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Home / Find the Exact Solution to the Exponential Equation

Find the Exact Solution to the Exponential Equation

1. Functions and Graphs

1.5 Exponential and Logarithmic Functions

Learning Objectives

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. We also define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions. (Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications of Integrations, and prove that the functions have the same properties with either definition.)

Exponential Functions

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

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

P(1)=P_0+0.02P_0=P_0(1+0.02)=P_0(1.02).

Its population after 2 years is

P(2)=P(1)+0.02P(1)=P(1)(1.02)=P_0(1.02)^2.

In general, its population after t years is

P(t)=P_0(1.02)^t,

which is an exponential function. More generally, any function of the form f(x)=b^x, where b>0, \, b \ne 1, is an exponential function with base b and exponent x. Exponential functions have constant bases and variable exponents. Note that a function of the form f(x)=x^b for some constant b 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 y=x^2 and y=2^x. In (Figure), we see that both 2^x and x^2 approach infinity as x \to \infty. Eventually, however, 2^x becomes larger than x^2 and grows more rapidly as x \to \infty. In the opposite direction, as x \to −\infty, \, x^2 \to \infty, whereas 2^x \to 0. The line y=0 is a horizontal asymptote for y=2^x.

Values of x^2 and 2^x
\mathbf{x} -3 -2 -1 0 1 2 3 4 5 6
\mathbf{x^2} 9 4 1 0 1 4 9 16 25 36
\mathbf{2^x} 1/8 1/4 1/2 1 2 4 8 16 32 64

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

Evaluating Exponential Functions

Recall the properties of exponents: If x is a positive integer, then we define b^x=b·b \cdots 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}=27. However, how is b^x defined if x is an irrational number? For example, what do we mean by 2^{\sqrt{2}}? This is too complex a question for us to answer fully right now; however, we can make an approximation. In (Figure), we list some rational numbers approaching \sqrt{2}, and the values of 2^x for each rational number x are presented as well. We claim that if we choose rational numbers x getting closer and closer to \sqrt{2}, the values of 2^x get closer and closer to some number L. We define that number L to be 2^{\sqrt{2}}.

Values of 2^x for a List of Rational Numbers Approximating \sqrt{2}
\mathbf{x} 1.4 1.41 1.414 1.4142 1.41421 1.414213
\mathbf{2^x} 2.639 2.65737 2.66475 2.665119 2.665138 2.665143

Bacterial Growth

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

Solution

f(4)=900; \, f(10)=24,300.

The Number e

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 P dollars in a savings account with an annual interest rate r, compounded annually. The amount of money after 1 year is

A(1)=P+rP=P(1+r).

The amount of money after 2 years is

A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P(1+r)^2.

More generally, the amount after t years is

A(t)=P(1+r)^t.

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

A(\frac{1}{2})=P+(\frac{r}{2})P=P(1+(\frac{r}{2})).

The amount of money after 1 year is

A(1)=A(\frac{1}{2})+(\frac{r}{2})A(\frac{1}{2})=P(1+\frac{r}{2})+\frac{r}{2}(P(1+\frac{r}{2}))=P(1+\frac{r}{2})^2.

After t years, the amount of money in the account is

A(t)=P(1+\frac{r}{2})^{2t}.

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

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

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

(1+\frac{r}{n})^{nt}=(1+\frac{1}{m})^{mrt},

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

Values of (1+\frac{1}{m})^m as m \to \infty
\mathbf{m} 10 100 1000 10,000 100,000 1,000,000
\mathbf{(1+\frac{1}{m})^m} 2.5937 2.7048 2.71692 2.71815 2.718268 2.718280

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

e \approx 2.718282.

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.

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

Since e>1, we know e^x is increasing on (−\infty ,\infty). In (Figure), we show a graph of f(x)=e^x along with a tangent line to the graph of at x=0. We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of f at x=a is a line that passes through the point (a,f(a)) and has the same "slope" as f at that point. 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 simplest exponential function to use in many instances.

Compounding Interest

If \$750 is invested in an account at an annual interest rate of 4\%, compounded continuously, find a formula for the amount of money in the account after t years. Find the amount of money after 30 years.

[reveal-answer q="505690″]Show Answer[/reveal-answer]
[hidden-answer a="505690″]A(t)=750e^{0.04t}. After 30 years, there will be approximately \$2,490.09.

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 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 if and only if b^y=x.

For example,

\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 \, \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.

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

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

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

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)).

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

Rule: 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}

Solving Equations Involving Exponential Functions

Solution

  1. 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.

  2. 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.

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

Solution

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

Solving Equations Involving Logarithmic Functions

Solution

  1. 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=1/e^4.

  2. 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}.

  3. 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, 2/x^5=1, which implies x=\sqrt[5]{2}. We should then check for any extraneous solutions.

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

Solution

x=\frac{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.

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}.

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

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)·(\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^{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.

□

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.

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

Chapter Opener: The Richter Scale for Earthquakes

A photograph of an earthquake fault.
Figure 6: (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 = 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}(\frac{A_1}{A_2})=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}(\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.

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

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 e^x and e^{−x}. 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 y-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

A photograph of a spider web collecting dew drops.
Figure 7. The shape of a strand of silk in a spider's web can be described in terms of a hyperbolic function. The same shape applies to a chain or cable hanging from two supports with only its own weight. (credit: "Mtpaley", Wikimedia Commons)

Definition

Hyperbolic cosine

\cosh x=\large \frac{e^x+e^{−x}}{2}

Hyperbolic sine

\sinh x=\large \frac{e^x-e^{−x}}{2}

Hyperbolic tangent

\tanh x=\large \frac{\sinh x}{\cosh x} \normalsize = \large \frac{e^x-e^{−x}}{e^x+e^{−x}}

Hyperbolic cosecant

\text{csch} \, x=\large \frac{1}{\sinh x} \normalsize = \large \frac{2}{e^x-e^{−x}}

Hyperbolic secant

\text{sech} \, x=\large \frac{1}{\cosh x} \normalsize = \large \frac{2}{e^x+e^{−x}}

Hyperbolic cotangent

\coth x=\large \frac{\cosh x}{\sinh x} \normalsize = \large \frac{e^x+e^{−x}}{e^x-e^{−x}}

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 \cosh(x) 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 h(x)=a \cosh(x/a)+c for certain constants a and c.

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

\cosh^2 t-\sinh^2 t=\large \frac{e^{2t}+2+e^{-2t}}{4}-\frac{e^{2t}-2+e^{-2t}}{4} \normalsize =1.

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

Identities Involving Hyperbolic Functions

The identity \cosh^2 t-\sinh^2 t, shown in (Figure), is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.

Evaluating Hyperbolic Functions

Simplify \cosh(2 \ln x).

Solution

(x^2+x^{-2})/2

Inverse Hyperbolic Functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except \cosh x and \text{sech} \, x. If we restrict the domains of these two functions to the interval [0,\infty), then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

Definition

Inverse Hyperbolic Functions

\begin{array}{cccc}\sinh^{-1} x=\text{arcsinh } x=\ln(x+\sqrt{x^2+1})\hfill & & & \cosh^{-1} x=\text{arccosh } x=\ln(x+\sqrt{x^2-1})\hfill \\ \tanh^{-1} x=\text{arctanh } x=\frac{1}{2}\ln\big(\frac{1+x}{1-x}\big)\hfill & & & \coth^{-1} x=\text{arccot } x=\frac{1}{2}\ln\big(\frac{x+1}{x-1}\big)\hfill \\ \text{sech}^{-1} x=\text{arcsech } x=\ln\big(\frac{1+\sqrt{1-x^2}}{x}\big)\hfill & & & \text{csch}^{-1} x=\text{arccsch } x=\ln\big(\frac{1}{x}+\frac{\sqrt{1+x^2}}{|x|}\big)\hfill \end{array}

Let's look at how to derive the first equation. The others follow similarly. Suppose y=\sinh^{-1} x. Then, x=\sinh y and, by the definition of the hyperbolic sine function, x=\frac{e^y-e^{−y}}{2}. Therefore,

e^y-2x-e^{−y}=0.

Multiplying this equation by e^y, we obtain

e^{2y}-2xe^y-1=0.

This can be solved like a quadratic equation, with the solution

e^y=\large \frac{2x \pm \sqrt{4x^2+4}}{2} \normalsize =x \pm \sqrt{x^2+1}.

Since e^y>0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

y=\ln(x+\sqrt{x^2+1}).

Evaluating Inverse Hyperbolic Functions

Evaluate each of the following expressions.

\sinh^{-1}(2)

\tanh^{-1}(1/4)

Solution

\sinh^{-1}(2)=\ln(2+\sqrt{2^2+1})=\ln(2+\sqrt{5}) \approx 1.4436

\tanh^{-1}(1/4)=\frac{1}{2}\ln(\frac{1+1/4}{1-1/4})=\frac{1}{2}\ln(\frac{5/4}{3/4})=\frac{1}{2}\ln(\frac{5}{3}) \approx 0.2554

Evaluate \tanh^{-1}(1/2).

Solution

\frac{1}{2}\ln(3) \approx 0.5493.

Key Concepts

For the following exercises, evaluate the given exponential functions as indicated, accurate to two significant digits after the decimal.

1. f(x)=5^x a. x=3 b. x=\frac{1}{2} c. x=\sqrt{2}

Solution

a. 125 b. 2.24 c. 9.74

3. f(x)=10^x a. x=-2 b. x=4 c. x=\frac{5}{3}

Solution

a. 0.01 b. 10,000 c. 46.42

For the following exercises, match the exponential equation to the correct graph.

  1. y=4^{−x}
  2. y=3^{x-1}
  3. y=2^{x+1}
  4. y=(\frac{1}{2})^x+2
  5. y=−3^{−x}
  6. y=1-5^x

5. An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -2 to 8. The graph is of a decreasing curved function. The function decreases until it approaches the line

6. An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -9 to 2. The graph is of a function that starts slightly below the line

7. An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, (1/3)). Another point of the graph is at (1, 1).

8. An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved decreasing function that decreases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, 1). Another point of the graph is at (-1, 4).

9. An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that increases until it comes close the x axis without touching it. There is no x intercept and the y intercept is at the point (0, -1). Another point of the graph is at (-1, -3).

10. An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, 2). Another point of the graph is at (-1, 1).

For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

11. f(x)=e^x+2

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the line

Solution

Domain: all real numbers, Range: (2,\infty), Horizontal asymptote at y=2

12. f(x)=−2^x

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a function that starts slightly below the x axis and begins decreasing rapidly in a curve. There is no x intercept and y intercept is at the point (0, -1).

13. f(x)=3^{x+1}

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the x axis and begins increasing rapidly. There is no x intercept and the y intercept is at the point (0, 3). Another point of the graph is at (-1, 1).

Solution

Domain: all real numbers, Range: (0,\infty), Horizontal asymptote at y=0

14. f(x)=4^x-1

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that starts slightly above the line

15. f(x)=1-2^{−x}

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved increasing function that increases until it comes close the line

Solution

Domain: all real numbers, Range: (−\infty ,1), Horizontal asymptote at y=1

16. f(x)=5^{x+1}+2

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -1 to 9. The graph is of a curved increasing function that starts slightly above the line

17. f(x)=e^{−x}-1

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of a curved decreasing function that decreases until it comes close the line

Solution

Domain: all real numbers, Range: (-1,\infty ), Horizontal asymptote at y=-1

For the following exercises, write the equation in equivalent exponential form.

18. \log_3 81=4

19. \log_8 2=\frac{1}{3}

Solution

8^{1/3}=2

20. \log_5 1=0

21. \log_5 25=2

Solution

5^2=25

22. \log 0.1=-1

23. \ln(\frac{1}{e^3})=-3

Solution

e^{-3}=\frac{1}{e^3}

24. \log_9 3=0.5

25. \ln 1=0

Solution

e^0=1

For the following exercises, write the equation in equivalent logarithmic form.

26. 2^3=8

27. 4^{-2}=\frac{1}{16}

Solution

\log_4(\frac{1}{16})=-2

28. 10^2=100

29. 9^0=1

Solution

\log_9 1=0

30. (\frac{1}{3})^3=\frac{1}{27}

31. \sqrt[3]{64}=4

Solution

\log_{64} 4=\frac{1}{3}

32. e^x=y

33. 9^y=150

Solution

\log_9 150=y

34. b^3=45

35. 4^{-3/2}=0.125

Solution

\log_4 0.125=-\frac{3}{2}

For the following exercises, use the given graphs of the logarithmic functions to determine their domain, range, and vertical asymptote.

36. f(x)=3+\ln x

An image of a graph. The x axis runs from -10 to 10 and the y axis runs from -10 to 10. The graph is of an increasing curved function which starts slightly to the right of the y axis. There is no y intercept and the x intercept is at the approximate point (0.05, 0). Another point on the graph is (1, 3).

37. f(x)=\ln(x-1)

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of an increasing curved function which starts slightly to the right of the vertical line

38. f(x)=\ln(−x)

An image of a graph. The x axis runs from -9 to 1 and the y axis runs from -5 to 5. The graph is a curved decreasing function that approaches the y axis without touching it. There is no y intercept and the x intercept is at (-1, 0).

39. f(x)=1-\ln x

An image of a graph. The x axis runs from -1 to 9 and the y axis runs from -5 to 5. The graph is of a decreasing curved function which starts slightly to the right of the y axis. There is no y intercept and the x intercept is at the point (e, 0).

40. f(x)=\log x-1

An image of a graph. The x axis runs from -5 to 20 and the y axis runs from -10 to 5. The graph is of an increasing curved function which starts slightly to the right of the y axis. There is no y intercept and the x intercept is at the point (10, 0).

41. f(x)=\ln(x+1)

An image of a graph. The x axis runs from -5 to 5 and the y axis runs from -5 to 5. The graph is of an increasing curved function which starts slightly to the right of the vertical line

For the following exercises, use properties of logarithms to write the expressions as a sum, difference, and/or product of logarithms.

42. \log x^4y

43. \log_3 \frac{9a^3}{b}

Solution

2+3\log_3 a-\log_3 b

44. \ln a\sqrt[3]{b}

45. \log_5 \sqrt{125xy^3}

Solution

\frac{3}{2}+\frac{1}{2}\log_5 x+\frac{3}{2}\log_5 y

46. \log_4 \frac{\sqrt[3]{xy}}{64}

47. \ln(\frac{6}{\sqrt{e^3}})

Solution

-\frac{3}{2}+\ln 6

For the following exercises, solve the exponential equation exactly.

48. 5^x=125

49. e^{3x}-15=0

Solution

\frac{\ln 15}{3}

50. 8^x=4

51. 4^{x+1}-32=0

Solution

\frac{3}{2}

52. 3^{x/14}=\frac{1}{10}

53. 10^x=7.21

Solution

\log 7.21

54. 4·2^{3x}-20=0

55. 7^{3x-2}=11

Solution

\frac{2}{3}+\frac{\log 11}{3\log 7}

For the following exercises, solve the logarithmic equation exactly, if possible.

56. \log_3 x=0

57. \log_5 x=-2

Solution

x=\frac{1}{25}

58. \log_4 (x+5)=0

59. \log(2x-7)=0

Solution

x=4

60. \ln\sqrt{x+3}=2

61. \log_6 (x+9)+\log_6 x=2

Solution

x=3

62. \log_4(x+2)-\log_4(x-1)=0

63. \ln x+\ln (x-2)=\ln 4

Solution

1+\sqrt{5}

For the following exercises, use the change-of-base formula and either base 10 or base e to evaluate the given expressions. Answer in exact form and in approximate form, rounding to four decimal places.

64. \log_5 47

65. \log_7 82

Solution

\frac{\ln 82}{\ln 7} \approx 2.2646

66. \log_6 103

67. \log_{0.5} 211

Solution

\frac{\ln 211}{\ln 0.5} \approx -7.7211

68. {\text{log}}_{2}\pi

69. \log_{0.2} 0.452

Solution

\frac{\ln 0.452}{\ln 0.2} \approx 0.4934

70.Rewrite the following expressions in terms of exponentials and simplify.

a. 2\cosh(\ln x)

b. \cosh 4x+\sinh 4x

c. \cosh 2x-\sinh 2x

d. \ln(\cosh x+\sinh x)+\ln(\cosh x-\sinh x)

71. [T] The number of bacteria N in a culture after t days can be modeled by the function N(t)=1300·2^{t/4}. Find the number of bacteria present after 15 days.

Solution

\approx 17,491

73. [T] The accumulated amount A of a $100,000 investment whose interest compounds continuously for t years is given by A(t)=100,000·e^{0.055t}. Find the amount A accumulated in 5 years.

Solution

Approximately $131,653 is accumulated in 5 years.

74. [T] An investment is compounded monthly, quarterly, or yearly and is given by the function A=P(1+\frac{j}{n})^{nt}, where A is the value of the investment at time t, \, P is the initial principle that was invested, j is the annual interest rate, and n is the number of time the interest is compounded per year. Given a yearly interest rate of 3.5% and an initial principle of $100,000, find the amount A accumulated in 5 years for interest that is compounded a. daily, b., monthly, c. quarterly, and d. yearly.

Solution

i. a. pH = 8 b. Base ii. a. pH = 3 b. Acid iii. a. pH = 4 b. Acid

77. [T] According to the World Bank, at the end of 2013 (t=0) the U.S. population was 316 million and was increasing according to the following model:

P(t)=316e^{0.0074t},

where P is measured in millions of people and t is measured in years after 2013.

  1. Based on this model, what will be the population of the United States in 2020?
  2. Determine when the U.S. population will be twice what it is in 2013.

Solution

a. \approx 333 million b. 94 years from 2013, or in 2107

79. [T] A bacterial colony grown in a lab is known to double in number in 12 hours. Suppose, initially, there are 1000 bacteria present.

  1. Use the exponential function Q=Q_0e^{kt} to determine the value k, which is the growth rate of the bacteria. Round to four decimal places.
  2. Determine approximately how long it takes for 200,000 bacteria to grow.

Solution

a. k \approx 0.0578 b. \approx 92 hours

80. [T] The rabbit population on a game reserve doubles every 6 months. Suppose there were 120 rabbits initially.

  1. Use the exponential function P=P_0 a^t to determine the growth rate constant a. Round to four decimal places.
  2. Use the function in part a. to determine approximately how long it takes for the rabbit population to reach 3500.

81. [T] The 1906 earthquake in San Francisco had a magnitude of 8.3 on the Richter scale. At the same time, in Japan, an earthquake with magnitude 4.9 caused only minor damage. Approximately how much more energy was released by the San Francisco earthquake than by the Japanese earthquake?

Solution

The San Francisco earthquake had 10^{3.4} or \approx 2512 times more energy than the Japan earthquake.

Find the Exact Solution to the Exponential Equation

Source: https://opentextbc.ca/calculusv1openstax/chapter/exponential-and-logarithmic-functions/