Solve Exponential and Logarithmic Equations

Lynn Marecek

Exponential and Logarithmic Functions

Solve Exponential and Logarithmic Equations

Learning Objectives

By the end of this section, you will be able to:

Solve logarithmic equations using the properties of logarithms
Solve exponential equations using logarithms
Use exponential models in applications

Before you get started, take this readiness quiz.

Solve: ${x}^{2}=16.$

If you missed this problem, review (Figure).
Solve: ${x}^{2}-5x+6=0.$

If you missed this problem, review (Figure).
Solve: $x\left(x+6\right)=2x+5.$

If you missed this problem, review (Figure).

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.

If our equation has two logarithms we can use a property that says that if ${\text{log}}_{a}M={\text{log}}_{a}N$ then it is true that $M=N.$ This is the One-to-One Property of Logarithmic Equations.

One-to-One Property of Logarithmic Equations

For $M>0,N>0,\phantom{\rule{0.2em}{0ex}}\text{a}\text{>}0,$ and $\text{a}\ne 1$ is any real number:

$\text{If}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M={\text{log}}_{a}N,\phantom{\rule{0.2em}{0ex}}\text{then}\phantom{\rule{0.2em}{0ex}}M=N.$

To use this property, we must be certain that both sides of the equation are written with the same base.

Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.

Solve: $2{\text{log}}_{5}x={\text{log}}_{5}81.$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill 2\phantom{\rule{0.2em}{0ex}}{\mathrm{log}}_{5}x& =\hfill & {\mathrm{log}}_{5}81\hfill \\ \text{Use the Power Property.}\hfill & & & & & \hfill {\mathrm{log}}_{5}{x}^{2}& =\hfill & {\mathrm{log}}_{5}81\hfill \\ \text{Use the One-to-One Property, if}\phantom{\rule{0.2em}{0ex}}{\mathrm{log}}_{a}M={\mathrm{log}}_{a}N,\hfill & & & & & \hfill {x}^{2}& =\hfill & 81\hfill \\ \text{then}\phantom{\rule{0.2em}{0ex}}M=N.\hfill & & & & & \\ \text{Solve using the Square Root Property.}\hfill & & & & & \hfill x& =\hfill & \text{±}9\hfill \end{array}\hfill \\ \text{We eliminate}\phantom{\rule{0.2em}{0ex}}x=-9\phantom{\rule{0.2em}{0ex}}\text{as we cannot take the logarithm}\phantom{\rule{3em}{0ex}}x=9,\phantom{\rule{0.2em}{0ex}}\overline{)x=-9}\hfill & & & & & \\ \text{of a negative number.}\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=9\hfill & & & & & \hfill 2{\mathrm{log}}_{5}x& =\hfill & {\mathrm{log}}_{5}81\hfill \\ & & & & & \hfill 2{\mathrm{log}}_{5}9& \stackrel{?}{=}\hfill & {\mathrm{log}}_{5}81\hfill \\ & & & & & \hfill {\mathrm{log}}_{5}{9}^{2}& \stackrel{?}{=}\hfill & {\mathrm{log}}_{5}81\hfill \\ & & & & & \hfill {\mathrm{log}}_{5}81& =\hfill & {\mathrm{log}}_{5}81✓\hfill \end{array}\hfill & & & & & \end{array}$

Solve: $2{\text{log}}_{3}x={\text{log}}_{3}36$

$x=6$

Solve: $3\text{log}\phantom{\rule{0.2em}{0ex}}x=\text{log}64$

$x=4$

Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.

Solve: ${\text{log}}_{3}x+{\text{log}}_{3}\left(x-8\right)=2.$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill {\mathrm{log}}_{3}x+{\mathrm{log}}_{3}\left(x-8\right)& =\hfill & 2\hfill \\ \text{Use the Product Property,}\phantom{\rule{0.2em}{0ex}}{\mathrm{log}}_{a}M+{\mathrm{log}}_{a}N={\mathrm{log}}_{a}M\cdot N.\hfill & & & & & \hfill {\mathrm{log}}_{3}x\left(x-8\right)& =\hfill & 2\hfill \\ \text{Rewrite in exponential form.}\hfill & & & & & \hfill {3}^{2}& =\hfill & x\left(x-8\right)\hfill \\ \text{Simplify.}\hfill & & & & & \hfill 9& =\hfill & {x}^{2}-8x\hfill \\ \text{Subtract 9 from each side.}\hfill & & & & & \hfill 0& =\hfill & {x}^{2}-8x-9\hfill \\ \text{Factor.}\hfill & & & & & \hfill 0& =\hfill & \left(x-9\right)\left(x+1\right)\hfill \\ \text{Use the Zero-Product Property.}\hfill & & & & & \hfill x-9& =\hfill & 0,\phantom{\rule{1em}{0ex}}x+1=0\hfill \end{array}\hfill \\ \text{Solve each equation.}\phantom{\rule{22em}{0ex}}x=9,\phantom{\rule{7em}{0ex}}\overline{)x=-1}\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=-1\hfill & & & & & \hfill {\mathrm{log}}_{3}x+{\mathrm{log}}_{3}\left(x-8\right)& =\hfill & 2\hfill \\ & & & & & \hfill {\mathrm{log}}_{3}\left(-1\right)+{\mathrm{log}}_{3}\left(-1-8\right)& \stackrel{?}{=}\hfill & 2\hfill \end{array}\hfill & & & & & \\ \text{We cannot take the log of a negative number.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=9\hfill & & & & & \hfill {\mathrm{log}}_{3}x+{\mathrm{log}}_{3}\left(x-8\right)& =\hfill & 2\hfill \\ & & & & & \hfill {\mathrm{log}}_{3}9+{\mathrm{log}}_{3}\left(9-8\right)& \stackrel{?}{=}\hfill & 2\hfill \\ & & & & & \hfill 2+0& \stackrel{?}{=}\hfill & 2\hfill \\ & & & & & \hfill 2& =\hfill & 2✓\hfill \end{array}\hfill & & & & & \end{array}$

Solve: ${\text{log}}_{2}x+{\text{log}}_{2}\left(x-2\right)=3$

$x=4$

Solve: ${\text{log}}_{2}x+{\text{log}}_{2}\left(x-6\right)=4$

$x=8$

When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.

Solve: ${\text{log}}_{4}\left(x+6\right)-{\text{log}}_{4}\left(2x+5\right)=\text{−}{\text{log}}_{4}x.$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill {\text{log}}_{4}\left(x+6\right)-{\text{log}}_{4}\left(2x+5\right)& =\hfill & \text{−}{\text{log}}_{4}x\hfill \\ \begin{array}{c}\text{Use the Quotient Property on the left side and the Power}\hfill \\ \text{Property on the right.}\hfill \end{array}\hfill & & & & & \hfill {\text{log}}_{4}\left(\frac{x+6}{2x+5}\right)& =\hfill & {\text{log}}_{4}{x}^{-1}\hfill \\ \text{Rewrite}\phantom{\rule{0.2em}{0ex}}{x}^{-1}=\frac{1}{x}.\hfill & & & & & \hfill {\text{log}}_{4}\left(\frac{x+6}{2x+5}\right)& =\hfill & {\text{log}}_{4}\frac{1}{x}\hfill \\ \begin{array}{c}\text{Use the One-to-One Property, if}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M={\text{log}}_{a}N,\hfill \\ \text{then}\phantom{\rule{0.2em}{0ex}}M=N.\hfill \end{array}\hfill & & & & & \hfill \frac{x+6}{2x+5}& =\hfill & \frac{1}{x}\hfill \\ \text{Solve the rational equation.}\hfill & & & & & \hfill x\left(x+6\right)& =\hfill & 2x+5\hfill \\ \text{Distribute.}\hfill & & & & & \hfill {x}^{2}+6x& =\hfill & 2x+5\hfill \\ \text{Write in standard form.}\hfill & & & & & \hfill {x}^{2}+4x-5& =\hfill & 0\hfill \\ \text{Factor.}\hfill & & & & & \hfill \left(x+5\right)\left(x-1\right)& =\hfill & 0\hfill \end{array}\hfill \\ \text{Use the Zero-Product Property.}\phantom{\rule{17em}{0ex}}x+5=0,\phantom{\rule{2em}{0ex}}x-1=0\hfill & & & & & \\ \text{Solve each equation.}\phantom{\rule{22.9em}{0ex}}\overline{)x=-5},\phantom{\rule{2.8em}{0ex}}x=1\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \text{We leave the check for you.}\hfill & & & & & \end{array}$

Solve: $\text{log}\left(x+2\right)-\text{log}\left(4x+3\right)=\text{−}\text{log}\phantom{\rule{0.2em}{0ex}}x.$

$x=3$

Solve: $\text{log}\left(x-2\right)-\text{log}\left(4x+16\right)=\text{log}\frac{1}{x}.$

$x=8$

Solve Exponential Equations Using Logarithms

In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal.

It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated.

Solve ${5}^{x}=11.$ Find the exact answer and then approximate it to three decimal places.

$\begin{array}{cccccccc}& & & & & \hfill {5}^{x}& =\hfill & 11\hfill \\ \text{Since the exponential is isolated, take the logarithm of both sides.}\hfill & & & & & \hfill \text{log}{5}^{x}& =\hfill & \text{log}11\hfill \\ \text{Use the Power Property to get the}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{as a factor, not an exponent.}\hfill & & & & & \hfill x\text{log}5& =\hfill & \text{log}11\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}x.\phantom{\rule{0.2em}{0ex}}\text{Find the exact answer.}\hfill & & & & & \hfill x& =\hfill & \frac{\text{log}11}{\text{log}5}\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill x& \approx \hfill & 1.490\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{5}^{1}=5\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{5}^{2}=25,\phantom{\rule{0.2em}{0ex}}\text{does it makes sense that}\phantom{\rule{0.2em}{0ex}}{5}^{1.490}\approx 11?\hfill & & & & & \end{array}$

Solve ${7}^{x}=43.$ Find the exact answer and then approximate it to three decimal places.

$x=\frac{\text{log}43}{\text{log}7}\approx 1.933$

Solve ${8}^{x}=98.$ Find the exact answer and then approximate it to three decimal places.

$x=\frac{\text{log}98}{\text{log}8}\approx 2.205$

When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base e, we use the natural logarithm.

Solve $3{e}^{x+2}=24.$ Find the exact answer and then approximate it to three decimal places.

$\begin{array}{cccccccc}& & & & & \hfill 3{e}^{x+2}& =\hfill & 24\hfill \\ \text{Isolate the exponential by dividing both sides by 3.}\hfill & & & & & \hfill {e}^{x+2}& =\hfill & 8\hfill \\ \text{Take the natural logarithm of both sides.}\hfill & & & & & \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{x+2}& =\hfill & \text{ln}8\hfill \\ \text{Use the Power Property to get the}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{as a factor, not an exponent.}\hfill & & & & & \hfill \left(x+2\right)\text{ln}\phantom{\rule{0.2em}{0ex}}e& =\hfill & \text{ln}8\hfill \\ \text{Use the property}\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}e=1\phantom{\rule{0.2em}{0ex}}\text{to simplify.}\hfill & & & & & \hfill x+2& =\hfill & \text{ln}8\hfill \\ \text{Solve the equation. Find the exact answer.}\hfill & & & & & \hfill x& =\hfill & \text{ln}8-2\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill x& \approx \hfill & 0.079\hfill \end{array}$

Solve $2{e}^{x-2}=18.$ Find the exact answer and then approximate it to three decimal places.

$x=\text{ln}9+2\approx 4.197$

Solve $5{e}^{2x}=25.$ Find the exact answer and then approximate it to three decimal places.

$x=\frac{\text{ln}5}{2}\approx 0.805$

Use Exponential Models in Applications

In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications.

We will again use the Compound Interest Formulas and so we list them here for reference.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A is:

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\begin{array}{}\\ \\ \phantom{\rule{5em}{0ex}}A=P{\left(1+\frac{r}{n}\right)}^{nt}\hfill & & & & & \text{when compounded}\phantom{\rule{0.2em}{0ex}}n\phantom{\rule{0.2em}{0ex}}\text{times a year.}\hfill \\ \phantom{\rule{5em}{0ex}}A=P{e}^{rt}\hfill & & & & & \text{when compounded continuously.}\hfill \end{array}

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Jermael’s parents put ?10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth ?50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?

*** QuickLaTeX cannot compile formula:
\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill A& =\hfill & \text{?}50,000\hfill \\ & & & & & \hfill P& =\hfill & \text{?}10,000\hfill \\ \text{Identify the variables in the formula.}\hfill & & & & & \hfill r& =\hfill & ?\hfill \\ & & & & & \hfill t& =\hfill & 17\phantom{\rule{0.2em}{0ex}}\text{years}\hfill \\ & & & & & \hfill A& =\hfill & P{e}^{rt}\hfill \\ \text{Substitute the values into the formula.}\hfill & & & & & \hfill 50,000& =\hfill & 10,000{e}^{r·17}\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}r.\text{Divide each side by 10,000.}\hfill & & & & & \hfill 5& =\hfill & {e}^{17r}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \hfill \text{ln}5& =\hfill & \text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{17r}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \hfill \text{ln}5& =\hfill & 17r\text{ln}\phantom{\rule{0.2em}{0ex}}e\hfill \\ \text{Simplify.}\hfill & & & & & \hfill \text{ln}5& =\hfill & 17r\hfill \\ \text{Divide each side by 17.}\hfill & & & & & \hfill \frac{\text{ln}5}{17}& =\hfill & r\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill r& \approx \hfill & 0.095\hfill \\ \text{Convert to a percentage.}\hfill & & & & & \hfill r& \approx \hfill & 9.5\text{%}\hfill \end{array}\hfill \\ \\ \phantom{\rule{20em}{0ex}}\text{They need the rate of growth to be approximately}\phantom{\rule{0.2em}{0ex}}9.5\text{%}.\hfill \end{array}

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Hector invests $\text{?}10,000$ at age 21. He hopes the investments will be worth $\text{?}150,000$ when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?

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Rachel invests $\text{?}15,000$ at age 25. She hopes the investments will be worth $\text{?}90,000$ when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

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r\approx 11.9\text{%}

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We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula $A={A}_{0}{e}^{kt}.$ Exponential growth has a positive rate of growth or growth constant, $k$ , and exponential decay has a negative rate of growth or decay constant, k.

Exponential Growth and Decay

For an original amount, ${A}_{0},$ that grows or decays at a rate, k, for a certain time, t, the final amount, A, is:

$A={A}_{0}{e}^{kt}$

We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.

Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria.

$\begin{array}{cccccc}\text{Identify the variables in the formula.}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & 300\hfill \\ \hfill {A}_{0}& =\hfill & 100\hfill \\ \hfill k& =\hfill & ?\hfill \\ \hfill t& =\hfill & 3\phantom{\rule{0.2em}{0ex}}\text{hours}\hfill \\ \hfill A& =\hfill & {A}_{0}{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute the values in the formula.}\hfill & & & & & \phantom{\rule{0.1em}{0ex}}300=100{e}^{k·3}\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}k.\phantom{\rule{0.2em}{0ex}}\text{Divide each side by 100.}\hfill & & & & & \phantom{\rule{1.15em}{0ex}}3={e}^{3k}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \phantom{\rule{0.4em}{0ex}}\text{ln}3=\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{3k}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \phantom{\rule{0.4em}{0ex}}\text{ln}3=3k\text{ln}\phantom{\rule{0.2em}{0ex}}e\hfill \\ \text{Simplify.}\hfill & & & & & \phantom{\rule{0.4em}{0ex}}\text{ln}3=3k\hfill \\ \text{Divide each side by 3.}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}\frac{\text{ln}3}{3}=k\hfill \\ \text{Approximate the answer.}\hfill & & & & & \phantom{\rule{1.15em}{0ex}}k\approx 0.366\hfill \\ \\ \\ \begin{array}{c}\text{We use this rate of growth to predict the number of}\hfill \\ \text{bacteria there will be in 24 hours.}\hfill \end{array}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & ?\hfill \\ \hfill {A}_{0}& =\hfill & 100\hfill \\ \hfill k& =\hfill & \frac{\text{ln}3}{3}\hfill \\ \hfill t& =\hfill & 24\phantom{\rule{0.2em}{0ex}}\text{hours}\hfill \\ \hfill A& =\hfill & {A}_{0}{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute in the values.}\hfill & & & & & A=100{e}^{\frac{\text{ln}3}{3}·24}\hfill \\ \text{Evaluate.}\hfill & & & & & A\approx 656,100\hfill \\ & & & & & \text{At this rate of growth, they can expect 656,100 bacteria.}\hfill \end{array}$

Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

There will be 62,500 bacteria.

Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?

There will be 5,870,061 bacteria.

Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance.

Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions.

The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?

This problem requires two main steps. First we must find the decay constant k. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k. Then we use that value of k to help us find the amount of sample that will be left in 500 years.

$\begin{array}{cccccc}\text{Identify the variables in the formula.}\hfill & & & & & \phantom{\rule{0.2em}{0ex}}\begin{array}{ccc}\hfill A& =\hfill & 50\hfill \\ \hfill {A}_{0}& =\hfill & 100\hfill \\ \hfill k& =\hfill & ?\hfill \\ \hfill t& =\hfill & 1590\text{years}\hfill \\ \hfill A& =\hfill & {A}_{0}{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute the values in the formula.}\hfill & & & & & \phantom{\rule{0.88em}{0ex}}50=100{e}^{k·1590}\hfill \\ \text{Solve for}\phantom{\rule{0.2em}{0ex}}k.\phantom{\rule{0.2em}{0ex}}\text{Divide each side by 100.}\hfill & & & & & \phantom{\rule{0.7em}{0ex}}0.5={e}^{1590k}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \text{ln}0.5=\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{1590k}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \text{ln}0.5=1590k\text{ln}\phantom{\rule{0.2em}{0ex}}e\hfill \\ \text{Simplify.}\hfill & & & & & \text{ln}0.5=1590k\hfill \\ \text{Divide each side by 1590.}\hfill & & & & & \frac{\text{ln}0.5}{1590}=k\phantom{\rule{0.2em}{0ex}}\text{exact answer}\hfill \\ \\ \\ \begin{array}{c}\text{We use this rate of growth to predict the amount}\hfill \\ \text{that will be left in 500 years.}\hfill \end{array}\hfill & & & & & \phantom{\rule{0.5em}{0ex}}\begin{array}{ccc}\hfill A& =\hfill & ?\hfill \\ \hfill {A}_{0}& =\hfill & 100\hfill \\ \hfill k& =\hfill & \frac{\text{ln}0.5}{1590}\hfill \\ \hfill t& =\hfill & 500\text{years}\hfill \\ \hfill A& =\hfill & {A}_{0}{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute in the values.}\hfill & & & & & A=100{e}^{\frac{\text{ln}0.5}{1590}·500}\hfill \\ \text{Evaluate.}\hfill & & & & & A\approx 80.4\phantom{\rule{0.2em}{0ex}}\text{mg}\hfill \\ & & & & & \begin{array}{c}\text{In 500 years there would be}\hfill \\ \text{approximately 80.4 mg remaining.}\hfill \end{array}\hfill \end{array}$

The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?

There will be 6.43 mg left.

The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?

There will be 31.5 mg left.

Access these online resources for additional instruction and practice with solving exponential and logarithmic equations.

Key Concepts

One-to-One Property of Logarithmic Equations: For $M>0,N>0,\phantom{\rule{0.2em}{0ex}}\text{a}\text{>}0,$ and $\text{a}\ne 1$ is any real number:

$\text{If}\phantom{\rule{0.2em}{0ex}}{\text{log}}_{a}M={\text{log}}_{a}N,\phantom{\rule{0.2em}{0ex}}\text{then}\phantom{\rule{0.2em}{0ex}}M=N.$

Compound Interest:

For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:

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\begin{array}{}\\ \\ \phantom{\rule{5em}{0ex}}A=P{\left(1+\frac{r}{n}\right)}^{nt}\hfill & & & & & \text{when compounded}\phantom{\rule{0.2em}{0ex}}n\phantom{\rule{0.2em}{0ex}}\text{times a year.}\hfill \\ \phantom{\rule{5em}{0ex}}A=P{e}^{rt}\hfill & & & & & \text{when compounded continuously.}\hfill \end{array}

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Exponential Growth and Decay: For an original amount, ${A}_{0}$ that grows or decays at a rate, r, for a certain time t, the final amount, A, is $A={A}_{0}{e}^{rt}.$

Section Exercises

Practice Makes Perfect

Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

${\text{log}}_{4}64=2{\text{log}}_{4}x$

$\text{log}49=2\text{log}\phantom{\rule{0.2em}{0ex}}x$

$x=7$

$3{\text{log}}_{3}x={\text{log}}_{3}27$

$3{\text{log}}_{6}x={\text{log}}_{6}64$

$x=4$

${\text{log}}_{5}\left(4x-2\right)={\text{log}}_{5}10$

${\text{log}}_{3}\left({x}^{2}+3\right)={\text{log}}_{3}4x$

$x=1,$ $x=3$

${\text{log}}_{3}x+{\text{log}}_{3}x=2$

${\text{log}}_{4}x+{\text{log}}_{4}x=3$

$x=8$

${\text{log}}_{2}x+{\text{log}}_{2}\left(x-3\right)=2$

${\text{log}}_{3}x+{\text{log}}_{3}\left(x+6\right)=3$

$x=3$

$\text{log}\phantom{\rule{0.2em}{0ex}}x+\text{log}\left(x+3\right)=1$

$\text{log}\phantom{\rule{0.2em}{0ex}}x+\text{log}\left(x-15\right)=2$

$x=20$

$\text{log}\left(x+4\right)-\text{log}\left(5x+12\right)=\text{−}\text{log}\phantom{\rule{0.2em}{0ex}}x$

$\text{log}\left(x-1\right)-\text{log}\left(x+3\right)=\text{log}\frac{1}{x}$

$x=3$

${\text{log}}_{5}\left(x+3\right)+{\text{log}}_{5}\left(x-6\right)={\text{log}}_{5}10$

${\text{log}}_{5}\left(x+1\right)+{\text{log}}_{5}\left(x-5\right)={\text{log}}_{5}7$

$x=6$

${\text{log}}_{3}\left(2x-1\right)={\text{log}}_{3}\left(x+3\right)+{\text{log}}_{3}3$

$\text{log}\left(5x+1\right)=\text{log}\left(x+3\right)+\text{log}2$

$x=\frac{5}{3}$

Solve Exponential Equations Using Logarithms

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

${3}^{x}=89$

${2}^{x}=74$

$x=\frac{\text{log}74}{\text{log}2}\approx 6.209$

${5}^{x}=110$

${4}^{x}=112$

$x=\frac{\text{log}112}{\text{log}4}\approx 3.404$

${e}^{x}=16$

${e}^{x}=8$

$x=\text{ln}8\approx 2.079$

${\left(\frac{1}{2}\right)}^{x}=6$

${\left(\frac{1}{3}\right)}^{x}=8$

$x=\frac{\text{log}8}{\text{log}\frac{1}{3}}\approx \text{−}1.893$

$4{e}^{x+1}=16$

$3{e}^{x+2}=9$

$x=\text{ln}3-2\approx \text{−}0.901$

$6{e}^{2x}=24$

$2{e}^{3x}=32$

$x=\frac{\text{ln}16}{3}\approx 0.924$

$\frac{1}{4}{e}^{x}=3$

$\frac{1}{3}{e}^{x}=2$

$x=\text{ln}6\approx 1.792$

${e}^{x+1}+2=16$

${e}^{x-1}+4=12$

$x=\text{ln}8+1\approx 3.079$

In the following exercises, solve each equation.

${3}^{3x+1}=81$

${6}^{4x-17}=216$

$x=5$

$\frac{{e}^{{x}^{2}}}{{e}^{14}}={e}^{5x}$

$\frac{{e}^{{x}^{2}}}{{e}^{x}}={e}^{20}$

$x=-4,x=5$

${\text{log}}_{a}64=2$

${\text{log}}_{a}81=4$

$a=3$

$\text{ln}\phantom{\rule{0.2em}{0ex}}x=-8$

$\text{ln}\phantom{\rule{0.2em}{0ex}}x=9$

$x={e}^{9}$

${\text{log}}_{5}\left(3x-8\right)=2$

${\text{log}}_{4}\left(7x+15\right)=3$

$x=7$

$\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{5x}=30$

$\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{6x}=18$

$x=3$

$3\text{log}\phantom{\rule{0.2em}{0ex}}x=\text{log}125$

$7{\text{log}}_{3}x={\text{log}}_{3}128$

$x=2$

${\text{log}}_{6}x+{\text{log}}_{6}\left(x-5\right)={\text{log}}_{6}24$

${\text{log}}_{9}x+{\text{log}}_{9}\left(x-4\right)={\text{log}}_{9}12$

$x=6$

${\text{log}}_{2}\left(x+2\right)-{\text{log}}_{2}\left(2x+9\right)=\text{−}{\text{log}}_{2}x$

${\text{log}}_{6}\left(x+1\right)-{\text{log}}_{6}\left(4x+10\right)={\text{log}}_{6}\frac{1}{x}$

$x=5$

In the following exercises, solve for x, giving an exact answer as well as an approximation to three decimal places.

${6}^{x}=91$

${\left(\frac{1}{2}\right)}^{x}=10$

$x=\frac{\text{log}10}{\text{log}\frac{1}{2}}\approx \text{−}3.322$

$7{e}^{x-3}=35$

$8{e}^{x+5}=56$

$x=\text{ln}7-5\approx \text{−}3.054$

Use Exponential Models in Applications

In the following exercises, solve.

Sung Lee invests $\text{?}5,000$ at age 18. He hopes the investments will be worth $\text{?}10,000$ when he turns 25. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?

Alice invests $\text{?}15,000$ at age 30 from the signing bonus of her new job. She hopes the investments will be worth $\text{?}30,000$ when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

$6.9\text{%}$

Coralee invests $\text{?}5,000$ in an account that compounds interest monthly and earns

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How long will it take for her money to double?

Simone invests $\text{?}8,000$ in an account that compounds interest quarterly and earns $5\text{%}.$ How long will it take for his money to double?

13.9 years

Researchers recorded that a certain bacteria population declined from 100,000 to 100 in 24 hours. At this rate of decay, how many bacteria will there be in 16 hours?

Researchers recorded that a certain bacteria population declined from 800,000 to 500,000 in 6 hours after the administration of medication. At this rate of decay, how many bacteria will there be in 24 hours?

122,070 bacteria

A virus takes 6 days to double its original population $\left(A=2{A}_{0}\right).$ How long will it take to triple its population?

A bacteria doubles its original population in 24 hours $\left(A=2{A}_{0}\right).$ How big will its population be in 72 hours?

8 times as large as the original population

Carbon-14 is used for archeological carbon dating. Its half-life is 5,730 years. How much of a 100-gram sample of Carbon-14 will be left in 1000 years?

Radioactive technetium-99m is often used in diagnostic medicine as it has a relatively short half-life but lasts long enough to get the needed testing done on the patient. If its half-life is 6 hours, how much of the radioactive material form a 0.5 ml injection will be in the body in 24 hours?

0.03 ml

Writing Exercises

Explain the method you would use to solve these equations: ${3}^{x+1}=81,$ ${3}^{x+1}=75.$ Does your method require logarithms for both equations? Why or why not?

What is the difference between the equation for exponential growth versus the equation for exponential decay?

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads solve logarithmic equations using the properties of logarithms, solve exponential equations using logarithms, and use exponential models in applications. The rest of the cells are blank.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Finding Composite and Inverse Functions

Find and Evaluate Composite Functions

In the following exercises, for each pair of functions, find ⓐ (f ∘ g)(x), ⓑ (g ∘ f)(x), and ⓒ (f · g)(x).

$f\left(x\right)=7x-2$ and

$g\left(x\right)=5x+1$

$f\left(x\right)=4x$ and

$g\left(x\right)={x}^{2}+3x$

ⓐ $4{x}^{2}+12x$ ⓑ $16{x}^{2}+12x$ ⓒ $4{x}^{3}+12{x}^{2}$

In the following exercises, evaluate the composition.

For functions

$f\left(x\right)=3{x}^{2}+2$ and

$g\left(x\right)=4x-3,$ find

ⓐ $\left(f\circ g\right)\left(-3\right)$

ⓑ $\left(g\circ f\right)\left(-2\right)$

For functions

$f\left(x\right)=2{x}^{3}+5$ and

$g\left(x\right)=3{x}^{2}-7,$ find

ⓐ $\left(f\circ g\right)\left(-1\right)$

ⓑ $\left(g\circ f\right)\left(-2\right)$

ⓐ $-123$ ⓑ 356 ⓒ 41

Determine Whether a Function is One-to-One

In the following exercises, for each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

$\left\{\left(-3,-5\right),\left(-2,-4\right),\left(-1,-3\right),\left(0,-2\right),$

$\left(-1,-1\right),\left(-2,0\right),\left(-3,1\right)\right\}$

$\left\{\left(-3,0\right),\left(-2,-2\right),\left(-1,0\right),\left(0,1\right),$

$\left(1,2\right),\left(2,1\right),\left(3,-1\right)\right\}$

Function; not one-to-one

$\left\{\left(-3,3\right),\left(-2,1\right),\left(-1,-1\right),\left(0,-3\right),$

$\left(1,-5\right),\left(2,-4\right),\left(3,-2\right)\right\}$

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ

This figure shows a line from (negative 6, negative 2) up to (negative 1, 3) and then down from there to (6, negative 4).

ⓑ

This figure shows a line from (6, 5) down to (0, negative 1) and then down from there to (5, negative 6).

ⓐ Function; not one-to-one ⓑ Not a function

ⓐ

This figure shows a curved line from (negative 6, negative 2) up to the origin and then continuing up from there to (6, 2).

ⓑ

This figure shows a circle of radius 2 with center at the origin.

Find the Inverse of a Function

In the following exercise, find the inverse of the function. Determine the domain and range of the inverse function.

$\left\{\left(-3,10\right),\left(-2,5\right),\left(-1,2\right),\left(0,1\right)\right\}$

Inverse function: $\left\{\left(10,-3\right),\left(5,-2\right),\left(2,-1\right),\left(1,0\right)\right\}.$ Domain: $\left\{1,2,5,10\right\}.$ Range: $\left\{-3,-2,-1,0\right\}.$

In the following exercise, graph the inverse of the one-to-one function shown.

This figure shows a line segment from (negative 4, negative 2) up to (negative 2, 1) then up to (2, 2) and then up to (3, 4).

In the following exercises, verify that the functions are inverse functions.

$f\left(x\right)=3x+7$ and

$g\left(x\right)=\frac{x-7}{3}$

$g\left(f\left(x\right)\right)=x,$ and $f\left(g\left(x\right)\right)=x,$ so they are inverses.

$f\left(x\right)=2x+9$ and

$g\left(x\right)=\frac{x+9}{2}$

In the following exercises, find the inverse of each function.

$f\left(x\right)=6x-11$

${f}^{-1}\left(x\right)=\frac{x+11}{6}$

$f\left(x\right)={x}^{3}+13$

$f\left(x\right)=\frac{1}{x+5}$

${f}^{-1}\left(x\right)=\frac{1}{x}-5$

$f\left(x\right)=\sqrt[5]{x-1}$

Evaluate and Graph Exponential Functions

Graph Exponential Functions

In the following exercises, graph each of the following functions.

$f\left(x\right)={4}^{x}$

This figure shows an exponential line passing through the points (negative 1, 1 over 4), (0, 1), and (1, 4).

$f\left(x\right)={\left(\frac{1}{5}\right)}^{x}$

$g\left(x\right)={\left(0.75\right)}^{x}$

This figure shows an exponential line passing through the points (negative 1, 4 over 3), (0, 1), and (1, 3 over 4).

$g\left(x\right)={3}^{x+2}$

$f\left(x\right)={\left(2.3\right)}^{x}-3$

This figure shows an exponential line passing through the points (negative 1, negative 59 over 23), (0, negative 2), and (1, negative7 over 10).

$f\left(x\right)={e}^{x}+5$

$f\left(x\right)=\text{−}{e}^{x}$

This figure shows an exponential line passing through the points (negative 1, negative 1 over e), (0, negative 1), and (1, negative e).

Solve Exponential Equations

In the following exercises, solve each equation.

${3}^{5x-6}=81$

${2}^{{x}^{2}}=16$

$x=-2,x=2$

${9}^{x}=27$

${5}^{{x}^{2}+2x}=\frac{1}{5}$

$x=-1$

${e}^{4x}·{e}^{7}={e}^{19}$

$\frac{{e}^{{x}^{2}}}{{e}^{15}}={e}^{2x}$

$x=-3,x=5$

Use Exponential Models in Applications

In the following exercises, solve.

Felix invested $\text{?}12,000$ in a savings account. If the interest rate is $4\text{%}$ how much will be in the account in 12 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

Sayed deposits $\text{?}20,000$ in an investment account. What will be the value of his investment in 30 years if the investment is earning $7\text{%}$ per year and is compounded continuously?

$\text{?}163,323.40$

A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. She starts her experiment with 150 of the bacteria that grows at a rate of

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per hour. She will check on the bacteria every 24 hours. How many bacteria will he find in 24 hours?

In the last five years the population of the United States has grown at a rate of $0.7\text{%}$ per year to about 318,900,000. If this rate continues, what will be the population in 5 more years?

330,259,000

Evaluate and Graph Logarithmic Functions

Convert Between Exponential and Logarithmic Form

In the following exercises, convert from exponential to logarithmic form.

${5}^{4}=625$

${10}^{-3}=\frac{1}{1,000}$

$\text{log}\frac{1}{1,000}=-3$

${63}^{\frac{1}{5}}=\sqrt[5]{63}$

${e}^{y}=16$

$\text{ln}16=y$

In the following exercises, convert each logarithmic equation to exponential form.

$7={\text{log}}_{2}128$

$5=\text{log}100,000$

$100000={10}^{5}$

$4=\text{ln}\phantom{\rule{0.2em}{0ex}}x$

Evaluate Logarithmic Functions

In the following exercises, solve for x.

${\text{log}}_{x}125=3$

$x=5$

${\text{log}}_{7}x=-2$

${\text{log}}_{\frac{1}{2}}\frac{1}{16}=x$

$x=4$

In the following exercises, find the exact value of each logarithm without using a calculator.

${\text{log}}_{2}32$

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

0

${\text{log}}_{3}\frac{1}{9}$

Graph Logarithmic Functions

In the following exercises, graph each logarithmic function.

$y={\text{log}}_{5}x$

This figure shows a logarithmic line passing through the points (1 over 5, negative 1), (1, 0), and (5, 1).

$y={\text{log}}_{\frac{1}{4}}x$

$y={\text{log}}_{0.8}x$

This figure shows a logarithmic line passing through the points (4 over 5, 1), (1, 0), and (5 over 4, negative 1).

Solve Logarithmic Equations

In the following exercises, solve each logarithmic equation.

${\text{log}}_{a}36=5$

$\text{ln}\phantom{\rule{0.2em}{0ex}}x=-3$

$x={e}^{-3}$

${\text{log}}_{2}\left(5x-7\right)=3$

$\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{3x}=24$

$x=8$

$\text{log}\left({x}^{2}-21\right)=2$

Use Logarithmic Models in Applications

What is the decibel level of a train whistle with intensity ${10}^{-3}$ watts per square inch?

90 dB

Use the Properties of Logarithms

Use the Properties of Logarithms

In the following exercises, use the properties of logarithms to evaluate.

ⓐ ${\text{log}}_{7}1$ ⓑ ${\text{log}}_{12}12$

ⓐ ${5}^{{\text{log}}_{5}13}$ ⓑ ${\text{log}}_{3}{3}^{-9}$

ⓐ 13 ⓑ $-9$

ⓐ ${10}^{\text{log}\sqrt{5}}$ ⓑ $\text{log}{10}^{-3}$

ⓐ ${e}^{\text{ln}8}$ ⓑ $\text{ln}\phantom{\rule{0.2em}{0ex}}{e}^{5}$

ⓐ 8 ⓑ 5

In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

${\text{log}}_{4}\left(64xy\right)$

$\text{log}10,000m$

$4+\text{log}m$

In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

${\text{log}}_{7}\frac{49}{y}$

$\text{ln}\frac{{e}^{5}}{2}$

$5-\text{ln}2$

In the following exercises, use the Power Property of Logarithms to expand each logarithm. Simplify, if possible.

$\text{log}{x}^{-9}$

${\text{log}}_{4}\sqrt[7]{z}$

$\frac{1}{7}{\text{log}}_{4}z$

In the following exercises, use properties of logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

${\text{log}}_{3}\left(\sqrt{4}{x}^{7}{y}^{8}\right)$

${\mathrm{log}}_{5}\frac{8{a}^{2}{b}^{6}c}{{d}^{3}}$

${\text{log}}_{5}8+2{\text{log}}_{5}a+6{\text{log}}_{5}b$

$+\phantom{\rule{0.2em}{0ex}}{\text{log}}_{5}c-3{\text{log}}_{5}d$

$\text{ln}\frac{\sqrt{3{x}^{2}-{y}^{2}}}{{z}^{4}}$

${\text{log}}_{6}\sqrt[3]{\frac{7{x}^{2}}{6{y}^{3}{z}^{5}}}$

$\frac{1}{3}\left({\text{log}}_{6}7+2{\text{log}}_{6}x-1-3{\text{log}}_{6}y$

$-\phantom{\rule{0.2em}{0ex}}5{\text{log}}_{6}z\right)$

In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.

${\text{log}}_{2}56-{\text{log}}_{2}7$

$3{\text{log}}_{3}x+7{\text{log}}_{3}y$

${\text{log}}_{3}{x}^{3}{y}^{7}$

${\text{log}}_{5}\left({x}^{2}-16\right)-2{\text{log}}_{5}\left(x+4\right)$

$\frac{1}{4}\text{log}y-2\text{log}\left(y-3\right)$

$\text{log}\frac{\sqrt[4]{y}}{{\left(y-3\right)}^{2}}$

Use the Change-of-Base Formula

In the following exercises, rounding to three decimal places, approximate each logarithm.

${\text{log}}_{5}97$

${\text{log}}_{\sqrt{3}}16$

5.047

Solve Exponential and Logarithmic Equations

Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

$3{\text{log}}_{5}x={\text{log}}_{5}216$

${\text{log}}_{2}x+{\text{log}}_{2}\left(x-2\right)=3$

$x=4$

$\text{log}\left(x-1\right)-\text{log}\left(3x+5\right)=\text{−}\text{log}\phantom{\rule{0.2em}{0ex}}x$

${\text{log}}_{4}\left(x-2\right)+{\text{log}}_{4}\left(x+5\right)={\text{log}}_{4}8$

$x=3$

$\text{ln}\left(3x-2\right)=\text{ln}\left(x+4\right)+\text{ln}2$

Solve Exponential Equations Using Logarithms

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

${2}^{x}=101$

$x=\frac{\text{log}101}{\text{log}2}\approx 6.658$

${e}^{x}=23$

${\left(\frac{1}{3}\right)}^{x}=7$

$x=\frac{\text{log}7}{\text{log}\frac{1}{3}}\approx \text{−}1.771$

$7{e}^{x+3}=28$

${e}^{x-4}+8=23$

$x=\text{ln}15+4\approx 6.708$

Use Exponential Models in Applications

Jerome invests $\text{?}18,000$ at age 17. He hopes the investments will be worth $\text{?}30,000$ when he turns 26. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?

Elise invests $\text{?}4500$ in an account that compounds interest monthly and earns $6\text{%}.$ How long will it take for her money to double?

11.6 years

Researchers recorded that a certain bacteria population grew from 100 to 300 in 8 hours. At this rate of growth, how many bacteria will there be in 24 hours?

Mouse populations can double in 8 months $\left(A=2{A}_{0}\right).$ How long will it take for a mouse population to triple?

12.7 months

The half-life of radioactive iodine is 60 days. How much of a 50 mg sample will be left in 40 days?

Practice Test

For the functions, $f\left(x\right)=6x+1$ and $g\left(x\right)=8x-3,$ find ⓐ $\left(f\circ g\right)\left(x\right),$ ⓑ $\left(g\circ f\right)\left(x\right),$ and ⓒ $\left(f·g\right)\left(x\right).$

ⓐ $48x-17$ ⓑ $48x+5$

Determine if the following set of ordered pairs represents a function and if so, is the function one-to-one. $\left\{\left(-2,2\right),\left(-1,-3\right),\left(0,1\right),\left(1,-2\right),\left(2,-3\right)\right\}$

Determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ

This figure shows a parabola opening to the right with vertex (negative 3, 0).

ⓑ

This figure shows an exponential line passing through the points (negative 1, 1 over 2), (0, 1), and (1, 2).

ⓐ Not a function ⓑ One-to-one function

Graph, on the same coordinate system, the inverse of the one-to-one function shown.

This figure shows a line segment passing from the point (negative 3, 3) to (negative 1, 2) to (0, negative 2) to (2, negative 4).

Find the inverse of the function $f\left(x\right)={x}^{5}-9.$

${f}^{-1}\left(x\right)=\sqrt[5]{x+9}$

Graph the function $g\left(x\right)={2}^{x-3}.$

Solve the equation ${2}^{2x-4}=64.$

$x=5$

Solve the equation $\frac{{e}^{{x}^{2}}}{{e}^{4}}={e}^{3x}.$

Megan invested $\text{?}21,000$ in a savings account. If the interest rate is

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how much will be in the account in 8 years by each method of compounding?

ⓐ compound quarterly

ⓑ compound monthly

Convert the equation from exponential to logarithmic form: ${10}^{-2}=\frac{1}{100}.$

Convert the equation from logarithmic equation to exponential form: $3={\text{log}}_{7}343$

$343={7}^{3}$

Solve for x: ${\text{log}}_{5}x=-3$

Evaluate ${\text{log}}_{11}1.$

0

Evaluate ${\text{log}}_{4}\frac{1}{64}.$

Graph the function

$y={\text{log}}_{3}x.$

This figure shows a logarithmic line passing through (1 over 3, 1), (1, 0), and (3, 1).

Solve for x:

$\text{log}\left({x}^{2}-39\right)=1$

What is the decibel level of a small fan with intensity ${10}^{-8}$ watts per square inch?

40 dB

Evaluate each. ⓐ ${6}^{{\text{log}}_{6}17}$

ⓑ ${\text{log}}_{9}{9}^{-3}$

In the following exercises, use properties of logarithms to write each expression as a sum of logarithms, simplifying if possible.

${\text{log}}_{5}25ab$

$2+{\text{log}}_{5}a+{\text{log}}_{5}b$

$\text{ln}\frac{{e}^{12}}{8}$

${\text{log}}_{2}\sqrt[4]{\frac{5{x}^{3}}{16{y}^{2}{z}^{7}}}$

$\frac{1}{4}\left({\text{log}}_{2}5+3{\text{log}}_{2}x-4-2{\text{log}}_{2}y$

$-\phantom{\rule{0.2em}{0ex}}7{\text{log}}_{2}z\right)$

In the following exercises, use the Properties of Logarithms to condense the logarithm, simplifying if possible.

$5{\text{log}}_{4}x+3{\text{log}}_{4}y$

$\frac{1}{6}\text{log}\phantom{\rule{0.2em}{0ex}}x-3\text{log}\left(x+5\right)$

$\text{log}\frac{\sqrt[6]{x}}{{\left(x+5\right)}^{3}}$

Rounding to three decimal places, approximate ${\text{log}}_{4}73.$

Solve for x:

${\text{log}}_{7}\left(x+2\right)+{\text{log}}_{7}\left(x-3\right)={\text{log}}_{7}24$

$x=6$

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

${\left(\frac{1}{5}\right)}^{x}=9$

$5{e}^{x-4}=40$

$x=\text{ln}8+4\approx 6.079$

Jacob invests ?14,000 in an account that compounds interest quarterly and earns

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How long will it take for his money to double?

Researchers recorded that a certain bacteria population grew from 500 to 700 in 5 hours. At this rate of growth, how many bacteria will there be in 20 hours?

1,921 bacteria

A certain beetle population can double in 3 months $\left(A=2{A}_{0}\right).$ How long will it take for that beetle population to triple?

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Learning Objectives

Solve Logarithmic Equations Using the Properties of Logarithms

Solve Exponential Equations Using Logarithms

Use Exponential Models in Applications

Key Concepts

Section Exercises

Practice Makes Perfect

Writing Exercises

Self Check

Chapter Review Exercises

Finding Composite and Inverse Functions

Evaluate and Graph Exponential Functions

Evaluate and Graph Logarithmic Functions

Use the Properties of Logarithms

Solve Exponential and Logarithmic Equations

Practice Test

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