Logarithmic Functions

Section 1.8 Logarithmic Functions

Learning Objectives.

Identify the form of a logarithmic function.
Explain the relationship between exponential and logarithmic functions.
Describe how to calculate a logarithm to a different base.

In this section we examine logarithmic functions. We use the properties of these functions to solve equations involving exponential and logarithmic terms. Note: This the second half of section 1.5 in the original version of OpenStax Calculus.

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Subsection 1.8.1 Logarithmic Functions

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Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

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The exponential function

f (x) = b^{x}

is one-to-one, with domain

(- \infty, \infty)

and range

(0, \infty) .

Therefore, it has an inverse function, called the logarithmic function with base

b .

For any

b > 0, b \neq 1,

the logarithmic function with base b, denoted

\log_{b},

has domain

(0, \infty)

and range

(- \infty, \infty),

and satisfies

\log_{b} (x) = y if and only if b^{y} = x .

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For example,

\begin{aligned} \log_{2} (8) = 3 & since 2^{3} = 8, \\ \log_{10} (\frac{1}{100}) = - 2 & since 10^{- 2} = \frac{1}{10^{2}} = \frac{1}{100}, \\ \log_{b} (1) = 0 & since b^{0} = 1 for any base b > 0. \end{aligned}

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Furthermore, since

y = \log_{b} (x)

and

y = b^{x}

are inverse functions,

\log_{b} (b^{x}) = x and b^{\log_{b} (x)} = x .

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The most commonly used logarithmic function is the function

\log_{e} .

Since this function uses natural

e

as its base, it is called the natural logarithm. Here we use the notation

\ln (x)

\ln x

to mean

\log_{e} (x) .

For example,

\ln (e) = \log_{e} (e) = 1, \ln (e^{3}) = \log_{e} (e^{3}) = 3, \ln (1) = \log_{e} (1) = 0.

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Since the functions

f (x) = e^{x}

and

g (x) = \ln (x)

are inverses of each other,

\ln (e^{x}) = x and e^{\ln x} = x,

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and their graphs are symmetric about the line

y = x

(Figure 1.122).

"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -3 to 4. The graph is of two functions. The first function is “f(x) = e to power of x”, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1) and there is no x intercept. The second function is “f(x) = ln(x)”, an increasing curved function. The x intercept is at the point (1, 0) and there is no y intercept. A dotted line with label “y = x” is also plotted on the graph, to show that the functions are mirror images over this line." — 🔗
Figure 1.122. The functions $y = e^{x}$ and $y = \ln (x)$ are inverses of each other, so their graphs are symmetric about the line $y = x .$

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Another frequently used logarithmic function is the function

\log_{10} .

It is called the common logarithm. Here we use the notation

\log (x)

\log x

to mean

\log_{10} (x) .

For example,

\log (10) = \log_{10} (10) = 1, \log (1000) = \log_{10} (10^{3}) = 3, \log (1) = \log_{10} (1) = 0.

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Note 1.123. Media.

At this site ¹ you can see an example of a base-10 logarithmic scale.

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In general, for any base

b > 0, b \neq 1,

the function

g (x) = \log_{b} (x)

is symmetric about the line

y = x

with the function

f (x) = b^{x} .

Using this fact and the graphs of the exponential functions, we graph functions

\log_{b}

for several values of

b > 1

(Figure 1.124).

"An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of three functions. All three functions a log functions that are increasing curved functions that start slightly to the right of the y axis and have an x intercept at (1, 0). The first function is “y = log base 10 (x)”, the second function is “f(x) = ln(x)”, and the third function is “y = log base 2 (x)”. The third function increases the most rapidly, the second function increases next most rapidly, and the third function increases the slowest." — 🔗
Figure 1.124. Graphs of $y = \log_{b} (x)$ are depicted for $b = 2, e, 10.$

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Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

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Note 1.125. Rule: Properties of Logarithms.

If $a, b, c > 0, b \neq 1,$ and $r$ is any real number, then

\begin{aligned} 1. \log_{b} (a c) = \log_{b} (a) + \log_{b} (c) & (Product property) \\ 2. \log_{b} (\frac{a}{c}) = \log_{b} (a) - \log_{b} (c) & (Quotient property) \\ 3. \log_{b} (a^{r}) = r \log_{b} (a) & (Power property) \end{aligned}

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Example 1.126. Solving Equations Involving Exponential Functions.

Solve each of the following equations for $x .$

$5^{x} = 2$
$e^{x} + 6 e^{- x} = 5$

Solution.

Applying the natural logarithm function to both sides of the equation, we have
$\ln 5^{x} = \ln 2.$
Using the power property of logarithms,
$x \ln 5 = \ln 2.$
Therefore, $x = \ln 2 / \ln 5.$
Multiplying both sides of the equation by $e^{x},$ we arrive at the equation
$e^{2 x} + 6 = 5 e^{x} .$
Rewriting this equation as
$e^{2 x} - 5 e^{x} + 6 = 0,$
we can then rewrite it as a quadratic equation in $e^{x} .$
$(e^{x})^{2} - 5 (e^{x}) + 6 = 0.$
Now we can solve the quadratic equation. Factoring this equation, we obtain
$(e^{x} - 3) (e^{x} - 2) = 0.$
Therefore, the solutions satisfy $e^{x} = 3$ and $e^{x} = 2.$ Taking the natural logarithm of both sides gives us the solutions $x = \ln 3, \ln 2.$

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Checkpoint 1.127.

Solve $e^{- x} / (3 + e^{- x}) = 1 / 2.$

Hint.

First solve the equation for $e^{- x} .$

Solution.

$x = \frac{\ln 3}{2}$

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Example 1.128. Solving Equations Involving Logarithmic Functions.

Solve each of the following equations for $x .$

$\ln (\frac{1}{x}) = 4$
$\log_{10} \sqrt{x} + \log_{10} x = 2$
$\ln (2 x) - 3 \ln (x^{2}) = 0$

Solution.

By the definition of the natural logarithm function,
$\ln (\frac{1}{x}) = 4 if and only if e^{4} = \frac{1}{2} .$
Therefore, the solution is $x = 1 / e^{4} .$
Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
$\log_{10} \sqrt{x} + \log_{10} x = \log_{10} x \sqrt{x} = \log_{10} x^{3 / 2} = \frac{3}{2} \log_{10} x .$
Therefore, the equation can be rewritten as
$\frac{3}{2} \log_{10} x = 2 or \log_{10} x = \frac{4}{3} .$
The solution is $x = 10^{4 / 3} = 10 \sqrt[3]{10} .$
Using the power property of logarithmic functions, we can rewrite the equation as $\ln (2 x) - \ln (x^{6}) = 0.$ Using the quotient property, this becomes
$\ln (\frac{2}{x^{5}}) = 0.$
Therefore, $2 / x^{5} = 1,$ which implies $x = \sqrt[5]{2} .$ We should then check for any extraneous solutions.

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Checkpoint 1.129.

Solve $\ln (x^{3}) - 4 > \ln (x) = 1.$

Hint.

First use the power property, then use the product property of logarithms.

Solution.

$x = \frac{1}{e}$

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When evaluating a logarithmic function with a calculator, you may have noticed that the only options are

\log_{10}

or log, called the common logarithm, or ln, which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base

b .

If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.

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Note 1.130. Rule: Change-of-Base Formulas.

Let $a > 0, b > 0,$ and $a \neq 1, b \neq 1.$

$a^{x} = b^{\log_{b} a^{x}}$ for any real number $x .$ If $b = e,$ this equation reduces to $a^{x} = e^{\log_{e} a^{x}} = e^{x \ln a} .$
$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$ for any real number $x > 0.$ If $b = e,$ this equation reduces to $\log_{a} x = \frac{\ln x}{\ln a} .$

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Example 1.131. Changing Bases.

Use a calculating utility to evaluate $\log_{3} 7$ with the change-of-base formula presented earlier.

Solution.

Use the second equation with $a = 3$ and $e = 3.$

$\log_{3} 7 = \frac{\ln 7}{\ln 3} \approx 1.77124 .$

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Checkpoint 1.132.

Use the change-of-base formula and a calculating utility to evaluate $\log_{4} 6.$

Hint.

Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.

Solution.

$1.29248$

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Example 1.133. Chapter Opener: The Richter Scale for Earthquakes.

"A photograph of an earthquake fault." — Figure 1.134. (credit: modification of work by Robb Hannawacker, NPS)

In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of an earthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude $R_{1}$ on the Richter scale and a second earthquake with magnitude $R_{2}$ on the Richter scale. Suppose $R_{1} > R_{2},$ which means the earthquake of magnitude $R_{1}$ is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If $A_{1}$ is the amplitude measured for the first earthquake and $A_{2}$ is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

R_{1} - R_{2} = \log_{10} (\frac{A_{1}}{A_{2}}) .

Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,

8 - 7 = \log_{10} (\frac{A_{1}}{A_{2}}) .

Therefore,

\log_{10} (\frac{A_{1}}{A_{2}}) = 1,

which implies $A_{1} / A_{2} = 10$ or $A_{1} = 10 A_{2} .$ Since $A_{1}$ is 10 times the size of $A_{2},$ we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation

\log_{10} (\frac{A_{1}}{A_{2}}) = 8 - 6 = 2.

Therefore, $A_{1} = 100 A_{2} .$ That is, the first earthquake is 100 times more intense than the second earthquake.

How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?

Solution.

To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:

$9 - 7.3 = \log_{10} (\frac{A_{1}}{A_{2}}) .$

Therefore, $A_{1} / A_{2} = 10^{1} .7,$ and we conclude that the earthquake in Japan was approximately $50$ times more intense than the earthquake in Haiti.

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Checkpoint 1.135.

Compare the relative severity of a magnitude $8.4$ earthquake with a magnitude $7.4$ earthquake.

Hint.

$R_{1} - R_{2} = \log_{10} (A_{1} / A_{2}) .$

Solution.

The magnitude $8.4$ earthquake is roughly $10$ times as severe as the magnitude $7.4$ earthquake.

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Subsection 1.8.2 Key Concepts

The logarithmic function $y = \log_{b} (x)$ is the inverse of $y = b^{x} .$ Its domain is $(0, \infty)$ and its range is $(- \infty, \infty) .$
The natural logarithmic function is $y = \ln x = \log_{e} x .$
Given an exponential function or logarithmic function in base $a,$ we can make a change of base to convert this function to any base $b > 0, b \neq 1.$ We typically convert to base $e .$

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This book is a custom edition based on OpenStax Calculus Volume 1. You can download the original for free at https://openstax.org/details/books/calculus-volume-1.²

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Additional practice exercises are available in at the bottom on this section in OpenStax Calculus Volume 1: https://openstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functions

openstax.org/l/20_logscale

https://openstax.org/details/books/calculus-volume-1