} 5^{- 3}$
$3^{\log_{3} 5} - \log_{2} 2^{- 3}$
$4^{\log_{4} 6} - \log_{4} 4^{- 2}$
$10^{\log x} - e^{\ln y}$

In Exercises 55–70, find the domain of each function.

$f (x) = \log_{2} (x + 1)$
$g (x) = \log_{3} (x - 8)$
$f (x) = \log_{3} \sqrt{x - 1}$
$g (x) = \log_{4} \sqrt{3 - x}$
$f (x) = \log (x - 2) + \log (2 x - 1)$
$g (x) = \ln \sqrt{x + 5} - \ln (x + 1)$
$h (x) = \ln (x - 1) / \ln (2 - x)$
$f (x) = \ln (x - 3) + \ln (2 - x)$
$f (x) = \ln | x |$
$f (x) = \sqrt{\ln x}$
$f (x) = \log_{2} (\frac{x}{3} - 4)$
$f (x) = \log_{3} (2 - \frac{x}{5})$
$f (x) = \ln (\frac{x}{x + 1})$
$f (x) = \ln (\frac{x - 2}{x})$
$f (x) = \log_{3} (\frac{x - 2}{x + 1})$
$f (x) = \log_{2} (\frac{x + 3}{x - 2})$
Match each logarithmic function with one of the graphs labeled a–f.
1. $f (x) = \log x$
2. $f (x) = - \log | x |$
3. $f (x) = - \log (- x)$
4. $f (x) = \log (x - 1)$
5. $y = (\log x) - 1$
6. $f (x) = \log (- 1 - x)$
For each given graph, find a function of the form $f (x) = \log_{a} x$ that represents the graph.

In Exercises 73–84, graph the given function by using transformations on the appropriate basic graph of the form $y = \log_{a} x .$ State the domain and range of the function and the vertical asymptote of the graph.

$f (x) = \log_{4} (x + 3)$
$f (x) = \log_{3} (x - 2)$
$f (x) = \log_{1 / 2} (x - 1)$
$f (x) = \log_{1 / 3} (x + 2)$
$f (x) = - \log_{5} x$
$f (x) = \log_{1 / 5} (- x)$
$f (x) = 2 - \log_{4} (x - 1)$
$f (x) = 1 + \log_{3} (x + 2)$
$f (x) = 1 + \log_{1 / 5} (- x)$
$f (x) = 3 - \log_{1 / 2} x$
$f (x) = | \log_{3} x |$
$f (x) = \log_{3} | x |$

In Exercises 85–92, begin with the graph of $f (x) = \log_{2} x$ and use transformations to sketch the graph of each function. Find the domain and range of the function and the vertical asymptote of the graph.

$f (x) = \log_{2} (x - 1)$
$f (x) = \log_{2} (x + 2)$
$f (x) = \log_{2} (3 - x)$
$f (x) = - \log_{2} (x + 1)$
$f (x) = - \log_{2} x - 2$
$f (x) = \log_{2} (- x) - 1$
$f (x) = 2 + \log_{2} (3 - x)$
$f (x) = 4 - \log_{2} (3 - x)$

In Exercises 93–98, evaluate each expression.

$\log_{4} (\log_{3} 81)$
$\log_{4} [\log_{3} (\log_{2} 8)]$
$\log_{\sqrt{2}} 2$
$\log_{2 \sqrt{2}} 8$
$\log_{\sqrt{2}} 4$
$\log_{\sqrt{3}} 27$

In Exercises 99–106, begin with the graph of $y = \ln x$ and use transformations to sketch the graph of each function.

$f (x) = \ln (x + 2)$
$f (x) = \ln (x - 2)$
$f (x) = - \ln (x - 1)$
$f (x) = \ln (1 - x)$
$f (x) = 2 + \ln (- x)$
$f (x) = 1 - \ln (1 - x)$
$f (x) = 3 - 2 \ln x$
$f (x) = 1 - 3 \ln x$

In Exercises 107 and 108, determine what sequence of transformations were applied on the graph of $f (x) = \log_{2} x$ to produce the given graph.

Applying the Concepts

The cost of PCs. The graph shows the consumer price index for personal computers during 2003–2017.

Source: U.S. Bureau of Labor Statistics.

The function $f (t) = 221.69 - 66.9$ ln t models these data, where t represents the number of years after 2002. So 2003 corresponds to $t = 1$ .
1. Use this function to estimate the consumer price index for personal computers in 2008.
2. Use this function to estimate what year the consumer price index will reach a value of 30.
3. What trend does the graph follow?
Life expectancy. The graph shows the relation between life expectancy and the gross domestic product (GDP) per capita based on World Bank data for 2014. This empirical logarithmic relationship was first described by Samuel H. Preston in 1974.

Source: World Bank.

The function $f (x) = 11.066 + 6.587$ ln x models these data, where x represents GDP per capita expressed in U.S. dollars.
1. Use this function to estimate the life expectancy for a country with GDP per capita equal to $25,000.
2. Use this function to estimate the GDP per capita needed for the life expectancy of 70 years.
3. What trend does the graph follow?

In Exercises 111–124, use the model $A = A_{0} e^{k t}$ or $\ln \frac{A}{A_{0}} = k t .$

Doubling your money. How long would it take to double your money if you invested P dollars at the rate of 8% compounded continuously?
Investment goal. How long would it take to grow your investment from $10,000 to $120,000 at the rate of 10% compounded continuously?
Rate for doubling your money. At what annual rate of return, compounded continuously, would your investment double in six years?
Rate-of-investment goal. At what annual rate of return, compounded continuously, would your investment grow from $8000 to $50,000 in 25 years?
Population of the United States. The U.S. population in 2010 was 308 million. The U.S. Census Bureau reported that the population grew 9.7% since 2000 (the slowest growth rate since the depression).
1. What was the population of the United States in 2000?
2. What was the annual rate of growth between 2000 and 2010?
3. Assuming the same rate of growth, determine the year when the population of the United States will be 400 million.
Population of Canada. The population of Canada in 2010 was 35 million, and it grew 12.33% since 2000.
1. What was the population of Canada in 2000?
2. What was the annual rate of growth between 2000 and 2010?
3. Assuming the same rate of growth, determine the time from 2010 until Canada’s population doubles.
4. Find the time from 2010 until Canada’s population reaches 50 million.
Newton’s Law of Cooling. A thermometer is taken from a room at $75 ° F$ to the outdoors, where the temperature is $20 ° F .$ The reading on the thermometer drops to $50 ° F$ after one minute.
1. Find the reading on the thermometer after
  1. Five minutes.
  2. Ten minutes.
  3. One hour.
2. How long will it take for the reading to drop to $22 ° F ?$
Newton’s Law of Cooling. The last bit of ice in a picnic cooler has melted $(T_{0} = 32 ° F) .$ The temperature in the park is $85 ° F .$ After 30 minutes, the temperature in the cooler is $40 ° F .$ How long will it take for the temperature inside the cooler to reach $50 ° F ?$
Cooking salmon. A salmon filet initially at $50 ° F$ is cooked in an oven at a constant temperature of $400 ° F .$ After ten minutes, the temperature of the filet rises to $160 ° F .$ How long does it take until the salmon is cooked at $220 ° F ?$
The time of murder. A forensic specialist took the temperature of a victim’s body lying in a street at 2:10 a.m. and found it to be $85.7 ° F .$ At 2:40 a.m., the temperature of the body was $84.8 ° F .$ When was the murder committed if the air temperature during the night was $55 ° F ?$ [Remember, normal body temperature is $98.6 ° F .$ ]
Water contamination. A chemical is spilled into a reservoir of pure water. The concentration of chemical in the contaminated water is 4%. In one month, 20% of the water in the reservoir is replaced with clean water.
1. What will be the concentration of the contaminant one year from now?
2. For water to be safe for drinking, the concentration of this contaminant cannot exceed 0.01%. How long will it be before the water is safe for drinking?
Toxic chemicals in a lake. In a lake, one-fourth of the water is replaced by clean water every year. Sixteen thousand cubic meters of soluble toxic chemical spill takes place in the lake. Let T(n) represent the amount of toxin left after n years.
1. Find a formula for T(n).
2. How much toxin will be left after 12 years?
3. When will 80% of the toxin be eliminated?
Sheep population. A herd of sheep doubles in size every three years. There are now 1500 sheep in the herd.
1. Find an exponential function of the form $P = P_{0} e^{r t}$ to model the herd’s growth.
2. How many sheep will be in the herd seven years from now (round your answer)?
3. In how many years will the herd have 15,000 sheep?
Shark population. A school of sharks is losing one-ninth of its population every two years. The colony currently has 150 sharks.
1. Find an exponential function of the form $P = P_{0} e^{- r t}$ to model the school’s decline.
2. How many sharks will be in the school 4 years from now?
3. In how many years (to the nearest tenth) will the colony have 35 sharks?
Benford’s Law. The law states that the probability that the first decimal-digit of a raw data sample (from 1 to 9) is given by $P_{m} = \log (m + 1) - \log m .$ That is, about $(100 P_{m}) %$ of the data can be expected to have m as the first digit.
1. What percent of the data can be expected to have 3 as the first digit?
2. Find $P_{1} + P_{2} + \dots_{} + P_{9} .$ Interpret your result.
Prime Number Theorem. A natural number $p \geq 2$ is prime if the only positive divisors of p are 1 and p. An important topic in number theory involves counting prime numbers {2, 3, 5, 7, 11, 13, …}. For any natural number $n \geq 2,$ the quantity of prime numbers less than or equal to n is denoted by $π (n)$ (read “pie of n”). For example, $π (2) = 1, π (3) = 2, π (4) = 2, π (5) = 3,$ and $π (6) = 3 .$ The Prime Number Theorem states that for large $n, π (n)$ can be approximated by $\frac{n}{\ln (n)} .$
1. Find $π (20)$ by listing the primes that are less than or equal to 20.
2. Find $π (50)$ by listing primes and compare your answer with $\frac{50}{\ln (50)} .$
3. Estimate $π (1, 000, 000)$ by using the Prime Number Theorem.

Beyond the Basics

Finding the domain. Find the domain of each function.
1. $f (x) = \log_{2} (\log_{3} x)$
2. $f (x) = \log (\ln (x - 1))$
3. $f (x) = \ln (\log (x - 1))$
4. $f (x) = \log (\log (\log (x - 1)))$
Finding the inverse. Find the inverse of each function in Exercise 127.

In Exercises 129 and 130, describe an order for the sequence of transformations on the graph of $y = \log x$ to produce the graph of the given equation.

$y = - 3 \log (\frac{1}{2} x - 2) + 4$
$y = - 4 \log (2 - 3 x) + 1$
Present value of an investment. Recall that if P dollars is invested in an account at an interest rate r compounded continuously, then the amount A (called the future value of P) in the account t years from now will be $A = P e^{r t} .$ Solving the equation for P, we get $P = A e^{- r t} .$ In this formulation, P is called the present value of the investment.
1. Find the present value of $100,000 at 7% compounded continuously for 20 years.
2. Find the interest rate r compounded continuously that is needed to have $50,000 be the present value of $75,000 in ten years.
Your uncle is 40 years old, and he wants to have an annual pension of $50,000 each year at age 65. What is the present value of his pension if the money can be invested at all times at a continuously compounded interest rate of
1. 5%?
2. 8%?
3. 10%?

Critical Thinking / Discussion / Writing

In Exercises 133 and 134, evaluate each expression without using a calculator.

$2^{\log_{2} 3} - 3^{\log_{3} 2}$
${(\log_{3} 4 + \log_{2} 9)}^{2} - {(\log_{3} 4 - \log_{2} 9)}^{2}$
Solve for $x : \log_{3} [\log_{4} (\log_{2} x)] = 0$
Write as a piecewise function.
1. $f (x) = | \log x |$
2. $g (x) = | \ln (x - 1) | + | \ln (x - 2) |$
Inequalities involving logarithms.
1. If a is positive, is the statement “ $a < b$ if and only if $\log a < \log b$ ” always true?
2. What property of logarithmic functions is used whenever the statement in part (a) is true?
Write a summary of the types of graphs (with sketches) for the logarithmic functions of the form $y = c \log_{a} x, c \neq 0, a > 0,$ and $a \neq 1 .$

Getting Ready for the Next Section

In Exercises 139–144, write each expression in the form $a^{n}$ where n is a positive integer.

$a^{2} \cdot a^{7}$
${(a^{2})}^{3}$
$\sqrt{a^{8}}$
$\sqrt[3]{a^{6}}$
${(\frac{243}{32})}^{4 / 5}$
$\sqrt[5]{{(32)}^{- 3}}$

In Exercises 145 and 146, perform the indicated operations and write the answer in scientific notation.

$(4.7 \times 10^{7}) (8.1 \times 10^{5})$
$\frac{7.2 \times 10^{6}}{2.4 \times 10^{- 3}}$

In Exercises 147–150, verify that each equation is true by evaluating each side without using a calculator.

$\log_{3} 81 = \log_{3} 3 + \log_{3} 27$
$\log_{2} 8 = \log_{2} 128 - \log_{2} 16$
$\log_{2} 16 = 2 \log_{2} 4$
$\log_{4} 64 = \frac{\log_{2} 64}{\log_{2} 4}$

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Table of Contents for Section 4.2 Logarithmic Functions

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Table of Contents for
Section 4.2 Logarithmic Functions