# 10.6: Solve Exponential and Logarithmic Equations

- Page ID
- 5185

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 Example 6.46. - Solve: \(x^{2}−5x+6=0\).

If you missed this problem, review Example 6.45. - Solve: \(x(x+6)=2x+5\).

If you missed this problem, review Example 6.47.

## 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 \(\log _{a} M=\log _{a} N\) 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 \(M>0,N>0,a>0\), and \(a≠1\) is any real number:

If \(\log _{a} M=\log _{a} N,\) then \(M=N\).

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: \(2 \log _{5} x=\log _{5} 81\).

**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.

\(x=9, \cancel{x=-9}\)

Check. \(x=9\)

\(\begin{aligned}2 \log _{5} x&=\log _{5} 81 \\ 2 \log _{5} 9 &\stackrel{?}{=} \log _{5} 81 \\ \log _{5} 9^{2} & \stackrel{?}{=}\log _{5} 81 \\ \log _{5} 81 & =\log _{5} 81\end{aligned}\)

Solve: \(2 \log _{3} x=\log _{3} 36\)

**Answer**-
\(x=6\)

Solve: \(3 \log x=\log 64\)

**Answer**-
\(x=4\)

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

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

**Solution**:

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

Use the Product Property, \(\log _{a} M+\log _{a} N=\log _{a} M \cdot N\).

\(\log _{3} x(x-8)=2\)

Rewrite in exponential form.

\(3^{2}=x(x-8)\)

Simplify.

\(9=x^{2}-8 x\)

Subtract \(9\) from each side.

\(0=x^{2}-8 x-9\)

Factor.

\(0=(x-9)(x+1)\)

Use the Zero-Product-Property

\(x-9=0, \quad x+1=0\)

Solve each equation.

\(x=9, \quad \cancel{x=-1}\)

Check. \(x=-1\)

\(\begin{aligned} \log _{3} x+\log _{3}(x-8)&=2 \\ \log _{3}(-1)+\log _{3}(-1-8) &\stackrel{?}{=}2\end{aligned}\)

We cannot take the log of a negative number.

Check. \(x=9\)

\(\begin{aligned} \log _{3} x+\log _{3}(x-8) &=2 \\ \log _{3} 9+\log _{3}(9-8) & \stackrel{?}{=} 2 \\ 2+0 &\stackrel{?}{=}2 \\ 2 &=2 \end{aligned}\)

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

**Answer**-
\(x=4\)

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

**Answer**-
\(x=8\)

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

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

**Solution**:

\(\log _{4}(x+6)-\log _{4}(2 x+5)=-\log _{4} x\)

Use the Quotient Property on the left side and the PowerProperty on the right.

\(\log _{4}\left(\frac{x+6}{2 x+5}\right)=\log _{4} x^{-1}\)

Rewrite \(x^{-1}=\frac{1}{x}\).

\(\log _{4}\left(\frac{x+6}{2 x+5}\right)=\log _{4} \frac{1}{x}\)

Use the One-to-One Property, if \(\log _{a} M=\log _{a} N\), then \(M=N\).

\(\frac{x+6}{2 x+5}=\frac{1}{x}\)

Solve the rational equation.

\(x(x+6)=2 x+5\)

Distribute.

\(x^{2}+6 x=2 x+5\)

Write in standard form.

\(x^{2}+4 x-5=0\)

Factor.

\((x+5)(x-1)=0\)

Use the Zero-Product-Property.

\(x+5=0, \quad x-1=0\)

Solve each equation.

\(\cancel{x=-5}, \quad x=1\)

Check.

We leave the check for you.

Solve: \(\log (x+2)-\log (4 x+3)=-\log x\).

**Answer**-
\(x=3\)

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

**Answer**-
\(x=8\)

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

**Solution**:

\(5^{x}=11\)

Since the exponential is isolated, take the logarithm of both sides.

\(\log 5^{x}=\log 11\)

Use the Power Property to get the \(x\) as a factor, not an exponent.

\(x \log 5=\log 11\)

Solve for \(x\). Find the exact answer.

\(x=\frac{\log 11}{\log 5}\)

Approximate the answer.

\(x \approx 1.490\)

Since \(5^{1}=5\) and \(5^{2}=25\), does it makes sense that \(5^{1.490}≈11\)?

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

**Answer**-
\(x=\frac{\log 43}{\log 7} \approx 1.933\)

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

**Answer**-
\(x=\frac{\log 98}{\log 8} \approx 2.205\)

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 \(3e^{x+2}=24\). 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.

\(\ln e^{x+2}=\ln 8\)

Use the Power Property to get the \(x\) as a factor, not an exponent.

\((x+2) \ln e=\ln 8\)

Use the property \(\ln e=1\) to simplify.

\(x+2=\ln 8\)

Solve the equation. Find the exact answer.

\(x=\ln 8-2\)

Approximate the answer.

\(x \approx 0.079\)

Solve \(2e^{x−2}=18\). Find the exact answer and then approximate it to three decimal places.

**Answer**-
\(x=\ln 9+2 \approx 4.197\)

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

**Answer**-
\(x=\frac{\ln 5}{2} \approx 0.805\)

## 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:

\(\begin{array}{ll}{A=P\left(1+\frac{r}{n}\right)^{n t}} & {\text { when compounded } n \text { times a year. }} \\ {A=P e^{r t}} & {\text { when compounded continuously. }}\end{array}\)

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**:

Identify the variables in the formula.

\(\begin{aligned} A &=\$ 50,000 \\ P &=\$ 10,000 \\ r &=? \\ t &=17 \text { years } \\ A &=P e^{r t} \end{aligned}\)

Substitute the values into the formula.

\(50,000=10,000 e^{r \cdot 17}\)

Solve for \(r\). Divide each side by \(10,000\).

\(5=e^{17 r}\)

Take the natural log of each side.

\(\ln 5=\ln e^{17 r}\)

Use the Power Property.

\(\ln 5=17 r \ln e\)

Simplify.

\(\ln 5=17 r\)

Divide each side by \(17\).

\(\frac{\ln 5}{17}=r\)

Approximate the answer.

\(r \approx 0.095\)

Convert to a percentage.

\(r \approx 9.5 \%\)

They need the rate of growth to be approximately \(9.5\)%.

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

**Answer**-
\(r \approx 9.3 \%\)

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

**Answer**-
\(r \approx 11.9 \%\)

We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula \(A=A_{0} e^{k t}\). 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:

\(A=A_{0} e^{k t}\)

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?

**Solution**:

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.

\(\begin{aligned} A &=300 \\ A_{0} &=100 \\ k &=? \\ t &=3 \text { hours } \\ A &=A_{0} e^{k t} \end{aligned}\)

Substitute the values in the formula.

\(300=100 e^{k \cdot 3}\)

Solve for \(k\). Divide each side by \(100\).

\(3=e^{3 k}\)

Take the natural log of each side.

\(\ln 3=\ln e^{3 k}\)

Use the Power Property.

\(\ln 3=3 k \ln e\)

Simplify.

\(\ln 3=3 k\)

Divide each side by \(3\).

\(\frac{\ln 3}{3}=k\)

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.

\(\begin{aligned} A &=? \\ A_{0} &=100 \\ k &=\frac{\ln 3}{3} \\ t &=24 \text { hours } \\ A &=A_{0} e^{k t} \end{aligned}\)

Substitute in the values.

\(A=100 e^{\frac{\ln 3}{3} \cdot 24}\)

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?

**Answer**-
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?

**Answer**-
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?

**Solution**:

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.

\(\begin{aligned} A &=50 \\ A_{0} &=100 \\ k &=? \\ t &=1590 \text { years } \\ A &=A_{0} e^{k t} \end{aligned}\)

Substitute the values in the formula.

\(50=100 e^{k \cdot 1590}\)

Solve for \(k\). Divide each side by \(100\).

\(0.5=e^{1590 k}\)

Take the natural log of each side.

\(\ln 0.5=\ln e^{1590 k}\)

Use the Power Property.

\(\ln 0.5=1590 k \ln e\)

Simplify.

\(\ln 0.5=1590 k\)

Divide each side by \(1590\).

\(\frac{\ln 0.5}{1590}=k\) exact answer

We use this rate of growth to predict the amount that will be left in \(500\) years.

\(\begin{aligned} A &=? \\ A_{0} &=100 \\ k &=\frac{\ln 0.5}{1590} \\ t &=500\: \mathrm{years} \\ A &=A_{0} e^{k t} \end{aligned}\)

Substitute in the values.

\(A=100 e^{\frac{1 \mathrm{n} 0.5}{1500} \cdot 500}\)

Evaluate.

\(A \approx 80.4 \mathrm{mg}\)

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?

**Answer**-
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?

**Answer**-
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 \(M>0, N>0, a>0\), and \(a≠1\) is any real number:If \(\log _{a} M=\log _{a} N,\) then \(M=N\)

**Compound Interest:**

For a principal, \(P\), invested at an interest rate, \(r\), for \(t\) years, the new balance, \(A\), is:\(\begin{array}{ll}{A} & {=P\left(1+\frac{r}{n}\right)^{n t}} & {\text { when compounded } n \text { times a year. }} \\ {A} & {=P e^{r t}} & {\text { when compounded continuously. }}\end{array}\)

**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 \(A=A_{0} e^{r t}\).