2.7.15: Exponential and Logarithmic Functions

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Page ID: 75447

Edwin “Jed” Herman & Gilbert Strang
OpenStax

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Learning Objectives

Identify the form of an exponential function.
Explain the difference between the graphs of $x^{b}$ and $b^{x}$.
Recognize the significance of the number $e$.
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 exponential and logarithmic functions. We use the properties of these functions to solve equations involving exponential or logarithmic terms, and we study the meaning and importance of the number $e$.

Exponential Functions

Recall the properties of exponents: If $x$ is a positive integer, then we define $b^x=b⋅b⋯b$ (with $x$ factors of $b$). If $x$ is a negative integer, then $x=−y$ for some positive integer $y$, and we define $b^x=b^{−y}=1/b^y$. Also, $b^0$ is defined to be $1$. If $x$ is a rational number, then $x=p/q$, where $p$ and $q$ are integers and $b^x=b^{p/q}=\sqrt[q]{b^p}$. For example, $9^{3/2}=\sqrt{9^3}=\left(\sqrt{9}\right)^3=27$.

Exercise $\PageIndex{1}$

Given the exponential function $f(x)=100⋅3^{x/2}$, evaluate $f(4)$ and $f(10)$.

Answer

$f(4)=900$

$f(10)=24,300$.

For any base $b>0$, $b≠1$, the exponential function $f(x)=b^x$ is defined for all real numbers $x$ and $b^x>0$. Therefore, the domain of $f(x)=b^x$ is $(−∞,∞)$ and the range is $(0,∞)$. To graph $b^x$, we note that for $b>1$, $b^x$ is increasing on $(−∞,∞)$ and $b^x→∞$ as $x→∞$, whereas $b^x→0$ as $x→−∞$. On the other hand, if $0<b<1$, $f(x)=b^x$ is decreasing on $(−∞,∞)$ and $b^x→0$ as $x→∞$ whereas $b^x→∞$ as $x→−∞$ (Figure $\PageIndex{2}$).

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 four functions. The first function is “f(x) = 2 to the power of x”, an increasing curved function, which starts slightly above the x axis and begins increasing. The second function is “f(x) = 4 to the power of x”, an increasing curved function, which starts slightly above the x axis and begins increasing rapidly, more rapidly than the first function. The third function is “f(x) = (1/2) to the power of x”, a decreasing curved function with decreases until it gets close to the x axis without touching it. The third function is “f(x) = (1/4) to the power of x”, a decreasing curved function with decreases until it gets close to the x axis without touching it. It decrases at a faster rate than the third function. — Figure $\PageIndex{2}$: If $b>1$, then $b^x$ is increasing on $(−∞,∞)$. If $0<b<1$, then $b^x$ is decreasing on $(−∞,∞)$.

Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.

Laws of Exponents

For any constants $a>0$, $b>0$, and for all $x$ and $y,$

\[b^x⋅b^y=b^{x+y} \nonumber \]
\[\dfrac{b^x}{b^y}=b^{x−y} \nonumber \]
\[(b^x)^y=b^{xy} \nonumber \]
\[(ab)^x=a^xb^x \nonumber \]
\[\dfrac{a^x}{b^x}=\left(\dfrac{a}{b}\right)^x \nonumber \]

Example $\PageIndex{2}$: Using the Laws of Exponents

Use the laws of exponents to simplify each of the following expressions.

$\dfrac{(2x^{2/3})^3}{(4x^{−1/3})^2}$
$\dfrac{(x^3y^{−1})^2}{(xy^2)^{−2}}$

Soution

a. We can simplify as follows:

\[\dfrac{(2x^{2/3})^3}{(4x^{−1/3})^2}=\dfrac{2^3(x^{2/3})^3}{4^2(x^{−1/3})^2}= \dfrac{8x^2}{16x^{−2/3}} =\dfrac{x^2x^{2/3}}{2}=\dfrac{x^{8/3}}{2}. \nonumber \]

b. We can simplify as follows:

\[\dfrac{(x^3y^{−1})^2}{(xy^2)^{−2}}=\dfrac{(x^3)^2(y^{−1})^2}{x^{−2}(y^2)^{−2}}=\dfrac{x^6y^{−2}}{x^{−2}y^{−4}} =x^6x^2y^{−2}y^4=x^8y^2. \nonumber \]

Exercise $\PageIndex{2}$

Use the laws of exponents to simplify $\dfrac{6x^{−3}y^2}{12x^{−4}y^5}$.

Hint: $x^a/x^b=x^{a-b}$

Answer: $x/(2y^3)$

The Number e

A special type of exponential function appears frequently in real-world applications involves the Euler number $e$:

\[e≈2.718282. \nonumber \]

Leonhard Euler

The letter $e$ was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between $e$ and logarithmic functions. We still use the notation $e$ today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

Functions involving base $e$ arise often in applications, we call the function $f(x)=e^x$ the natural exponential function. Since $e>1$, we know $f(x) = e^x$ is increasing on $(−∞,∞)$. In Figure $\PageIndex{3}$, we show a graph of $f(x)=e^x$ along with a tangent line to the graph of $f$ at $x=0$. The function $f(x)=e^x$ is the only exponential function $b^x$ with tangent line at $x=0$ that has a slope of $1.$ As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.

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 the function “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). At this point, a line is drawn tangent to the function. This line has the label “slope = 1”. — Figure $\PageIndex{3}$: The graph of $f(x)=e^x$ has a tangent line with slope $1$ at $x=0$.

Logarithmic Functions

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 the pH scale in chemistry or decibels in sound levels.

The exponential function $f(x)=b^x$ is one-to-one, with domain $(−∞,∞)$ and range $(0,∞)$. Therefore, it has an inverse function, called the logarithmic function with base $b$. For any $b>0,\, b≠1$, the logarithmic function with base $b$, denoted $\log_b$, has domain $(0,∞)$ and range $(−∞,∞)$,and satisfies

\[\log_b(x)=y \nonumber \]

if and only if $b^y=x$.

For example,

\[\log_2(8)=3\nonumber \]

since $2^3=8$,

\[\log_{10}\left(\dfrac{1}{100}\right)=−2 \nonumber \]

since $10^{−2}=\dfrac{1}{10^2}=\dfrac{1}{100}$,

\[\log_b(1)=0 \nonumber \]

since $b^0=1$ for any base $b>0$.

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

\[\log_b(b^x)=x \nonumber \]

and

\[b^{\log_b(x)}=x. \nonumber \]

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 $\ln (x)$ or $\ln x$ to mean $\log_e(x)$. For example,

\[ \begin{align*} \ln (e) &=\log_e(e)=1 \\[4pt] \ln (e^3) &=\log_e(e^3)=3 \\[4pt] \ln (1) &=\log_e(1)=0. \end{align*}\]

Since the functions $f(x)=e^x$ and $g(x)=\ln (x)$ are inverses of each other,

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

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

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

In general, for any base $b>0$, $b≠1$, 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 $\PageIndex{5}$).

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 $\PageIndex{5}$: 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 $a,\,b,\,c>0,\,b≠1$, and $r$ is any real number, then

Product property

\[\log_b(ac)=\log_b(a)+\log_b(c) \label{productprop} \]

Quotient property

\[\log_b \left(\dfrac{a}{c} \right)=\log_b(a)−\log_b(c) \label{quotientprop} \]

Power property

\[\log_b(a^r)=r\log_b(a) \label{powerprop} \]

Example $\PageIndex{4}$: Solving Equations Involving Exponential Functions

Solve each of the following equations for $x$.

$5^x=2$
$e^x+6e^{−x}=5$

Solution

a. 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,

\[x= \dfrac{\ln 2}{\ln 5}. \nonumber \]

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

Exercise $\PageIndex{4}$

Solve

\[e^{2x}/(3+e^{2x})=1/2. \nonumber \]

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

Answer: $x=\dfrac{\ln 3}{2}$.

Example $\PageIndex{5}$: Solving Equations Involving Logarithmic Functions

Solve each of the following equations for $x$.

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

Solution

a. By the definition of the natural logarithm function,

$\ln \left(\dfrac{1}{x} \right)=4$

if and only if $e^4=\dfrac{1}{x}$.

Therefore, the solution is $x=1/e^4$.

b. Using the product (Equation \ref{productprop}) and power (Equation \ref{powerprop}) properties of logarithmic functions, rewrite the left-hand side of the equation as

\[\begin{align*} \log_{10}\sqrt{x} + \log_{10}x &= \log_{10} x \sqrt{x} \\[4pt] &= \log_{10}x^{3/2} \\[4pt] &= \dfrac{3}{2}\log_{10}x. \end{align*}\]

Therefore, the equation can be rewritten as

$\dfrac{3}{2}\log_{10}x=2$

$\log_{10}x=\dfrac{4}{3}$.

The solution is $x=10^{4/3}=10\sqrt[3]{10}$.

c. Using the power property (Equation \ref{powerprop}) of logarithmic functions, we can rewrite the equation as $\ln (2x)−\ln (x^6)=0$.

Using the quotient property (Equation \ref{quotientprop}), this becomes

$\ln \left(\dfrac{2}{x^5}\right)=0$

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

Exercise $\PageIndex{5}$

Solve $\ln (x^3)−4\ln (x)=1$.

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

Answer: $x=\dfrac{1}{e}$

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.

Rule: Change-of-Base Formulas

Let $a>0,\,b>0$, and $a≠1,\,b≠1$.

1. $a^x=b^{x \log_ba}$ for any real number $x$.

If $b=e$, this equation reduces to $a^x=e^{x \log_ea}=e^{x \ln a}$.

2. $\log_ax=\dfrac{\log_bx}{\log_ba}$ for any real number $x>0$.

If $b=e$, this equation reduces to $\log_ax=\dfrac{\ln x}{\ln a}$.

Example $\PageIndex{6}$: Changing Bases

Use a calculating utility to evaluate $\log_37$ with the change-of-base formula presented earlier.

Solution

Use the second equation with $a=3$ and $b=e$: $\log_37=\dfrac{\ln 7}{\ln 3}≈1.77124$.

Exercise $\PageIndex{6}$

Use the change-of-base formula and a calculating utility to evaluate $\log_46$.

Hint: Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.
Answer: $\log_46 = \dfrac{\ln 6}{\ln 4} \approx 1.29248$

Example $\PageIndex{7}$: The Richter Scale for Earthquakes

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 photograph of an earthquake fault. — Figure $\PageIndex{6}$: (credit: modification of work by Robb Hannawacker, NPS)

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:

$R_1−R_2=\log_{10}\left(\dfrac{A1}{A2}\right)$.

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

$8−7=\log_{10}\left(\dfrac{A1}{A2}\right)$.

Therefore,

$\log_{10}\left(\dfrac{A1}{A2}\right)=1$,

which implies $A_1/A_2=10$ or $A_1=10A_2$. 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

$\log_{10}\left(\dfrac{A1}{A2}\right)=8−6=2$.

Therefore, $A_1=100A_2$.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:

$9−7.3=\log_{10}\left(\dfrac{A1}{A2}\right)$.

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.

Exercise $\PageIndex{7}$

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

Hint: $R_1−R_2=\log_{10}(A1/A2)$.
Answer: The magnitude $8.4$ earthquake is roughly $10$ times as severe as the magnitude $7.4$ earthquake.

Use the Properties of Logarithms

Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.

The first two properties derive from the definition of logarithms. Since $a^{0}=1$, we can convert this to logarithmic form and get $\log _{a} 1=0$. Also, since $a^{1}=a$, we get $\log _{a} a=1$.

Definition $\PageIndex{1}$

Properties of Logarithms

$\log _{a} 1=0 \quad \log _{a} a=1$

In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.

Example $\PageIndex{1}$

Evaluate using the properties of logarithms:

$\log _{8} 1$
$\log _{6} 6$

Solution:

$\log _{8} 1$

Use the property, $\log _{a} 1=0$.

$0 \quad \log _{8} 1=0$

$\log _{6} 6$

Use the property, $\log _{a} a=1$.

$1 \quad \log _{6} 6=1$