4.7: Exponential and Logarithmic Equations

Learning Objectives

• Use like bases to solve exponential equations.
• Use logarithms to solve exponential equations.
• Use the definition of a logarithm to solve logarithmic equations.
• Use the one-to-one property of logarithms to solve logarithmic equations.
• Solve applied problems involving exponential and logarithmic equations.

In 1859, an Australian landowner named Thomas Austin released $24$ rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Figure $\PageIndex{1}$: Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr)

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential and logarithmic equations.

Exponential Equations

Exponential equations are equations that contain variables in the exponent.

Some examples of exponential equations: $$2^x=2^{3x-7} \qquad \qquad 8^{t-3}+4=7\nonumber$$

There are two main techniques to solving exponential equations. The first technique is for solving equations that involve like bases. The second technique is for solving equations that we are unable to write with the same base.

Using Like Bases to Solve Exponential Equations

Consider the equation $3^{4x−7}=3^{2x}$. Notice that the bases are the same. To have equivalency, the exponents must be equal: $4x-7=2x$. Solving this, we get $x=\dfrac{7}{2}$.

The one-to-one property of exponential functions tells us that, for any real numbers $b$, $S$, and $T$, where $b>0$, $b≠1$,

$b^S=b^T$ if and only if $S=T$.

Example $\PageIndex{1}$

Solve $2^{x−1}=2^{2x−4}$

Solution

$$\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}$$

You Try $\PageIndex{1}$

Solve $5^{2x}=5^{3x+2}$

Answer

$x=−2$

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation $256=4^{x−5}$. We can rewrite both sides of this equation as a power of $2$. Then we apply the rules of exponents, along with the one-to-one property, to solve for $x$:

$$\begin{align*} 16&= 2^{x-5}\\ 2^4&= 2^{x-5} \qquad &&\text{Rewrite 16 as a power with base 2}\\ 4&= x-5 \qquad &&\text{Apply the one-to-one property of exponents}\\ 9&= x \qquad &&\text{Add 5 to both sides} \end{align*}$$

Example $\PageIndex{2}$

Solve $8^{x+2}={16}^{x+1}$

Solution

$$\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for x} \end{align*}$$

You Try $\PageIndex{2}$

Solve $5^{2x}={25}^{3x+2}$

Answer

$x=−1$

Example $\PageIndex{3}$

Solve $2^{5x}=\sqrt{2}$

Solution

$$\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for x} \end{align*}$$

You Try $\PageIndex{3}$

Solve $5^x=\sqrt{5}$

Answer

$x=\dfrac{1}{2}$

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. The one-to-one property for logarithms tells us that, for real numbers $a>0$ and $b>0$, $\log(a)=\log(b)$ is equivalent to $a=b$. This means that we may apply logarithms with the same base on both sides of an exponential equation. It is most common to apply the natural logarithm on both sides but any logarithms with the same base applied to both sides may be used.

Example $\PageIndex{4}$

Solve $5^{x+2}=4^x$

Solution

$$\begin{align*} 5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\[5pt] \ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\[5pt] (x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\[5pt] x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\[5pt] x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\[5pt] x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\[5pt] x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\[5pt] x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x}\\[5pt] x&=\dfrac{-\ln25}{\ln{5}-\ln{4}} \qquad &&\text{simplify using Laws of Logarithms}\end{align*}$$

Example $\PageIndex{5}$

Solve $2^x=3^x$

Solution

$$\begin{align*} \ln2^x&= \ln3^x \qquad &&\text{Take ln of both sides}\\[4pt] x \ln2&= x \ln3 \qquad &&\text{move exponents to the front}\\[4pt] x \ln2-x\ln3&= 0 \qquad &&\text{get like terms on one side}\\[4pt] x(\ln2-\ln3)&= 0 \qquad &&\text{factor out x}\\[4pt] x&=\dfrac{0}{\ln2 - \ln3} \qquad &&\text{divide both sides by} \ln2 - \ln3\\[4pt] x&= 0 \end{align*}$$

You Try $\PageIndex{4}$

Solve $2^x=3^{x+1}$

Answer

$x=\dfrac{\ln3}{\ln2-\ln3 \left (\dfrac{2}{3} \right )}$ or $x=\dfrac{\ln3}{\ln2-\ln3 }$

Example $\PageIndex{6}$

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 $5^{2}=25$, it makes sense that $5^{1.490}≈11$.

Note in the previous example that $y=5^x$ is a basic exponential function. We know what the graph looks like and how the function values will start to become large quite quickly for larger x-values since this is an increasing exponential function of base 5, it seems reasonable that the function value of 11 would occur at a smaller x-value. Also, knowing that there is are points on the graph at $(0,1)$ and at $(1,5)$, this would indicate that the function would equal 11 for an x-value that is is just a bit bigger than 1.

You Try $\PageIndex{5}$

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

Answer

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

One common type of exponential equations are those with base $e$. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base $e$ on either side, we can use the natural logarithm to solve it.

Example $\PageIndex{7}$

Solve $100=20e^{2t}$

Solution

$$\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) &&\text{ and } e^x &&\text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}$$

Analysis

Using laws of logs, we can also write this answer in the form $t=\ln\sqrt{5}$. If we want a decimal approximation of the answer, we use a calculator.

You Try $\PageIndex{6}$

Solve $3e^{0.5t}=11$

Answer

$t=2\ln \left (\dfrac{11}{3} \right )$ or $\ln{ \left (\dfrac{11}{3} \right )}^2$

Example $\PageIndex{8}$

Solve $4e^{2x}+5=12$

Solution

$$\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides}\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}$$

You Try $\PageIndex{7}$

Solve $3+e^{2t}=7e^{2t}$

Answer

$t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)$

Example $\PageIndex{9}$

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.

$x \approx 0.079$

You Try $\PageIndex{8}$

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

Answer

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

You may be wondering if all exponential equations have solutions. The answer is not all exponential equations have solutions.

The equation $3^{x+1}=−2$ has no solution. There is no real value of $x$ that will make the equation a true statement because any power of a positive number is positive.

Figure $\PageIndex{2}$ shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Figure $\PageIndex{2}$

Another example of an equation that has no solution is $2=−3e^t$.

Figure $\PageIndex{3}$ shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Figure $\PageIndex{3}$: Graph shows that the functions do not intersect

Checking for Extraneous Solutions When Solving Exponential Equations

Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example $\PageIndex{10}$

Solve $e^{2x}−e^x=56$

Solution

$$\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero since this equation is of quadratic form}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\end{align*}$$

Solving for each factor, $$\begin{align*}e^x+7= 0 \qquad \text{or} \qquad e^x-8= 0 \\ e^x= -7 \qquad \text{or} \qquad e^x= 8 \end{align*}$$ $e^x=-7$ has no solution since $x=\ln(-7)$ is undefined. Solving $e^x= 8$, $$x= \ln8\nonumber$$

You Try $\PageIndex{9}$

Solve $e^{2x}=e^x+2$

Answer

$x=\ln2$

Logarithmic Equations

Logarithmic equations are equations that contain logarithms with variables.

Some examples of logarithmic equations: $$\log_5x=8 \qquad \qquad \ln(t)-\ln(t-2)=\ln11\nonumber$$

There are two main techniques to solving logarithmic equations. The first technique is for solving equations that contain only logarithmic terms with the same base. The second technique is for solving equations that contain logarithmic terms and constants.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers $x>0$, $S>0$, $T>0$ and any positive real number $b$, where $b≠1$,

${\log}_bS={\log}_bT$ if and only if $S=T$.

For example,

If ${\log}_2(x−1)={\log}_2(8)$, then $x−1=8$.

So, if $x−1=8$, then we can solve for $x$,and we get $x=9$. To check, we can substitute $x=9$ into the original equation: ${\log}_2(9−1)={\log}_2(8)=3$. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation $\log(3x−2)−\log(2)=\log(x+4)$. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for $x$:

$$\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\[4pt] \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad &&\text{Apply the quotient rule of logarithms}\\[4pt] \dfrac{3x-2}{2}&= x+4 \qquad &&\text{Apply the one to one property of a logarithm}\\[4pt] 3x-2&= 2x+8 \qquad &&\text{Multiply both sides of the equation by 2}\\[4pt] x&= 10 \qquad &&\text{Subtract 2x and add 2} \end{align*}$$

To check the result, substitute $x=10$ into $\log(3x−2)−\log(2)=\log(x+4)$.

$$\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\[4pt] \log(28)-\log(2)&= \log(14)\\[4pt] \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad &&\text{The solution checks} \end{align*}$$

Example $\PageIndex{11}$

Solve $\ln(x^2)=\ln(2x+3)$

Solution

$$\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad &&\text{Use the one-to-one property of the logarithm}\\[4pt] x^2-2x-3&= 0 \qquad &&\text{Get zero on one side before factoring}\\[4pt] (x-3)(x+1)&= 0 \qquad &&\text{Factor}\end{align*}$$

Solving for each factor, $$x-3= 0 \qquad \text{ or } \qquad x+1=0 \nonumber$$ $$x=3 \qquad \text{ or } \qquad x= -1 \nonumber$$

Analysis

There are two solutions: $3$ or $−1$. The solution $−1$ is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

You Try $\PageIndex{10}$

Solve $\ln(x^2)=\ln1$

Answer

$x=1$ or $x=−1$

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation ${\log}_b(x)=y$ is equivalent to the exponential equation $b^y=x$. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation ${\log}_2(2)+{\log}_2(3x−5)=3$. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for $x$:

$$\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\[4pt] {\log}_2(2(3x-5))&= 3 \qquad &&\text{Apply the product rule of logarithms}\\[4pt] {\log}_2(6x-10)&= 3 \qquad &&\text{Distribute}\\[4pt] 2^3&= 6x-10 \qquad &&\text{Convert to exponential form}\\[4pt] 8&= 6x-10 \qquad &&\text{Calculate } 2^3\\[4pt] 18&= 6x \qquad &&\text{Add 10 to both sides}\\[4pt] x&= 3 \qquad &&\text{Divide by 6} \end{align*}$$

Example $\PageIndex{12}$

Solve $2\ln x+3=7$

Solution

$$\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad &&\text{Subtract 3}\\ \ln x&= 2 \qquad &&\text{Divide by 2}\\ x&= e^2 \qquad &&\text{Rewrite in exponential form} \end{align*}$$

You Try $\PageIndex{11}$

Solve $6+\ln x=10$

Answer

$x=e^4$

Example $\PageIndex{13}$

Solve $2\ln(6x)=7$

Solution

$$\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad &&\text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad &&\text{convert to exponential form}\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad &&\text{Divide by 6} \end{align*}$$

You Try $\PageIndex{12}$

Solve $2\ln(x+1)=10$

Answer

$x=e^5−1$

Example $\PageIndex{14}$

Solve $\ln x=3$

Solution

$$\begin{align*} \ln x&= 3\\ x&= e^3 \end{align*}$$

Figure $\PageIndex{4}$ represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to $20$. In other words $e^3≈20$. A calculator gives a better approximation: $e^3≈20.0855$.

Figure $\PageIndex{4}$: The graphs of $y=\ln x$ and $y=3$ cross at the point $(e^3,3)$, which is approximately $(20.0855, 3)$.

You Try $\PageIndex{13}$

Use a graphing calculator to estimate the approximate solution to the logarithmic equation $2^x=1000$ to $2$ decimal places.

Answer

$x≈9.97$

Checking for Extraneous Solutions When Solving Logarithmic Equations

It is possible to have extraneous solutions when solving logarithmic equations since the domain of a logarithm is $(0,\infty)$. Check your answers to see if they are extraneous or not.

Example $\PageIndex{15}$

Solve $\ln{x}+\ln(2x-5) = \ln{3}$

Solution

$\begin{align*} \ln{x}+\ln(2x-5) &= \ln3 \\[4pt] \ln(x(2x-5)) &= \ln3 \\[4pt] \ln(2x^2-5x) &= \ln3 \\[4pt] 2x^2-5x &= 3 \\[4pt] 2x^2-5x-3 &= 0 \\[4pt] (2x+1)(x-3) &=0 \\[4pt] 2x+1=0 \qquad x-3&=0 \\[4pt] x=\dfrac{1}{2} \qquad x&=3 \\[4pt] \end{align*}$

Example $\PageIndex{16}$

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 Power Property on the right.

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

Rewrite $x^{-1}=\dfrac{1}{x}$.

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

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

$\dfrac{x+6}{2 x+5}=\dfrac{1}{x}$

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.

Example $\PageIndex{17}$

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. $x=9$

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

You Try $\PageIndex{14}$

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

Answer

$x=6$

You Try $\PageIndex{15}$

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

Answer

$x=3$

You Try $\PageIndex{16}$

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

Answer

$x=8$

Solving Applied Problems

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

Example $\PageIndex{18}$

Suppose $$2000$ is invested in an account which offers $7.125 \%$ compounded monthly.

1. Express the amount $A$ in the account as a function of the term of the investment $t$ in years.
2. How much is in the account after $5$ years?
3. How long will it take for the initial investment to double?
4. Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year.

Solution

1. Substituting $P = 2000$, $r = 0.07125$, and $n = 12$ (since interest is compounded monthly into the compound interest formula yields $$A(t) = 2000\left(1 + \dfrac{0.07125}{12}\right)^{12t}=2000 (1.0059375)^{12t}\nonumber$$
2. Since $t$ represents the length of the investment in years, we substitute $t=5$ into $A(t)$ to find $$A(5) = 2000 (1.0059375)^{12(5)} \approx 2852.92\nonumber$$ After $5$ years, we have approximately $$2852.92$.
3. Our initial investment is $$2000$, so to find the time it takes this to double, we need to find $t$ when $A(t) = 4000$. We get $2000 (1.0059375)^{12t}=4000$, or $(1.0059375)^{12t}=2$. Taking natural logs of both sides, we get $$t = \dfrac{\ln(2)}{12 \ln(1.0059375)} \approx 9.75.\nonumber$$ Hence, it takes approximately $9$ years $9$ months for the investment to double.
4. To find the average rate of change of $A$ from the end of the fourth year to the end of the fifth year, we compute $$\frac{A(5)-A(4)}{5-4} \approx 195.63\nonumber$$ Similarly, the average rate of change of $A$ from the end of the thirty-fourth year to the end of the thirty-fifth year is $$\frac{A(35)-A(34)}{35-34} \approx 1648.21\nonumber$$ This means that the value of the investment is increasing at a rate of approximately $$195.63$ per year between the end of the fourth and fifth years, while that rate jumps to $$1648.21$ per year between the end of the thirty-fourth and thirty-fifth years. So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases. (in fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really defines exponential functions and we refer the reader to a course in Calculus.