5.2 Logarithmic Functions

Logarithms and Exponents

Before we look at logarithmic functions, let's quickly review exponents and logs. For a particular base, let's say 5, taking a logarithm is the opposite operation for raising to a power. For example, if we raise base 5 to a power of $2$ , we get

$5^{2} = 25 and thus \log_{5} (25) = 2$

We can say that a logarithm is actually an exponent. Asking for the log base 5 of 25 is asking "what power of 5, or what exponent on base 5 will give me 25?"

For a more thorough review of logarithms you can refer to Section 4.3.1 .

Which of these could you use to estimate the value of $log sub 5 ⁡ 378$ ?

_____

step1

Which of these could you use to estimate the value of $log sub 5 ⁡ open parenthesis 378 close parenthesis$ ?

Find multiples of 5.
Find the fifth root of 378.
Find powers of 5.
Divide 378 by 5.

Now we'll consider functions defined in terms of logarithms, or logarithmic functions. For example,

$f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$

is a logarithmic function. In order to understand logarithmic functions better, we first investigate how they are related to more familiar functions, the exponential functions.

Inverse of the Exponential Function

Inverse functions are really a generalization of inverse operations. For example, raising to the $n$ th power and taking $n$ th roots are inverse operations. In fact, we use the following rule to define cube roots:

$\sqrt[3]{b} = a if and only if a^{3} = b$

Compare this rule to the definition of inverse functions from Inverse Functions.

In this case, $g (x) = \sqrt[3]{x}$ and $f (x) = x^{3}$ , and the equations above tell us that the two functions $f (x) = x^{3}$ and $g (x) = \sqrt[3]{x}$ are inverse functions.

In Exponential Functions, we saw that a similar rule relates the operations of raising a base $b$ to a power and taking a base $b$ logarithm, because they are inverse operations.

We can now define the logarithmic function, $g (x) = \log_{b} (x)$ , that takes the log base $b$ of its input values. The conversion formulas tell us that the log function, $g (x) = \log_{b} (x)$ , is the inverse of the exponential function, $f (x) = b^{x}$ .

Graphs of Logarithmic Functions

What does the graph of a log function look like? We can use exponential functions to help us.

We can obtain a table of values for $g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ by making a table for $f (x) = 2^{x}$ and then interchanging the columns, as shown in the tables below. You can see that the graphs of $f (x) = 2^{x}$ and $g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ , shown in the figure, are symmetric about the line $y = x$ .

$x$	$f (x) = 2^{x}$
$minus 2$	$\frac{1}{4}$
$minus 1$	$\frac{1}{2}$
$0$	$1$
$1$	$2$
$2$	$4$

$x$	$g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$
$\frac{1}{4}$	$minus 2$
$\frac{1}{2}$	$minus 1$
$1$	$0$
$2$	$1$
$4$	$2$

The same procedure works for graphing log functions with any base: If we want to find values for the function $y = \log_{b} (x)$ , we can find the values for the exponential function $y = b^{x}$ , and then interchange the $x$ and $y$ values in each ordered pair.

Example 2

Graph the function $f (x) = 10^{x}$ and its inverse $g open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ on the same axes.

We start by making a table of values for the function $f (x) = 10^{x}$ . We can make a table of values for the inverse function, $g open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ , by interchanging the components of each ordered pair in the table for $f$ .

$x$	$f (x)$
$minus 2$	$0.01$
$minus 1$	$0.1$
$0$	$1$
$1$	$10$
$2$	$100$

$x$	$g (x)$
$0.01$	$minus 2$
$0.1$	$minus 1$
$1$	$0$
$10$	$1$
$100$	$2$

We plot each set of points and connect them with smooth curves to obtain the graphs shown above.

Make a table of values and graph the function $h open parenthesis x close parenthesis equals log sub 4 ⁡ x$ .

$x$	$\frac{1}{4}$	$1$	$2$	$4$	$16$
$log sub 4 ⁡ x$	_____	_____	_____	_____	_____

$x$	$\frac{1}{4}$	$1$	$2$	$4$	$16$
$log sub 4 ⁡ x$	$a$	$b$	$c$	$d$	$e$

A graph is below.

Complete the table of values and graph the function $h open parenthesis x close parenthesis equals log sub 4 ⁡ open parenthesis x close parenthesis$ .

$x$	$\frac{1}{4}$	$1$	$2$	$4$	$16$
$log sub 4 ⁡ open parenthesis x close parenthesis$

$x$	$\frac{1}{4}$	$1$	$2$	$4$	$16$
$log sub 4 ⁡ open parenthesis x close parenthesis$	$minus 1$	$0$	$\frac{1}{2}$	$1$	$2$

What is the $y$ -intercept of the graph of $y equals log sub 5 ⁡ x$ ?

_____

There is none: the graph of $y equals log sub 5 ⁡ open parenthesis x close parenthesis$ has no $y$ -intercept.

What is the $y$ -intercept of the graph of $y equals log sub 5 ⁡ open parenthesis x close parenthesis$ ?

$open parenthesis 0, 1 close parenthesis$
$open parenthesis 1, 0 close parenthesis$
$open parenthesis 0, 5 close parenthesis$
There is none.

You can also see that while an exponential growth function increases very rapidly for positive input values, its inverse, the logarithmic function, grows extremely slowly.

In addition, the logarithmic function $y = \log_{b} (x)$ has the following properties.

The domain of the function $g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis$ is

_____

The domain of the function $g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis$ is all positive numbers.

The domain of the function $g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis$ is

all real numbers.
all multiples of 3.
all non-negative numbers.
all positive numbers.

Why does the function $f (x) = \log (x)$ grow so slowly?

_____

Why does the function $f (x) = \log (x)$ grow so slowly?

Modeling with Logarithmic Functions

We can use the LOG key on a calculator to evaluate the function $f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ .

Which statement is true?

_____

We cannot take the log of a negative number.

Which statement is true?

The log of a number is never negative.
We cannot take the log of a negative number.
The log of a fraction is called a common log.
The log of 0 is 1.

The formula $T = \frac{\log (2) \cdot t_{i}}{3 \log (D_{f} / D_{0})}$ is used by X-ray technicians to calculate the doubling time of a malignant tumor. $D sub 0$ is the diameter of the tumor when first detected, $D_{f}$ is its diameter at the next reading, and $t_{i}$ is the time interval between readings, in days. Calculate the doubling time of the following tumor: its diameter when first detected was 1 cm, and 7 days later its diameter was 1.05 cm.

_____ days

33 days

The formula

$T = \frac{\log (2) \cdot t_{i}}{3 \log (D_{f} / D_{0})}$

is used by X-ray technicians to calculate the doubling time of a malignant tumor. $D sub 0$ is the diameter of the tumor when first detected, $D_{f}$ is its diameter at the next reading, and $t_{i}$ is the time interval between readings, in days.

Calculate the doubling time of the following tumor: its diameter when first detected was 1 cm, and 7 days later its diameter was 1.05 cm.

$T = \frac{\log (2) \cdot 7}{3 \log (1.05 / 1)} =$ 33 days

Logarithmic functions are useful for modeling increasing functions that slow down as the input increases.

Example 4

Life expectancy at birth is the average number of years a newborn child is expected to live. In 1900, the average life expectancy at birth in the U.S. was 47.3 years, and in 1910 it had risen to 50.0 years. During rest of the twentieth century, life expectancy was modeled by the formula

$L (x) = 19.13 + 28.34 \log (x)$

where $x$ is the number of years after 1900.

Graph the life expectancy function for the years 1910 to 2000.
The life expectancy in 1950 was 68.2 years. What does the function $L (x)$ predict for life expectancy in 1950?
According to the model, how much did life expectancy increase between 1920 and 1930? How much did it increase between 1990 and 2000?

We can make a table of values for $L (x)$ and plot points to obtain the graph below, which also shows the actual data points for life expectancy for the decades from 1910 to 2000.
$x$ $10$ $20$ $30$ $40$ $50$ $60$ $70$ $80$ $90$ $100$
$L (x)$ $47.5$ $56.1$ $61.1$ $64.6$ $67.4$ $69.6$ $71.5$ $73.1$ $74.6$ $75.9$
We substitute $x equals 50$ into the function to find
$L (50) = 19.13 + 28.34 \log (50) \approx 67.4$
The function predicts a life expectancy of 67.4 years in 1950.
Between 1920 and 1930, life expectancy increased from 56.1 to 61.1, or 5 years. Between 1990 and 2000 it increased from 74.6 to 75.9, or 1.3 years.

In the previous Example, we see that although life expectancy has been increasing over time, it has been slowing down or leveling off. In fact, life expectancy in the US actually declined slightly from 78.94 in 2013 to 78.81 in 2018. (What factors may have contributed to this decline?) By 2024 it had rebounded to 79.25. It remains to be seen how well the model predicts life expectancy in the 21st century.

The CDC (Centers for Disease Control and Prevention) provides Growth Charts for the average height and weight of children from age 2 to 20. The average height of girl children is given in centimeters by

$\begin{matrix} H (t) = 49.29 + 91.3 \log t \end{matrix}$

where $t$ is age in years.

Graph the height function for $2 less than or equal to t less than or equal to 20$ .
Use the height function to complete the table.
$t$ $2$ $5$ $10$ $15$ $20$
$H (t)$ _____ _____ _____ _____ _____
How much is a girl's height expected to increase between the ages of 5 and 10? _____ cm.
Between the ages of 15 and 20? _____ cm.

$t$	$2$	$5$	$10$	$15$	$20$
$H (t)$	_____	_____	_____	_____	_____

A graph is below.
$t$ $2$ $5$ $10$ $15$ $20$
$H (t)$ H0 H1 H2 H3 H4
c1 cm, c2 cm

$t$	$2$	$5$	$10$	$15$	$20$
$H (t)$	H0	H1	H2	H3	H4

Graph for part (a):

$H (t) = 49.29 + 91.3 \log (t)$

where $t$ is age in years.

Graph the height function for $2 less than or equal to t less than or equal to 20$ .

Use the height function to complete the table.

$t$	$2$	$5$	$10$	$15$	$20$
$H (t)$

How much is a girl's height expected to increase between the ages of 5 and 10? Between the ages of 15 and 20?

$t$ $2$ $5$ $10$ $15$ $20$
$H (t)$ $77$ $113$ $141$ $157$ $168$
28 cm, 11 cm

$t$	$2$	$5$	$10$	$15$	$20$
$H (t)$	$77$	$113$	$141$	$157$	$168$

Logarithmic Equations

A logarithmic equation is one in which the variable appears inside of a logarithm. For example,

$log sub 4 ⁡ open parenthesis x close parenthesis equals 3$

is a log equation. To solve a log equation, we can use the conversion equations to rewrite the equation in exponential form.

Example 5

Solve for $x$ .

$2 (\log_{3} (x)) - 1 = 4$
$\log_{10} (2 x + 100) = 3$

We isolate the logarithm, then rewrite the equation in exponential form:
$\begin{aligned} 2 (\log_{3} (x)) & = 5 & Divide both sides by 5. \\ \log_{3} (x) & = \frac{5}{2} & Convert to exponential form. \\ x & = 3^{5 / 2} \end{aligned}$
First, we convert the equation to exponential form.
$2 x + 100 = 10^{3} = 1000$
Now we can solve for $x$ to find $2 x equals 900$ , or $x equals 450$ .

Solve for the unknown value in each equation.

$\log_{b} (2) = \frac{1}{2}$
$b =$ _____
$\log_{3} (2 x - 1) = 4$
$x =$ _____

$b equals 4$
$x equals 41$

Solve for the unknown value in each equation.

$\log_{b} (2) = \frac{1}{2}$
$\log_{3} (2 x - 1) = 4$

$b equals 4$
$x equals 41$

Imagine the graph of $f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ . How far must you travel along the $x$ -axis until the $y$ -coordinate reaches a height of 5.25?

Answer: Until $x =$ _____

Do not enter commas, that is, enter "10000" rather than "10,000".

$x equals 177, 827.941$

If an equation contains more than one log, we must first combine any expressions involving logs into a single logarithm.

Example 7

Solve $\log_{10} (x + 1) + \log_{10} (x - 2) = 1$ .

We use Property (1) of logarithms (see Properties of Logarithms) to rewrite the left-hand side as a single logarithm:

$\log_{10} (x + 1) (x - 2) = 1$

Once the left-hand side is expressed as a single logarithm, we can rewrite the equation in exponential form as

$(x + 1) (x - 2) = 10^{1}$

Simplifying the right side gives us a quadratic equation to solve.

$\begin{aligned} x^{2} - x - 2 & = 10 & Subtract 10 from both sides. \\ x^{2} - x - 12 & = 0 & Factor the left side. \\ (x - 4) (x + 3) & = 0 & Apply the zero-factor principle. \end{aligned}$

We find $x equals 4$ or $x = - 3$ . But let us check both of these values in the original equation.

For $x equals 4$ we have

$\log_{10} (4 + 1) (4 - 2) = \log_{10} (5 \cdot 2) = \log_{10} (10) = 1$

so $x equals 4$ is a solution.

But $x = - 3$ is not a solution of the original equation, because neither $log sub 10 ⁡ open parenthesis x plus 1 close parenthesis$ nor $log sub 10 ⁡ open parenthesis x minus 2 close parenthesis$ is defined for $x = - 3$ . (Remember that we cannot take a logarithm of a negative number or zero.) We say that the apparent solution $x = - 3$ is extraneous, and the only solution of the original equation is $4$ .

Extraneous solutions can arise whenever we solve a logarithmic equation, especially if there is more than one apparent solution. Therefore, we should always check that a possible solution does not cause one of the logarithms to be undefined. Here are guidelines for solving a logarithmic equation.

Which of these is the first step in solving the equation $\log (x) + \log (x - 1) = 2$ ?

_____

$\log (x (x - 1)) = 2$

Which of these is the first step in solving the equation $\log x + \log (x - 1) = 2$ ?

$\log (2 x - 1) = 2$
$\log (x) = 2 - \log (x - 1)$
$\log (x (x - 1)) = 2$
$x + x - 1 = 2^{x}$

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

$x =$ _____ [Separate multiple solutions with commas when appropriate.]

Rewrite the left side as a single logarithm.

Rewrite the equation in exponential form.

Solve for $x$ .

Check for extraneous solutions.

$x equals 4$

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

Follow the steps:

Rewrite the left side as a single logarithm.

Rewrite the equation in exponential form.

Solve for $x$ .

Check for extraneous solutions.

The solution is $x equals 4$ .

We cannot take a logarithm of _____.
After solving a logarithmic equation, we must check for _____.
If an equation contains more than one log, we must first combine them into _____.
If there is only one log involved, we write the equation in _____ form.

a negative number or zero
extraneous solutions
a single logarithm
exponential form

Fill in the blanks to complete each statement.

We cannot take a logarithm of ______.
After solving a logarithmic equation, we must check for ______.
If an equation contains more than one log, we must first combine them into ______.
If there is only one log involved, we rewrite the equation in ______ form.

What is an extraneous solution?

_____

What is an extraneous solution?

More About Inverse Functions

Let's take a closer look at the relationship between functions and their inverse functions. In Section 5.1 we saw that an inverse function "undoes" the effects of the function, and vice versa. In other words, if we apply a function and then its inverse to an input, we return to that input. For example, consider the function $f (x) = x^{3}$ and its inverse function $g (x) = \sqrt[3]{x}$ . We'll start with an input of $x equals 5$ , first apply the function $f$ , and then apply the function $g$ to the output. In function notation, that operation looks like this.

$g (f (5)) = g (5^{3}) = g (125) = \sqrt[3]{125} = 5$

We start by applying the innermost function, namely $f$ , to get $5 cubed equals 125$ , and then apply $g$ , to get $\sqrt[3]{125} = 5$ . We have returned to out original input.

In Example of Section 5.1 we found that the inverse of the function $f (t) = 6 + 2 t$ is $g (t) = \frac{t - 6}{2} .$

Show that $g open parenthesis f open parenthesis 7 close parenthesis close parenthesis equals 7.$
Show that $f open parenthesis g open parenthesis 30 close parenthesis close parenthesis equals 30.$

In each case, start by evaluating the innermost function.

These examples illustrate a general rule about inverse functions.

Because a logarithmic function is the inverse of the exponential function with the same base, each undoes the effect of the other. For example, the function $g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ is the inverse of $f (x) = 2^{x}$ . So, if we start with $x equals 3$ , apply $f$ , and then apply $g$ to the result, we return to the original number, 3.

$x = 3 \to_{exponential function}^{Apply the} f (3) = 2^{3} = 8 \to_{log function}^{Apply the} g (8) = \log_{2} 8 = 3_{number}^{Original}$

And because $f (x) = b^{x}$ and $g (x) = \log_{b} (x)$ are inverse functions, we can write these operations in one expression as

$g (f (3)) = \log_{2} (2^{3}) = 3$

We evaluate the expression starting with the inside function, $f (3) = 2^{3} = 8$ , and then compute the log base 2 of the result. Applying the exponential function and then the log function with the same base returne us to the original input.

Because the log and the exponential are inverse functions, similar calculations hold for any value of $x$ and any base $b greater than 0$ , so that $\log_{b} (b^{x}) = x$ .

Example 8

Simplify each expression.

$log sub 4 ⁡ open parenthesis 4 to the power 6 close parenthesis$
$\log_{8} (8^{2 a + 3})$

In this expression, we start with $6$ , apply the exponential function with base $4$ , and then take a logarithm base $4$ . Because the logarithm is the inverse of the exponential function, we return to the original number, $6$ .
$x = 6 \to_{exponential function}^{Apply the} 4^{6} \to_{log function}^{Apply the} \log_{4} (4^{6}) = 6_{number}^{Original}$
The input of the exponential function is the expression $2 a plus 3$ . Because the bases of the log and the exponential function are both $8$ , they are inverse functions, and applying them in succession returns us to the original input. Thus, $\log_{8} (8^{2 a + 3}) = 2 a + 3$ .

Simplify each expression.

$\log_{10} (10^{6}) =$ _____
$\log_{w} (w^{x + 1}) =$ _____, for $w greater than 0, w not equal to 1$

$6$
$w plus 1$

Simplify each expression.

$log sub 10 ⁡ open parenthesis 10 to the power 6 close parenthesis$
$\log_{w} (w^{x + 1}),$ for $w greater than 0, w not equal to 1$

$6$
$w plus 1$

We can also apply the two functions in the opposite order. For example,

$2^{\log_{2} (8)} = 8$

To see that this equation is true, we simplify the exponent first. We start with $8$ , and apply the log base $2$ function. Because $log sub 2 ⁡ 8 equals 3$ , we have

$8 \to_{log function}^{Apply the} \log_{2} 8 = 3 \to_{exponential function}^{Apply the} = 2^{\log_{2} 8} = 2^{3} = 8_{number}^{Original}$

Using function notation, the caluclation above looks ike this.

$f (g (8)) = 2^{\log_{2} (8)} = 8$

(Remember the order of operations: do what's inside of parentheses first, to get $g open parenthesis 8 close parenthesis equals log sub 2 ⁡ open parenthesis 8 close parenthesis$ .) In other words, applying first the log function and then the exponential function returns the original input value.

Of course, a similar equation holds for any positive value of $x$ and any base $b greater than 0, b not equal to 1$ :

$b^{\log_{b} x} = x$

Example 9

Simplify each expression.

$10 to the power log sub 10 ⁡ open parenthesis 1000 close parenthesis$
$Q to the power log sub Q ⁡ open parenthesis 25 close parenthesis, for Q greater than 0, Q not equal to 1$

Think of simplifying the exponent first: $log sub 10 ⁡ open parenthesis 1000 close parenthesis equals 3$ . Then
$10^{\log_{10} (1000)} = 10^{3} = 1000$
We first found the exponent on 10 that gives 1000, then we raised 10 to that exponent (namely 3), so we get 1000.
The expression $log sub Q ⁡ 25$ asks for the exponent on $Q$ that gives $25$ . Let's call that exponent $k$ , so that $log sub Q ⁡ open parenthesis 25 close parenthesis equals k$ . Then $Q^{\log_{Q} (25)} = Q^{k}$ , which gives us $25$ .

Simplify each expression.

$4^{\log_{4} (64)} =$ _____
$2^{\log_{2} (x^{2} + 1)} =$ _____

$64$
$x squared plus 1$

Simplify each expression.

$4 to the power log sub 4 ⁡ open parenthesis 64 close parenthesis$
$2^{\log_{2} (x^{2} + 1)}$

$64$
$x squared plus 1$

We summarize these relationships as follows.

In Chapter 4 we solved exponential equations by using the conversion equations to rewrite them in logarithmic form. The fact that $\log_{b} b^{x} = x$ gives us another way to think of the solution; we can take the log of both sides of the equation. We'll use this method in the next Example. Recall from Inverse Function Notation in Section 5.1 that the inverse function for a function $f$ is often denoted by $f^{- 1}$ .

Example 10

Find the inverse of the function $f (x) = 2^{x - 3} - 4$ .
Graph $f$ and $f^{- 1}$ on the same grid.
State the domain and range of $f$ and of $f^{- 1}$ .

We write the function as $y = 2^{x - 3} - 4$ , and solve for $x$ in terms of $y$ . First, we isolate the power:
$\begin{aligned} y + 4 & = 2^{x - 3} & Take the log base 2 of each side. \\ \log_{2} (y + 4) & = \log_{2} {(2)}^{x - 3} & Simplify the right side. \\ \log_{2} (y + 4) & = x - 3 \\ 3 + \log_{2} (y + 4) & = x \end{aligned}$
The inverse function is $f^{- 1} (y) = 3 + \log_{2} (y + 4)$ . However, to graph both $f$ and $f^{- 1}$ on the same grid, we write the inverse function as $f^{- 1} (x) = 3 + \log_{2} (x + 4)$ .
To graph $f$ , we translate the graph of $y = 2^{x}$ by $3$ units to the right and $4$ units down. The graph of $f^{- 1}$ looks like the graph of $y equals log sub 2 ⁡ open parenthesis x close parenthesis$ , but shifted $4$ units to the left and $3$ units up. The graphs are shown below, along with the line $y = x$ .
The function $f$ is a translation of an exponential function, and its domain consists of all real numbers. Because the graph is shifted $4$ units down, the range of $f$ is $open parenthesis minus 4, ∞ close parenthesis$ .
Because the log of a negative number or zero is undefined, for $f^{- 1} (x) = 3 + \log_{2} (x + 4)$ , we must have $x plus 4 greater than 0$ , so the domain of $f^{- 1}$ is $x greater than minus 4$ . We can verify on the graph that the range of $f^{- 1}$ includes all real numbers. Thus,
$\begin{aligned} Domain (f) = all real numbers = Range (f^{- 1}) \\ Range (f) = (- 4, \infty) = Domain (f^{- 1}) \end{aligned}$

Find the inverse function for $f (x) = 2 \log (x + 1)$ .
$f^{- 1} (x) =$ _____
Graph $f$ and $f^{- 1}$ in the window
$\begin{aligned} Xmin & = - 6 & Xmax = 6 \\ Ymin & = - 4 & Ymax = 4 \end{aligned}$
State the domain and range of $f$ and $f^{- 1}$ . Use "inf" for $∞$ .
Domain of $f$ : _____; Range of $f$ : _____
Domain of $f^{- 1}$ : _____; Range of $f^{- 1}$ : _____

$f^{- 1} (x) = 10^{x / 2} - 1$
A graph is below.
Domain of $f$ : $open parenthesis minus 1, ∞ close parenthesis$ ; Range of $f$ : all real numbers; Domain of $f^{- 1}$ : all real numbers; Range of $f^{- 1}$ : $open parenthesis minus 1, ∞ close parenthesis$

Find the inverse function for $f (x) = 2 \log (x + 1)$ .
Graph $f$ and $f^{- 1}$ in the window
$\begin{aligned} Xmin & = - 6 & Xmax = 6 \\ Ymin & = - 4 & Ymax = 4 \end{aligned}$
State the domain and range of $f$ and $f^{- 1}$ .

$f^{- 1} (x) = 10^{x / 2} - 1$
Domain of $f$ : $open parenthesis minus 1, ∞ close parenthesis$ ; Range of $f$ : all real numbers; Domain of $f^{- 1}$ : all real numbers; Range of $f^{- 1}$ : $open parenthesis minus 1, ∞ close parenthesis$

Compare the graphs of $f (x) = \log \frac{1}{x}$ and $g (x) = - \log (x)$ , and explain.

_____

Compare the graphs of $f (x) = (\log \frac{1}{x})$ and $g (x) = - \log (x)$ , and explain.

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Logarithmic function
Logarithmic equation
Extraneous solution

CONCEPTS

We define the logarithmic function, $g (x) = \log_{b} (x)$ , which takes the log base $b$ of its input values. The log function $g (x) = \log_{b} x$ is the inverse of the exponential function $f (x) = b^{x}$ .
Note
Exponential and Logarithmic Functions
Because $f (x) = b^{x}$ and $g (x) = \log_{b} (x)$ are inverse functions for $b greater than 0, b not equal to 1$ ,
$\log_{b} b^{x} = x, for all x and b^{\log_{b} (x)} = x, for x > 0$
Note
Logarithmic Functions $y = \log_{b} (x)$
1. Domain: all positive real numbers
2. Range: all real numbers
3. $x$ -intercept: $open parenthesis 1, 0 close parenthesis$
4. $y$ -intercept: none
5. Vertical asymptote at $x equals 0$
6. The graphs of $y = \log_{b} (x)$ and $y = b^{x}$ are symmetric about the line $y = x$ .
A logarithmic equation is one in which the variable appears inside of a logarithm. We can solve logarithmic equations by converting to exponential form.
Note
Steps for Solving Logarithmic Equations
1. Use the properties of logarithms to combine all logs into one log.
2. Isolate the log on one side of the equation.
3. Convert the equation to exponential form.
4. Solve for the variable.
5. Check for extraneous solutions.

STUDY QUESTIONS

Can the output of the function $y = \log_{b} (x)$ be negative?
Francine says that $\log_{2} (\frac{1}{x}) = - \log_{2} (x)$ . Is she correct? Why or why not?
Sketch a typical logarithmic function.
Simplify:
1. $10 to the power log ⁡ open parenthesis 13 close parenthesis$
2. $7 to the power log sub 7 ⁡ open parenthesis 13 close parenthesis$
Why is the following attempt to solve the equation incorrect?
$\begin{aligned} Solve: \log (x) + \log (x + 1) & = 2 \\ x + x + 1 & = 10^{2} \end{aligned}$

SKILLS

Practice each skill in the Homework problems listed.

Evaluate log functions: #1–16, 27 and 28
Simplify expressions involving logs: #15 and 16, 19, and 20
Graph logarithmic functions and transformations of log functions: #1–4, 25–28
Find formulas for inverse functions: #17–24
Solve logarithmic equations: #29–54
Solve formulas involving logs: #55–60

Homework 5.2

In Problems 1–4,

Make tables of values for each exponential function and its inverse logarithmic function.
Graph both functions on the same set of axes.

$f (x) = 2^{x}$

$x$ $minus 1$ $0$ $1$ $2$
$2 to the power x$ $\frac{1}{2}$ $1$ $2$ $4$
$x$ $\frac{1}{2}$ $1$ $2$ $4$
$log sub 2 ⁡ x$ $minus 1$ $0$ $1$ $2$

$x$	$minus 1$	$0$	$1$	$2$
$2 to the power x$	$\frac{1}{2}$	$1$	$2$	$4$

$x$	$\frac{1}{2}$	$1$	$2$	$4$
$log sub 2 ⁡ x$	$minus 1$	$0$	$1$	$2$

$f (x) = 3^{x}$

$f (x) = {(\frac{1}{3})}^{x}$

$x$ $minus 2$ $minus 1$ $0$ $1$
${(\frac{1}{3})}^{x}$ $9$ $3$ $1$ $\frac{1}{3}$
$x$ $9$ $3$ $1$ $\frac{1}{3}$
$log sub 1 / 3 ⁡ x$ $minus 2$ $minus 1$ $0$ $1$

$x$	$minus 2$	$minus 1$	$0$	$1$
${(\frac{1}{3})}^{x}$	$9$	$3$	$1$	$\frac{1}{3}$

$x$	$9$	$3$	$1$	$\frac{1}{3}$
$log sub 1 / 3 ⁡ x$	$minus 2$	$minus 1$	$0$	$1$

$f (x) = {(\frac{1}{2})}^{x}$

How large must $x$ be before the graph of $y equals log sub 10 ⁡ open parenthesis x close parenthesis$ reaches a height of $4$ ?
How large must $x$ be before the graph of $y equals log sub 10 ⁡ open parenthesis x close parenthesis$ reaches a height of $8$ ?

$x equals 10, 000$
$x = 10^{8}$

How large must $x$ be before the graph of $y equals log sub 2 ⁡ open parenthesis x close parenthesis$ reaches a height of $5$ ?
How large must $x$ be before the graph of $y equals log sub 2 ⁡ open parenthesis x close parenthesis$ reaches a height of $10$ ?

For what values of $x$ is $y = \log_{10} (x) < - 2$ ?

$0 less than open parenthesis x close parenthesis less than 0.01$

For what values of $x$ is $y = \log_{2} (x) < - 3$ ?

In Problems 9–14, $f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ . Evaluate.

$f open parenthesis 487 close parenthesis plus f open parenthesis 206 close parenthesis$
$f open parenthesis 487 plus 206 close parenthesis$

$log ⁡ 100, 322 approximately equals 5.001$
$log ⁡ 693 approximately equals 2.841$

$f open parenthesis 93 close parenthesis plus f open parenthesis 1500 close parenthesis$
$f open parenthesis 93 plus 1500 close parenthesis$

$f open parenthesis minus 7 close parenthesis$
$6 f open parenthesis 28 close parenthesis$

$log ⁡ open parenthesis minus 7 close parenthesis$ is undefined.
$6 log ⁡ 28 approximately equals 8.683$

$f open parenthesis 0 close parenthesis$
$3 f open parenthesis 41 close parenthesis$

$18 minus 5 f open parenthesis 3 close parenthesis$
$\frac{2}{5 + f (0.6)}$

$15.614$
$0.419$

$15 minus 4 f open parenthesis 7 close parenthesis$
$\frac{3}{2 + f (0.2)}$

Let $f (x) = 3^{x}$ and $g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis$ .

Compute $f open parenthesis 4 close parenthesis$ .
Compute $g [ f open parenthesis 4 close parenthesis ]$ .
Explain why $\log_{3} (3^{x}) = x$ for any $x$ .
Compute $log sub 3 ⁡ open parenthesis 3 to the power 1.8 close parenthesis$ .
Simplify $log sub 3 ⁡ open parenthesis 3 to the power a close parenthesis$ .

$81$
$4$
Definition of logarithm base $3$
$1.8$
$a$

Let $f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$ and $g (x) = 2^{x}$ .

Compute $f open parenthesis 32 close parenthesis$ .
Compute $g [ f open parenthesis 32 close parenthesis ]$ .
Explain why $2^{\log_{2} (x)} = x$ for any $x greater than 0$ .
Compute $2 to the power log sub 2 ⁡ open parenthesis 6 close parenthesis$ .
Simplify $2 to the power log sub 2 ⁡ open parenthesis Q close parenthesis$ .

If $h open parenthesis r close parenthesis equals log sub 2 ⁡ open parenthesis r close parenthesis$ , find $h^{- 1} (8)$ .
If $H (w) = 3^{w}$ , find $H^{- 1} (\frac{1}{9})$ .

$2 to the power 8$
$minus 2$

If $g open parenthesis z close parenthesis equals log sub 3 ⁡ open parenthesis z close parenthesis$ , find $g^{- 1} (- 3)$ .
If $G (q) = 2^{q}$ , find $G^{- 1} (1)$ .

For Problems 19–20, simplify.

$10 to the power log ⁡ open parenthesis 2 k close parenthesis$
$10 to the power 3 log ⁡ open parenthesis x close parenthesis$
$(\sqrt{10})^{\log (x)}$
$log ⁡ open parenthesis 100 to the power m close parenthesis$

$2 k$
$x^{3}$
$\sqrt{x}$
$2 m$

$\log (10^{(1 - x)})$
$100 to the power log ⁡ open parenthesis 2 x close parenthesis$
$(0.1)^{\log (x - 1)}$
$log ⁡ open parenthesis 10 to the power log ⁡ open parenthesis 10 close parenthesis close parenthesis$

What is the domain of the function $f (x) = 4 + \log_{3} (x - 9)$ ?
Find a formula for $f^{- 1} (x)$ .

$open parenthesis 9, ∞ close parenthesis$
$f^{- 1} (x) = 3^{x - 4} + 9$

What is the domain of the function $f (x) = 1 - \log_{2} (16 - 4 x)$ ?
Find a formula for $f^{- 1} (x)$ .

Find the inverse of the function $f (x) = 100 - 4^{x + 2}$ .
Show that $f^{- 1}$ undoes the effect of $f$ on $x equals 1$ .
Show that $f$ undoes the effect of $f^{- 1}$ on $x equals 84$ .

$f^{- 1} (x) = \log_{4} (100 - x) - 2$
$f^{- 1} (f (1)) = f^{- 1} (36) = \log_{4} (64) - 2 = 1$
$f (f^{- 1} (84)) = f (0) = 100 - 4^{2} = 84$

Find the inverse of the function $f (x) = 5 + 2^{- x}$ .
Show that $f^{- 1}$ undoes the effect of $f$ on $x = - 2$ .
Show that $f$ undoes the effect of $f^{- 1}$ on $x equals 6$ .

For Problems 25–26, match each graph to its equation.

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

$y equals 5 log ⁡ open parenthesis x close parenthesis$
$y = \log (\frac{x}{2})$
$y = \log (\frac{1}{x})$
$y = \log (- x)$

In a psychology experiment, volunteers were asked to memorize a list of nonsense words, then 24 hours later were tested to see how many of the words they recalled. On average, the subjects had forgotten $20 %$ of the words. The researchers found that the more lists their volunteers memorized, the larger the fraction of words they were unable to recall. (Source: Underwood, Scientific American, vol. 210, no. 3)

Number of lists, $n$	$1$	$4$	$8$	$12$	$16$	$20$
Percent forgotten, $F$	$20$	$40$	$55$	$66$	$74$	$80$

Plot the data. What sort of function seems to fit the data points?
Psychologists often describe rates of forgetting by logarithmic functions. Graph the function
$f (n) = 16.6 + 46.3 \log (n)$
on the same graph with your data. Comment on the fit.
What happens to the function $f (n)$ as $n$ grows increasingly large? Does this behavior accurately reflect the situation being modeled?

The graph resembles a logarithmic function. The (translated) log function is close to the points but appears too steep at first and not steep enough after $n equals 15$ . Overall, it is a good fit.
$f$ grows (more and more slowly) without bound. $f$ will eventually exceed $100$ per cent, but no one can forget more than $100 %$ of what is learned.

The water velocity at any point in a stream or river is related to the logarithm of the depth at that point. For the Hoback River near Bondurant, Wyoming,

$v = 2.63 + 1.03 \log (d)$

where $v$ is the velocity of the water, in feet per second, and $d$ is the vertical distance from the stream bed, in feet, at that point. For Pole Creek near Pinedale, Wyoming,

$v = 1.96 + 0.65 \log (d)$

Both streams are $1.2$ feet deep at the locations mentioned. (Source: Leopold, Luna, Wolman, and Gordon, 1992)

Complete the table of values for each stream.
Distance from bed (feet) $0.2$ $0.4$ $0.6$ $0.8$ $1.0$ $1.2$
Velocity, Hoback
River, (ft/sec)
Velocity, Pole Creek (ft/sec)
If you double the distance from the bed, by how much does the velocity increase in each stream?
Plot both functions on the same graph.
The average velocity of the entire stream can be closely approximated as follows: Measure the velocity at $20 %$ of the total depth of the stream from the surface and at $80 %$ of the total depth, then average these two values. Find the average velocity for the Hoback River and for Pole Creek.

In Problems 29–30, $f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis$ . Solve for $x$ .

$f open parenthesis x close parenthesis equals 1.41$
$f (x) = - 1.69$
$f open parenthesis x close parenthesis equals 0.52$

$10^{1.41} \approx 25.704$
$10^{- 1.69} \approx 0.020417$
$10^{0.52} \approx 3.3113$

$f open parenthesis x close parenthesis equals 2.3$
$f (x) = - 1.3$
$f open parenthesis x close parenthesis equals 0.8$

For Problems 31–38, convert the logarithmic equation to exponential form.

$log sub 16 ⁡ open parenthesis 256 close parenthesis equals w$

$16^{w} = 256$

$log sub 9 ⁡ open parenthesis 729 close parenthesis equals y$

$\log_{b} (9) = - 2$

$b^{- 2} = 9$

$\log_{b} (8) = - 3$

$\log_{10} (A) = - 2.3$

$10^{- 2.3} = A$

$\log_{10} (C) = - 4.5$

$\log_{u} (v) = w$

$u^{w} = v$

$\log_{m} (n) = p$

For Problems 39–46, solve for the unknown value.

$log sub b ⁡ open parenthesis 8 close parenthesis equals 3$

$b equals 2$

$log sub b ⁡ open parenthesis 625 close parenthesis equals 4$

$\log_{b} (10) = \frac{1}{2}$

$b equals 100$

$\log_{b} (0.1) = - 1$

$\log_{2} (3 x - 1) = 5$

$x equals 11$

$\log_{5} (9 - 4 x) = 3$

$3 \log_{7} (x) + 5 = 7$

$x = 7^{2 / 3}$

$5 \log_{2} (x) + 6 = - 14$

For Problems 47–54, solve the logarithmic equation.

$\log_{10} (x) + \log_{10} (x + 21) = 2$

$x equals 4$

$\log_{10} (x + 3) + \log_{10} (x) = 1$

$\log_{8} (x + 5) - \log_{8} (2) = 1$

$x equals 11$

$\log_{10} (x - 1) - \log_{10} (4) = 2$

$\log_{10} (x + 2) + \log_{10} (x - 1) = 1$

$x equals 3$

$\log_{4} (x + 8) - \log_{4} (x + 2) = 2$

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

No solution

$\log_{10} (x + 3) - \log_{10} (x - 1) = 1$

For Problems 55–60, solve for the indicated variable.

$t = T \log_{10} (1 + \frac{A}{k})$ , for $A$

$A = k (10^{t / T} - 1)$

$\log_{10} (R) = \log_{10} (R_{0}) + k t$ , for $R$

$N equals N sub 0 log sub b ⁡ open parenthesis k s close parenthesis$ , for $s$

$s = \frac{b^{N / N_{0}}}{k}$

$T = \frac{H \log_{10} (\frac{N}{N_{0}})}{\log_{10} (\frac{1}{2})}$ , for $N$

$M = \sqrt{\frac{\log_{10} (H)}{k \log_{10} (H_{0})}}$ , for $H$

$H = (H_{0})^{k M^{2}}$

$h = a - \sqrt{\frac{\log_{10} (B)}{t}}$ , for $B$

Choose the graph for each function described below.

The area, $A$ , of a pentagon is a quadratic function of the length, $l$ , of its side.
The strength, $F$ , of a hurricane varies inversely with its speed, $s$ .
The price of food has increased by $3 %$ every year for a decade.
The magnitude, $M$ , of a star is a logarithmic function of its brightness, $I$ .
The speed of the train increased at a constant rate.
If you do not practice a foreign language, you lose $\frac{1}{8}$ of the words in your working vocabulary, $V$ , each year.

For each of the functions listed below, select the graph of its inverse function, if possible, from the figures labeled I–VI. (The inverse of one of the functions is not shown.)

$f (x) = 2^{x}$
$f open parenthesis x close parenthesis equals x squared, x greater than or equal to 0$
$f (x) = \frac{2}{x}$
$f (x) = \sqrt{x}$
$f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis$
$f (x) = {(\frac{1}{2})}^{x}$

For Problems 63–64, graph the function on the domain $[ minus 4, 4 ]$ and a suitable range. Which have inverses that are also functions?

$f (x) = 5 (2^{- x^{2}})$
$f (x) = 2^{x} + 2^{- x}$

No inverse function
No inverse function

$f (x) = 5 (\log (x))^{2} + 1$
$f (x) = 5 \log (x^{2} + 1)$

For Problems 65–68, graph the pair of functions on your calculator. Explain the result.

$f (x) = \log (2 x), g (x) = \log (2) + \log (x)$

The functions are equal.

$f (x) = \log (\frac{x}{3}), g (x) = \log (x) - \log (3)$

$f (x) = \log (\frac{1}{x}), g (x) = - \log (x)$

The functions are equal.

$f (x) = \log (x^{3}), g (x) = 3 \log x$

Complete the following table.

$x$	$x^{2}$	$log sub 10 ⁡ open parenthesis x close parenthesis$	$log sub 10 ⁡ open parenthesis x squared close parenthesis$
$1$
$2$
$3$
$4$
$5$
$6$

Do you notice a relationship between $log sub 10 ⁡ x$ and $log sub 10 ⁡ open parenthesis x squared close parenthesis$ ? State the relationship as an equation.

$x$	$x^{2}$	$log sub 10 ⁡ open parenthesis x close parenthesis$	$log sub 10 ⁡ open parenthesis x squared close parenthesis$
$1$	$1$	$0$	$0$
$2$	$4$	$0.301$	$0.602$
$3$	$9$	$0.477$	$0.954$
$4$	$16$	$0.602$	$1.204$
$5$	$25$	$0.699$	$1.398$
$6$	$36$	$0.778$	$1.556$

$log sub 10 ⁡ x squared equals 2 log sub 10 ⁡ open parenthesis x close parenthesis$

Complete the following table.

$x$	$\frac{1}{x}$	$log sub 10 ⁡ open parenthesis x close parenthesis$	$\log_{10} (\frac{1}{x})$
$1$
$2$
$3$
$4$
$5$
$6$

Do you notice a relationship between $log sub 10 ⁡ open parenthesis x close parenthesis$ and $\log_{10} (\frac{1}{x})$ ? State the relationship as an equation.

In Problems 71 and 72, you found relationships between $log sub 10 ⁡ open parenthesis x close parenthesis$ and $log sub 10 ⁡ open parenthesis x squared close parenthesis$ , and between $log sub 10 ⁡ open parenthesis x close parenthesis$ and $\log_{10} (\frac{1}{x})$ . Assuming that those relationships hold for any base, complete the following tables and use them to graph the given functions.

$x$	$y = \log_{e} (x)$
$1$	$0$
$2$	$0.693$
$4$
$16$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{16}$

$x$	$y = \log_{e} (x)$
$1$	$0$
$2$	$0.693$
$4$	$1.386$
$16$	$2.772$
$\frac{1}{2}$	$minus 0.693$
$\frac{1}{4}$	$minus 1.386$
$\frac{1}{16}$	$minus 2.772$

$x$	$y = \log_{f} (x)$
$1$	$0$
$2$	$0.431$
$4$
$16$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{16}$

Investigation

Interest Compounded Continuously

We learned in Properties of Logarithms that the amount, $A$ (principal plus interest), accumulated in an account with interest compounded $n$ times annually is

$A = P {(1 + \frac{r}{n})}^{n t}$

where $P$ is the principal invested, $r$ is the interest rate, and $t$ is the time period, in years.

Suppose you keep $$ 1000$ in an account that pays $8 %$ interest. How much is the amount $A$ after $1$ year if the interest is compounded twice a year? Four times a year?
$\begin{matrix} n = 2 : A = 1000 {(1 + \frac{0.08}{2})}^{2 (1)} = \\ n = 4 : A = 1000 {(1 + \frac{0.08}{4})}^{4 (1)} = \end{matrix}$
What happens to $A$ as we increase $n$ , the number of compounding periods per year? Fill in the table showing the amount in the account for different values of $n$ .
$n$ $A$
$1$ (annually) $1080$
$2$ (semiannually)
$4$ (quarterly)
$6$ (bimonthly)
$12$ (monthly)
$365$ (daily)
$1000$
$10, 000$
Plot the values in the table from $n equals 1$ to $n equals 12$ , and connect them with a smooth curve. Describe the curve: What is happening to the value of $A$ ?
In part (2), as you increased the value of n , the other parameters in the formula stayed the same. In other words, A is a function of n , given by A = 1000 ( 1 + 0.08 n ) n . Use your calculator to graph A on successively larger domains:
1. $Xmin = 0, Xmax = 12; Ymin = 1080, Ymax = 1084$
2. $Xmin = 0, Xmax = 50; Ymin = 1080, Ymax = 1084$
3. $Xmin = 0, Xmax = 365; Ymin = 1080, Ymax = 1084$
Use the Trace feature or the Table feature to evaluate $A$ for very large values of $n$ . Rounded to the nearest penny, what is the largest value of $A$ that you can find?
As $n$ increases, the values of $A$ approach a limiting value. Although $A$ continues to increase, it does so by smaller and smaller increments and will never exceed $$ 1083.29$ . When the number of compounding periods increases without bound, we call the limiting result continuous compounding.
Is there an easier way to compute $A$ under continues compounding? Yes! Compute $1000 e to the power 0.08$ on your calculator. (Press 2nd LN to enter $e^{x}$ .) Compare the value to your answer in part (5) for the limiting value. The number $e$ is called the natural base. We'll compute its value shortly.
Repeat your calculations for two other interest rates, 15 % and (an extremely unrealistic) 100 % , again for an investment of $ 1000 for 1 year. In each case, compare the limiting value of A , and compare to the value of 1000 e r .
1. $r equals 0.15$
  $n$ $A$
  $1$ $115$
  $2$
  $4$
  $6$
  $12$
  $3652$
  $1000$
  $10, 000$
  $1000 e^{0.15} =$
2. $r equals 1$
  $n$ $A$
  $1$ $200$
  $2$
  $4$
  $6$
  $12$
  $3652$
  $1000$
  $10, 000$
  $1000 e^{1} =$
In part (8b), you have computed an approximation for $1000 e$ . What is the value of $e$ , rounded to $5$ decimal places?
Complete the table of values. What does ${(1 + \frac{1}{n})}^{n}$ appear to approach as $n$ increases?
$n$ $100$ $1000$ $10, 000$ $100, 000$
${(1 + \frac{1}{n})}^{n}$

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.

Distance from bed (feet)	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$	$1.2$
Velocity, Hoback River, (ft/sec)
Velocity, Pole Creek (ft/sec)

$n$	$A$
$1$ (annually)	$1080$
$2$ (semiannually)
$4$ (quarterly)
$6$ (bimonthly)
$12$ (monthly)
$365$ (daily)
$1000$
$10, 000$