Exponential Functions

Exponential functions arise in many applications. One common example is population growth .

For example, if a population starts with [latex]P_0[/latex] individuals and then grows at an annual rate of [latex]2\%[/latex], its population after 1 year is

Its population after 2 years is

In general, its population after [latex]t[/latex] years is

which is an exponential function. More generally, any function of the form [latex]f(x)=b^x[/latex], where [latex]b < 0, \, b \ne 1[/latex], is an exponential function with base [latex]b[/latex] and exponent [latex]x[/latex]. Exponential functions have constant bases and variable exponents. Note that a function of the form [latex]f(x)=x^b[/latex] for some constant [latex]b[/latex] is not an exponential function but a power function.

To see the difference between an exponential function and a power function, we compare the functions [latex]y=x^2[/latex] and [latex]y=2^x[/latex]. In (Figure) , we see that both [latex]2^x[/latex] and [latex]x^2[/latex] approach infinity as [latex]x \to \infty[/latex]. Eventually, however, [latex]2^x[/latex] becomes larger than [latex]x^2[/latex] and grows more rapidly as [latex]x \to \infty[/latex]. In the opposite direction, as [latex]x \to -\infty, \, x^2 \to \infty[/latex], whereas [latex]2^x \to 0[/latex]. The line [latex]y=0[/latex] is a horizontal asymptote for [latex]y=2^x[/latex].

In (Figure) , we graph both [latex]y=x^2[/latex] and [latex]y=2^x[/latex] to show how the graphs differ.

Graphing Exponential Functions

For any base [latex]b < 0, \, b\ne 1[/latex], the exponential function [latex]f(x)=b^x[/latex] is defined for all real numbers [latex]x[/latex] and [latex]b^x < 0[/latex]. Therefore, the domain of [latex]f(x)=b^x[/latex] is [latex](-\infty ,\infty)[/latex] and the range is [latex](0,\infty)[/latex]. To graph [latex]b^x[/latex], we note that for [latex]b < 1, \, b^x[/latex] is increasing on [latex](-\infty ,\infty)[/latex] and [latex]b^x \to \infty[/latex] as [latex]x \to \infty[/latex], whereas [latex]b^x \to 0[/latex] as [latex]x \to -\infty[/latex]. On the other hand, if [latex]0 < b < 1, \, f(x)=b^x[/latex] is decreasing on [latex](-\infty ,\infty)[/latex] and [latex]b^x \to 0[/latex] as [latex]x \to \infty[/latex] whereas [latex]b^x \to \infty[/latex] as [latex]x \to -\infty[/latex] ( (Figure) ).

Visit this site for more exploration of the graphs of exponential functions.

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

Rule: Laws of Exponents

For any constants [latex]a > 0, \, b > 0[/latex], and for all [latex]x[/latex] and [latex]y[/latex],

1. [latex]b^x\cdot b^y=b^{x+y}[/latex]
2. [latex]\large\frac{b^x}{b^y} \normalsize = b^{x-y}[/latex]
3. [latex](b^x)^y=b^{xy}[/latex]
4. [latex](ab)^x=a^x b^x[/latex]
5. [latex]\large \frac{a^x}{b^x} \normalsize =(\frac{a}{b})^x[/latex]

Using the Laws of Exponents

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

1. [latex]\large \frac{(2x^{2/3})^3}{(4x^{-1/3})^2}[/latex]
2. [latex]\large \frac{(x^3 y^{-1})^2}{(xy^2)^{-2}}[/latex]

Solution

1. We can simplify as follows:
   [latex]\large \frac{(2x^{2/3})^3}{(4x^{-1/3})^2} \normalsize = \large \frac{2^3(x^{2/3})^3}{4^2(x^{-1/3})^2} \normalsize = \large \frac{8x^2}{16x^{-2/3}} \normalsize = \large \frac{x^2x^{2/3}}{2} \normalsize = \large \frac{x^{8/3}}{2}[/latex].
2. We can simplify as follows:
   [latex]\large \frac{(x^3y^{-1})^2}{(xy^2)^{-2}} \normalsize = \large \frac{(x^3)^2(y^{-1})^2}{x^{-2}(y^2)^{-2}} \normalsize = \large \frac{x^6y^{-2}}{x^{-2}y^{-4}} \normalsize = x^6x^2y^{-2}y^4 = x^8y^2[/latex].

Use the laws of exponents to simplify [latex]\frac{6x^{-3}y^2}{12x^{-4}y^5}[/latex].

Hint

[latex]\frac{x^a}{x^b}=x^{a-b}[/latex]

Solution

[latex]\frac{x}{2y^3}[/latex]

The Number  [latex]e[/latex]

A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests [latex]P[/latex] dollars in a savings account with an annual interest rate [latex]r[/latex], compounded annually. The amount of money after 1 year is

The amount of money after 2 years is

More generally, the amount after [latex]t[/latex] years is

If the money is compounded 2 times per year, the amount of money after half a year is

The amount of money after 1 year is

After [latex]t[/latex] years, the amount of money in the account is

More generally, if the money is compounded [latex]n[/latex] times per year, the amount of money in the account after [latex]t[/latex] years is given by the function

What happens as [latex]n\to \infty[/latex]? To answer this question, we let [latex]m=n/r[/latex] and write

and examine the behavior of [latex](1+1/m)^m[/latex] as [latex]m\to \infty[/latex], using a table of values ( (Figure) ).

Looking at this table, it appears that [latex](1+1/m)^m[/latex] is approaching a number between 2.7 and 2.8 as [latex]m\to \infty[/latex]. In fact, [latex](1+1/m)^m[/latex] does approach some number as [latex]m\to \infty[/latex]. We call this number [latex]e[/latex]. To six decimal places of accuracy,

The letter [latex]e[/latex] 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 [latex]e[/latex] and logarithmic functions. We still use the notation [latex]e[/latex] today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

Returning to our savings account example, we can conclude that if a person puts [latex]P[/latex] dollars in an account at an annual interest rate [latex]r[/latex], compounded continuously, then [latex]A(t)=Pe^{rt}[/latex]. This function may be familiar. Since functions involving base [latex]e[/latex] arise often in applications, we call the function [latex]f(x)=e^x[/latex] the natural exponential function . Not only is this function interesting because of the definition of the number [latex]e[/latex], but also, as discussed next, its graph has an important property.

Since [latex]e < 1[/latex], we know [latex]e^x[/latex] is increasing on [latex](-\infty ,\infty)[/latex]. In (Figure) , we show a graph of [latex]f(x)=e^x[/latex] along with a tangent line to the graph of at [latex]x=0[/latex]. We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of [latex]f[/latex] at [latex]x=a[/latex] is a line that passes through the point [latex](a,f(a))[/latex] and has the same “slope” as [latex]f[/latex] at that point. The function [latex]f(x)=e^x[/latex] is the only exponential function [latex]b^x[/latex] with tangent line at [latex]x=0[/latex] that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the simplest exponential function to use in many instances.

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 pH in chemistry or decibels in sound levels.

The exponential function [latex]f(x)=b^x[/latex] is one-to-one, with domain [latex](-\infty ,\infty)[/latex] and range [latex](0,\infty )[/latex]. Therefore, it has an inverse function, called the logarithmic function with base [latex]b[/latex]. For any [latex]b < 0, \, b \ne 1[/latex], the logarithmic function with base [latex]b[/latex], denoted [latex]\log_b[/latex], has domain [latex](0,\infty )[/latex] and range [latex](-\infty ,\infty )[/latex], and satisfies

For example,

Furthermore, since [latex]y=\log_b (x)[/latex] and [latex]y=b^x[/latex] are inverse functions,

The most commonly used logarithmic function is the function [latex]\log_e (x)[/latex]. Since this function uses natural [latex]e[/latex] as its base, it is called the natural logarithm . Here we use the notation [latex]\ln(x)[/latex] or [latex]\ln x[/latex] to mean [latex]\log_e (x)[/latex]. For example,

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

and their graphs are symmetric about the line [latex]y=x[/latex] ( (Figure) ).

In general, for any base [latex]b < 0, \, b\ne 1[/latex], the function [latex]g(x)=\log_b (x)[/latex] is symmetric about the line [latex]y=x[/latex] with the function [latex]f(x)=b^x[/latex]. Using this fact and the graphs of the exponential functions, we graph functions [latex]\log_b (x)[/latex] for several values of [latex]b < 1[/latex] ( (Figure) ).

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

When evaluating a logarithmic function with a calculator, you may have noticed that the only options are [latex]\log_{10}[/latex] 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 [latex]b[/latex]. 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.

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 [latex]b < 0, \, b\ne 1, \, \log_b (a^x)=x \log_b a[/latex]. Therefore,

[latex]b^{\log_b(a^x)}=b^{x \log_b a}[/latex].

In addition, we know that [latex]b^x[/latex] and [latex]\log_b (x)[/latex] are inverse functions. Therefore,

[latex]b^{\log_b (a^x)}=a^x[/latex].

Combining these last two equalities, we conclude that [latex]a^x=b^{x \log_b a}[/latex].

To prove the second property, we show that

[latex](\log_b a)\cdot (\log_a x)=\log_b x[/latex].

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

[latex]b^{uv}=(b^u)^v=a^v=x=b^w[/latex].

Therefore, [latex]b^{uv}=b^w[/latex]. Since exponential functions are one-to-one, we can conclude that [latex]u\cdot v=w[/latex].

□

Changing Bases

Use a calculating utility to evaluate [latex]\log_3 7[/latex] with the change-of-base formula presented earlier.

Solution

Use the second equation with [latex]a=3[/latex] and [latex]e=3[/latex]:

[latex]\log_3 7=\frac{\ln 7}{\ln 3} \approx 1.77124[/latex].

Use the change-of-base formula and a calculating utility to evaluate [latex]\log_4 6[/latex].

Hint

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

Solution

1.29248

Chapter Opener: 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 [latex]R_1[/latex] on the Richter scale and a second earthquake with magnitude [latex]R_2[/latex] on the Richter scale. Suppose [latex]R_1 < R_2[/latex], which means the earthquake of magnitude [latex]R_1[/latex] 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 [latex]A_1[/latex] is the amplitude measured for the first earthquake and [latex]A_2[/latex] is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

[latex]R_1 - R_2 = \log_{10}(\frac{A_1}{A_2})[/latex].

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

[latex]8-7=\log_{10}(\frac{A_1}{A_2})[/latex].

Therefore,

[latex]\log_{10}(\frac{A_1}{A_2})=1[/latex],

which implies [latex]A_1 / A_2 = 10[/latex] or [latex]A_1 = 10A_2[/latex]. Since [latex]A_1[/latex] is 10 times the size of [latex]A_2[/latex], 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

[latex]\log_{10}(\frac{A_1}{A_2})=8-6=2[/latex].

Therefore, [latex]A_1=100A_2[/latex]. 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:

[latex]9-7.3=\log_{10}(\frac{A_1}{A_2})[/latex].

Therefore, [latex]A_1 / A_2=10^{1.7}[/latex], and we conclude that the earthquake in Japan was approximately 50 times more intense than the earthquake in Haiti.

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

Hint

[latex]R_1-R_2=\log_{10}(A_1 / A_2)[/latex].

Solution:

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

Hyperbolic Functions

The hyperbolic functions are defined in terms of certain combinations of [latex]e^x[/latex] and [latex]e^{-x}[/latex]. These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary ( (Figure) ). If we introduce a coordinate system so that the low point of the chain lies along the [latex]y[/latex]-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh , sech , csch , and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.

Using the definition of [latex]\cosh(x)[/latex] and principles of physics, it can be shown that the height of a hanging chain, such as the one in (Figure) , can be described by the function [latex]h(x)=a \cosh(x/a)+c[/latex] for certain constants [latex]a[/latex] and [latex]c[/latex].

But why are these functions called hyperbolic functions ? To answer this question, consider the quantity [latex]\cosh^2 t-\sinh^2 t[/latex]. Using the definition of [latex]\cosh[/latex] and [latex]\sinh[/latex], we see that

This identity is the analog of the trigonometric identity [latex]\cos^2 t+\sin^2 t=1[/latex]. Here, given a value [latex]t[/latex], the point [latex](x,y)=(\cosh t,\sinh t)[/latex] lies on the unit hyperbola [latex]x^2-y^2=1[/latex] ( (Figure) ).

Graphs of Hyperbolic Functions

To graph [latex]\cosh x[/latex] and [latex]\sinh x[/latex], we make use of the fact that both functions approach [latex](1/2)e^x[/latex] as [latex]x \to \infty[/latex], since [latex]e^{-x} \to 0[/latex] as [latex]x \to \infty[/latex]. As [latex]x \to -\infty, \, \cosh x[/latex] approaches [latex]1/2e^{-x}[/latex], whereas [latex]\sinh x[/latex] approaches [latex]-1/2e^{-x}[/latex]. Therefore, using the graphs of [latex]1/2e^x, \, 1/2e^{-x}[/latex], and [latex]-1/2e^{-x}[/latex] as guides, we graph [latex]\cosh x[/latex] and [latex]\sinh x[/latex]. To graph [latex]\tanh x[/latex], we use the fact that [latex]\tanh(0)=1, \, -1<\tanh(x)<1[/latex] for all [latex]x, \, \tanh x \to 1[/latex] as [latex]x \to \infty[/latex], and [latex]\tanh x \to -1[/latex] as [latex]x \to -\infty[/latex]. The graphs of the other three hyperbolic functions can be sketched using the graphs of [latex]\cosh x, \, \sinh x[/latex], and [latex]\tanh x[/latex] ( (Figure) ).