Logarithmic Functions

Learning Outcomes

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

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Recall the definition of an inverse function.

Recall: Inverse Function

For any one-to-one function $f\left(x\right)=y$ , a function ${f}^{-1}\left(x\right)$ is an inverse function of $f$ if ${f}^{-1}\left(y\right)=x$ . This can also be written as ${f}^{-1}\left(f\left(x\right)\right)=x$ for all $x$ in the domain of $f$ . It also follows that $f\left({f}^{-1}\left(x\right)\right)=x$ for all $x$ in the domain of ${f}^{-1}$ if ${f}^{-1}$ is the inverse of $f$ .

The notation ${f}^{-1}$ is read “ $f$ inverse.” Like any other function, we can use any variable name as the input for ${f}^{-1}$ , so we will often write ${f}^{-1}\left(x\right)$ , which we read as $"f$ inverse of $x$ “.

Logarithmic functions 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.

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 $b>0, \, b \ne 1$ , the logarithmic function with base $b$ , denoted $\log_b$ , has domain $(0,\infty )$ and range $(−\infty ,\infty )$ , and satisfies

\log_{b} (x) = y

$\log_b(x)=y$ if and only if

b^{y} = x

$b^y=x$

For example,

\begin{array}{cccc} \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{array}

$\begin{array}{cccc} \log_2 (8)=3\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}2^3=8,\hfill \\ \log_{10} (\frac{1}{100})=-2\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}10^{-2}=\frac{1}{10^2}=\frac{1}{100},\hfill \\ \log_b (1)=0\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}b^0=1 \, \text{for any base} \, b>0.\hfill \end{array}$

Furthermore, since $y=\log_b (x)$ and $y=b^x$ are inverse functions,

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

$\log_b (b^x)=x \, \text{and} \, b^{\log_b (x)}=x$

The most commonly used logarithmic function is the function $\log_e (x)$ . Since this function uses natural $e$ as its base, it is called the natural logarithm. Here we use the notation $\ln(x)$ or $\ln x$ 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

$\ln (e)=\log_e (e)=1, \, \ln(e^3)=\log_e (e^3)=3, \, \ln(1)=\log_e (1)=0$

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

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

and their graphs are symmetric about the line $y=x$ (Figure 4).

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 4: The functions $y=e^x$ and $y=\ln(x)$ are inverses of each other, so their graphs are symmetric about the line $y=x$ .

Interactive

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

In general, for any base $b>0, \, b\ne 1$ , 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 $\log_b (x)$ for several values of $b>1$ (Figure 5).

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 5: Graphs of $y=\log_b (x)$ are depicted for $b=2, \, e, \, 10$ .

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

Properties of Logarithms

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

$\begin{array}{cccc}1.\phantom{\rule{2em}{0ex}}\log_b (ac)=\log_b (a)+\log_b (c)\hfill & & & \text{(Product property)}\hfill \\ 2.\phantom{\rule{2em}{0ex}}\log_b(\frac{a}{c})=\log_b (a) -\log_b (c)\hfill & & & \text{(Quotient property)}\hfill \\ 3.\phantom{\rule{2em}{0ex}}\log_b (a^r)=r \log_b (a)\hfill & & & \text{(Power property)}\hfill \end{array}$

Example: Solving Equations Involving Exponential Functions

Solve each of the following equations for $x$ .

$5^x=2$
$e^x+6e^{−x}=5$

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

Rewriting this equation as

$e^{2x}-5e^x+6=0$ ,

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

$x=\ln 3, \, \ln 2$

Watch the following video to see the worked solution to Example: Solving Equations Involving Exponential Functions

Closed Captioning and Transcript Information for Video

Try It

Solve $\dfrac{e^{2x}}{(3+e^{2x})}=\dfrac{1}{2}$ .

Hint

First solve the equation for $e^{2x}$ .

Show Solution

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

Example: Solving Equations Involving Logarithmic Functions

Solve each of the following equations for $x$ .

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

Show Solution

By the definition of the natural logarithm function,
$\ln\big(\frac{1}{x}\big)=4 \, \text{ if and only if } \, e^4=\frac{1}{x}$

Therefore, the solution is $x=\frac{1}{e^4}$ .
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

$\frac{3}{2} \log_{10} x = 2 \, \text{ or } \, \log_{10} x = \frac{4}{3}$

The solution is $x=10^{4/3}=10\sqrt[3]{10}$ .
Using the power property of logarithmic functions, we can rewrite the equation as $\ln(2x) - \ln(x^6) = 0$ .
Using the quotient property, this becomes

$\ln\big(\frac{2}{x^5}\big)=0$

Therefore, $\frac{2}{x^5}=1$ , which implies $x=\sqrt[5]{2}$ . We should then check for any extraneous solutions.

Watch the following video to see the worked solution to Example: Solving Equations Involving Logarithmic Functions

Closed Captioning and Transcript Information for Video

Try It

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

Hint

Show Solution

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

Try It

When evaluating a logarithmic function with a calculator, you may have noticed that the only options are $\log_{10}$ or log, called the , 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.

Try It

Change-of-Base Formulas

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

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

Proof

For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base $b>0, \, b\ne 1, \, \log_b (a^x)=x \log_b a$ . Therefore,

b^{\log_{b} (a^{x})} = b^{x \log_{b} a}

$b^{\log_b(a^x)}=b^{x \log_b a}$

In addition, we know that $b^x$ and $\log_b (x)$ are inverse functions. Therefore,

b^{\log_{b} (a^{x})} = a^{x}

$b^{\log_b (a^x)}=a^x$

Combining these last two equalities, we conclude that $a^x=b^{x \log_b a}$ .

To prove the second property, we show that

(\log_{b} a) \cdot (\log_{a} x) = \log_{b} x

$(\log_b a)·(\log_a x)=\log_b x$

Let $u=\log_b a, \, v=\log_a x$ , and $w=\log_b x$ . We will show that $u·v=w$ . By the definition of logarithmic functions, we know that $b^u=a, \, a^v=x$ , and $b^w=x$ . From the previous equations, we see that

b^{u v} = (b^{u})^{v} = a^{v} = x = b^{w}

$b^{uv}=(b^u)^v=a^v=x=b^w$

Therefore, $b^{uv}=b^w$ . Since exponential functions are one-to-one, we can conclude that $u·v=w$ .

$_\blacksquare$

Example: Changing Bases

Use a calculating utility to evaluate $\log_3 7$ with the change-of-base formula presented earlier.

Show Solution

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

$\log_3 7=\frac{\ln 7}{\ln 3} \approx 1.77124$ .

Try It

Use the change-of-base formula and a calculating utility to evaluate $\log_4 6$ .

Hint

Show Solution

Try It

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

Hint

$R_1-R_2=\log_{10}(A_1 / A_2)$ .

Show Solution

Module 1: Functions and Graphs