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Solve Exponential and Logarithmic Equations

Solve logarithmic equations using the properties of logarithms

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.

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.

Example

Try it.

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

Solution

	$\ 2\ {\log }_{5}x\ =\ {\log }_{5}81$
Use the Power Property.	$\ {\log }_{5}{x}^{2}\ =\ {\log }_{5}81$
Use the One-to-One Property, if ${\log }_{a}M={\log }_{a}N$, then $M=N$	$\ {x}^{2}\ =\ 81$.
Solve using the Square Root Property.	$\ x\ =\ \pm 9$
We eliminate $x=-9$ as we cannot take the logarithm of a negative number.	$\ x=9,\ x=-9$
Check.
$\begin{array}{llll} \\ x=9 & 2{\log }_{5}x & = & {\log }_{5}81 \\ & 2{\log }_{5}9 & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}{9}^{2} & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}81 & = & {\log }_{5}81✓\end{array}$

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

Example

Try it.

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

Solution

	${\log }_{3}x+{\log }_{3}(x-8)\ =\ 2$
Use the Product Property, ${\log }_{a}M+{\log }_{a}N={\log }_{a}M⋅N$.	$\ {\log }_{3}x(x-8)\ =\ 2$
Rewrite in exponential form.	$\ {3}^{2}\ =\ x(x-8)$
Simplify.	$\ 9\ =\ {x}^{2}-8x$
Subtract 9 from each side.	$\ 0\ =\ {x}^{2}-8x-9$
Factor.	$\ 0\ =\ (x-9)(x+1)$
Use the Zero-Product Property.	$\ x-9\ =\ 0,\ x+1=0$
Solve each equation.	$\ x=9,\ x=-1$
Check.
$\begin{array}{ll}x=-1 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2 \\ & {\log }_{3}(-1)+{\log }_{3}(-1-8)\ \overset{?}{=}\ 2\end{array}$
We cannot take the log of a negative number.
$\begin{array}{ll}x=9 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2\ \\ & {\log }_{3}9+{\log }_{3}(9-8)\ \overset{?}{=}\ 2\ \\ & 2+0\ \overset{?}{=}\ 2\ \\ & 2\ =\ 2✓\end{array}$

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

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.

Example

Try it.

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

Solution

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

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.

Example

Try it.

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

Solution

	$3{e}^{x+2}\ =\ 24$
Isolate the exponential by dividing both sides by 3.	${e}^{x+2}\ =\ 8$
Take the natural logarithm of both sides.	$\text{ln}\ {e}^{x+2}\ =\ \text{ln}8$
Use the Power Property to get the $x$ as a factor, not an exponent.	$(x+2)\text{ln}\ e\ =\ \text{ln}8$
Use the property $\text{ln}\ e=1$ to simplify.	$x+2\ =\ \text{ln}8$
Solve the equation. Find the exact answer.	$x\ =\ \text{ln}8-2$
Approximate the answer.	$x\ \approx \ 0.079$

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.

Example

Try it.

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?

Solution

	$\ A=\text{\$}50,000$
	$\ P=\text{\$}10,000$
Identify the variables in the formula	$\ r=?$
	$\ t=17\ \text{years}$
	$\ A=P{e}^{rt}$
Substitute the values into the formula.	$\ 50,000=10,000{e}^{r\cdot 17}$
Solve for $r.$ Divide each side by 10,000.	$\ 5={e}^{17r}$
Take the natural log of each side.	$\ \text{ln}5=\text{ln}\ {e}^{17r}$
Use the Power Property.	$\ \text{ln}5=17r\text{ln}\ e$
Simplify.	$\ \text{ln}5=17r$
Divide each side by 17.	$\ \frac{\text{ln}5}{17}=r$
Approximate the answer.	$\ r\approx 0.095$
Convert to a percentage.	$\ r\approx 9.5\text{\%}$
	They need the rate of growth to be approximately $9.5\text{\%}$.

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.

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.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

One-to-One Property of Logarithmic Equations: For $M>0,N>0,\ a\ >\ 0,$ and $\text{a}\ne 1$ is any real number:
\[\text{If}\ {\text{log}}_{a}M={\text{log}}_{a}N,\ \text{then}\ M=N.\]
Compound Interest:
For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:
\[\begin{array}{llllll} \\ \\ A=P{(1+\frac{r}{n})}^{nt} & & & & & \text{when compounded}\ n\ \text{times a year.} \\ A=P{e}^{rt} & & & & & \text{when compounded continuously.}\end{array}\]
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 $A={A}_{0}{e}^{rt}.$

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Àwọn Ìṣàmúlò-ètò (40)

Wá gbogbo àwọn ní pàtó nínú àwọn àkọ́lé ní ìbẹrẹ. Fi àwọn

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

Fi àwọn àgbèwọlé hàn

$x=4,\ x=-4$
Solve: ${x}^{2}-5x+6=0.$

Fi àwọn àgbèwọlé hàn

$x=2,\ x=3$
Solve: $x(x+6)=2x+5.$

Fi àwọn àgbèwọlé hàn

$x=-5,\ x=1$

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

Fi àwọn àgbèwọlé hàn

	$\ 2\ {\log }_{5}x\ =\ {\log }_{5}81$
Use the Power Property.	$\ {\log }_{5}{x}^{2}\ =\ {\log }_{5}81$
Use the One-to-One Property, if ${\log }_{a}M={\log }_{a}N$, then $M=N$	$\ {x}^{2}\ =\ 81$.
Solve using the Square Root Property.	$\ x\ =\ \pm 9$
We eliminate $x=-9$ as we cannot take the logarithm of a negative number.	$\ x=9,\ x=-9$
Check.
$\begin{array}{llll} \\ x=9 & 2{\log }_{5}x & = & {\log }_{5}81 \\ & 2{\log }_{5}9 & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}{9}^{2} & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}81 & = & {\log }_{5}81✓\end{array}$

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

Fi àwọn àgbèwọlé hàn

$x=6$
Solve: $3\text{log}\ x=\text{log}64$

Fi àwọn àgbèwọlé hàn

$x=4$

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

Fi àwọn àgbèwọlé hàn

	${\log }_{3}x+{\log }_{3}(x-8)\ =\ 2$
Use the Product Property, ${\log }_{a}M+{\log }_{a}N={\log }_{a}M⋅N$.	$\ {\log }_{3}x(x-8)\ =\ 2$
Rewrite in exponential form.	$\ {3}^{2}\ =\ x(x-8)$
Simplify.	$\ 9\ =\ {x}^{2}-8x$
Subtract 9 from each side.	$\ 0\ =\ {x}^{2}-8x-9$
Factor.	$\ 0\ =\ (x-9)(x+1)$
Use the Zero-Product Property.	$\ x-9\ =\ 0,\ x+1=0$
Solve each equation.	$\ x=9,\ x=-1$
Check.
$\begin{array}{ll}x=-1 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2 \\ & {\log }_{3}(-1)+{\log }_{3}(-1-8)\ \overset{?}{=}\ 2\end{array}$
We cannot take the log of a negative number.
$\begin{array}{ll}x=9 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2\ \\ & {\log }_{3}9+{\log }_{3}(9-8)\ \overset{?}{=}\ 2\ \\ & 2+0\ \overset{?}{=}\ 2\ \\ & 2\ =\ 2✓\end{array}$

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

Fi àwọn àgbèwọlé hàn

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

Fi àwọn àgbèwọlé hàn

$x=8$

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

Fi àwọn àgbèwọlé hàn

	${\text{log}}_{4}(x+6)-{\text{log}}_{4}(2x+5)=\text{-}{\text{log}}_{4}x$
Use the Quotient Property on the left side and the Power Property on the right.	${\text{log}}_{4}(\frac{x+6}{2x+5})={\text{log}}_{4}{x}^{-1}$
Rewrite ${x}^{-1}=\frac{1}{x}$.	${\text{log}}_{4}(\frac{x+6}{2x+5})={\text{log}}_{4}\frac{1}{x}$
Use the One-to-One Property, if ${\text{log}}_{a}M={\text{log}}_{a}N$, then $M=N$.	$\frac{x+6}{2x+5}=\frac{1}{x}$
Solve the rational equation.	$x(x+6)=2x+5$
Distribute.	${x}^{2}+6x=2x+5$
Write in standard form.	${x}^{2}+4x-5=0$
Factor.	$(x+5)(x-1)=0$
Use the Zero-Product Property.	$\ x+5=0,\ x-1=0$
Solve each equation.	$\ x=-5,\ x=1$
Check.
We leave the check for you.

Solve: $\text{log}(x+2)-\text{log}(4x+3)=\text{-}\text{log}\ x.$

Fi àwọn àgbèwọlé hàn

$x=3$
Solve: $\text{log}(x-2)-\text{log}(4x+16)=\text{log}\frac{1}{x}.$

Fi àwọn àgbèwọlé hàn

$x=8$

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

Fi àwọn àgbèwọlé hàn

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

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

Fi àwọn àgbèwọlé hàn

$x=\frac{\text{log}43}{\text{log}7}\approx 1.933$
Solve ${8}^{x}=98.$ Find the exact answer and then approximate it to three decimal places.

Fi àwọn àgbèwọlé hàn

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

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

Fi àwọn àgbèwọlé hàn

	$3{e}^{x+2}\ =\ 24$
Isolate the exponential by dividing both sides by 3.	${e}^{x+2}\ =\ 8$
Take the natural logarithm of both sides.	$\text{ln}\ {e}^{x+2}\ =\ \text{ln}8$
Use the Power Property to get the $x$ as a factor, not an exponent.	$(x+2)\text{ln}\ e\ =\ \text{ln}8$
Use the property $\text{ln}\ e=1$ to simplify.	$x+2\ =\ \text{ln}8$
Solve the equation. Find the exact answer.	$x\ =\ \text{ln}8-2$
Approximate the answer.	$x\ \approx \ 0.079$

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

Fi àwọn àgbèwọlé hàn

$x=\text{ln}9+2\approx 4.197$
Solve $5{e}^{2x}=25.$ Find the exact answer and then approximate it to three decimal places.

Fi àwọn àgbèwọlé hàn

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

Fi àwọn àgbèwọlé hàn

	$\ A=\text{\$}50,000$
	$\ P=\text{\$}10,000$
Identify the variables in the formula	$\ r=?$
	$\ t=17\ \text{years}$
	$\ A=P{e}^{rt}$
Substitute the values into the formula.	$\ 50,000=10,000{e}^{r\cdot 17}$
Solve for $r.$ Divide each side by 10,000.	$\ 5={e}^{17r}$
Take the natural log of each side.	$\ \text{ln}5=\text{ln}\ {e}^{17r}$
Use the Power Property.	$\ \text{ln}5=17r\text{ln}\ e$
Simplify.	$\ \text{ln}5=17r$
Divide each side by 17.	$\ \frac{\text{ln}5}{17}=r$
Approximate the answer.	$\ r\approx 0.095$
Convert to a percentage.	$\ r\approx 9.5\text{\%}$
	They need the rate of growth to be approximately $9.5\text{\%}$.

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?

Fi àwọn àgbèwọlé hàn

$r\approx 9.3\text{\%}$
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?

Fi àwọn àgbèwọlé hàn

$r\approx 11.9\text{\%}$

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?

Fi àwọn àgbèwọlé hàn

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.

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

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?

Fi àwọn àgbèwọlé hàn

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?

Fi àwọn àgbèwọlé hàn

There will be 47,700 bacteria.

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

Fi àwọn àgbèwọlé hàn

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.

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

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

Fi àwọn àgbèwọlé hàn

There will be 6.44 mg left.
The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?

Fi àwọn àgbèwọlé hàn

There will be 31.5 mg left.
${\text{log}}_{4}64=2{\text{log}}_{4}x$
$\text{log}49=2\text{log}\ x$

Fi àwọn àgbèwọlé hàn

$x=7$
$3{\text{log}}_{3}x={\text{log}}_{3}27$
$3{\text{log}}_{6}x={\text{log}}_{6}64$

Fi àwọn àgbèwọlé hàn

$x=4$
${\text{log}}_{5}(4x-2)={\text{log}}_{5}10$
${\text{log}}_{3}({x}^{2}+3)={\text{log}}_{3}4x$

Fi àwọn àgbèwọlé hàn

$x=1,$ $x=3$
${\text{log}}_{3}x+{\text{log}}_{3}x=2$
${\text{log}}_{4}x+{\text{log}}_{4}x=3$

Fi àwọn àgbèwọlé hàn

$x=8$
${\text{log}}_{2}x+{\text{log}}_{2}(x-3)=2$
${\text{log}}_{3}x+{\text{log}}_{3}(x+6)=3$

Fi àwọn àgbèwọlé hàn

$x=3$
$\text{log}\ x+\text{log}(x+3)=1$
$\text{log}\ x+\text{log}(x-15)=2$

Fi àwọn àgbèwọlé hàn

$x=20$
$\text{log}(x+4)-\text{log}(5x+12)=\text{-}\text{log}\ x$

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Bii o ṣe le: Solve Exponential and Logarithmic Equations

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

Àwọn Àtòjọ-ẹ̀yàn

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Àwọn ààyè ojú-ìwé yìí tí a fi pamọ́ láti OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Tí a fi pamọ́ síì níbẹ̀; àwọn àwọn àṣiṣe ní wa.

Diẹ̀ nínú Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

	\(\ 2\ {\log }_{5}x\ =\ {\log }_{5}81\)
Use the Power Property.	\(\ {\log }_{5}{x}^{2}\ =\ {\log }_{5}81\)
Use the One-to-One Property, if \({\log }_{a}M={\log }_{a}N\), then \(M=N\)	\(\ {x}^{2}\ =\ 81\).
Solve using the Square Root Property.	\(\ x\ =\ \pm 9\)
We eliminate \(x=-9\) as we cannot take the logarithm of a negative number.	\(\ x=9,\ x=-9\)
Check.
\(\begin{array}{llll} \\ x=9 & 2{\log }_{5}x & = & {\log }_{5}81 \\ & 2{\log }_{5}9 & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}{9}^{2} & \overset{?}{=} & {\log }_{5}81 \\ & {\log }_{5}81 & = & {\log }_{5}81✓\end{array}\)

	\({\log }_{3}x+{\log }_{3}(x-8)\ =\ 2\)
Use the Product Property, \({\log }_{a}M+{\log }_{a}N={\log }_{a}M⋅N\).	\(\ {\log }_{3}x(x-8)\ =\ 2\)
Rewrite in exponential form.	\(\ {3}^{2}\ =\ x(x-8)\)
Simplify.	\(\ 9\ =\ {x}^{2}-8x\)
Subtract 9 from each side.	\(\ 0\ =\ {x}^{2}-8x-9\)
Factor.	\(\ 0\ =\ (x-9)(x+1)\)
Use the Zero-Product Property.	\(\ x-9\ =\ 0,\ x+1=0\)
Solve each equation.	\(\ x=9,\ x=-1\)
Check.
\(\begin{array}{ll}x=-1 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2 \\ & {\log }_{3}(-1)+{\log }_{3}(-1-8)\ \overset{?}{=}\ 2\end{array}\)
We cannot take the log of a negative number.
\(\begin{array}{ll}x=9 & {\log }_{3}x+{\log }_{3}(x-8)\ =\ 2\ \\ & {\log }_{3}9+{\log }_{3}(9-8)\ \overset{?}{=}\ 2\ \\ & 2+0\ \overset{?}{=}\ 2\ \\ & 2\ =\ 2✓\end{array}\)

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

	\(3{e}^{x+2}\ =\ 24\)
Isolate the exponential by dividing both sides by 3.	\({e}^{x+2}\ =\ 8\)
Take the natural logarithm of both sides.	\(\text{ln}\ {e}^{x+2}\ =\ \text{ln}8\)
Use the Power Property to get the \(x\) as a factor, not an exponent.	\((x+2)\text{ln}\ e\ =\ \text{ln}8\)
Use the property \(\text{ln}\ e=1\) to simplify.	\(x+2\ =\ \text{ln}8\)
Solve the equation. Find the exact answer.	\(x\ =\ \text{ln}8-2\)
Approximate the answer.	\(x\ \approx \ 0.079\)

	\(\ A=\text{\$}50,000\)
	\(\ P=\text{\$}10,000\)
Identify the variables in the formula	\(\ r=?\)
	\(\ t=17\ \text{years}\)
	\(\ A=P{e}^{rt}\)
Substitute the values into the formula.	\(\ 50,000=10,000{e}^{r\cdot 17}\)
Solve for \(r.\) Divide each side by 10,000.	\(\ 5={e}^{17r}\)
Take the natural log of each side.	\(\ \text{ln}5=\text{ln}\ {e}^{17r}\)
Use the Power Property.	\(\ \text{ln}5=17r\text{ln}\ e\)
Simplify.	\(\ \text{ln}5=17r\)
Divide each side by 17.	\(\ \frac{\text{ln}5}{17}=r\)
Approximate the answer.	\(\ r\approx 0.095\)
Convert to a percentage.	\(\ r\approx 9.5\text{\%}\)
	They need the rate of growth to be approximately \(9.5\text{\%}\).

	\({\text{log}}_{4}(x+6)-{\text{log}}_{4}(2x+5)=\text{-}{\text{log}}_{4}x\)
Use the Quotient Property on the left side and the Power Property on the right.	\({\text{log}}_{4}(\frac{x+6}{2x+5})={\text{log}}_{4}{x}^{-1}\)
Rewrite \({x}^{-1}=\frac{1}{x}\).	\({\text{log}}_{4}(\frac{x+6}{2x+5})={\text{log}}_{4}\frac{1}{x}\)
Use the One-to-One Property, if \({\text{log}}_{a}M={\text{log}}_{a}N\), then \(M=N\).	\(\frac{x+6}{2x+5}=\frac{1}{x}\)
Solve the rational equation.	\(x(x+6)=2x+5\)
Distribute.	\({x}^{2}+6x=2x+5\)
Write in standard form.	\({x}^{2}+4x-5=0\)
Factor.	\((x+5)(x-1)=0\)
Use the Zero-Product Property.	\(\ x+5=0,\ x-1=0\)
Solve each equation.	\(\ x=-5,\ x=1\)
Check.
We leave the check for you.

Identify the variables in the formula.	\(\begin{array}{lll}A & = & 300 \\ {A}_{0} & = & 100 \\ k & = & ? \\ t & = & 3\ \text{hours} \\ A & = & {A}_{0}{e}^{kt}\end{array}\)
Substitute the values in the formula.	\(\ 300=100{e}^{k\cdot 3}\)
Solve for \(k\). Divide each side by 100.	\(\ 3={e}^{3k}\)
Take the natural log of each side.	\(\ \text{ln}3=\text{ln}\ {e}^{3k}\)
Use the Power Property.	\(\ \text{ln}3=3k\text{ln}\ e\)
Simplify.	\(\ \text{ln}3=3k\)
Divide each side by 3.	\(\ \frac{\text{ln}3}{3}=k\)
Approximate the answer.	\(\ k\approx 0.366\)
We use this rate of growth to predict the number of bacteria there will be in 24 hours.	\(\begin{array}{lll}A & = & ? \\ {A}_{0} & = & 100 \\ k & = & \frac{\text{ln}3}{3} \\ t & = & 24\ \text{hours} \\ A & = & {A}_{0}{e}^{kt}\end{array}\)
Substitute in the values.	\(A=100{e}^{\frac{\text{ln}3}{3}\cdot 24}\)
Evaluate.	\(A\approx 656,100\)
	At this rate of growth, they can expect 656,100 bacteria.

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