Exponential and Logarithmic Equations

Learning Outcomes

Solve an exponential equation with a common base.
Rewrite an exponential equation so all terms have a common base then solve.
Recognize when an exponential equation does not have a solution.
Use logarithms to solve exponential equations.
Solve a logarithmic equation algebraically.
Solve a logarithmic equation graphically.
Use the one-to-one property of logarithms to solve a logarithmic equation.
Solve a radioactive decay problem.

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.

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

The first technique we will introduce for solving exponential equations involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where $b>0,\text{ }b\ne 1$ , ${b}^{S}={b}^{T}$ if and only if S = T.

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.

For example, consider the equation ${3}^{4x - 7}=\frac{{3}^{2x}}{3}$ . To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x:

$\begin{array}{l}{3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{3}\hfill & \hfill \\ {3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{{3}^{1}}\hfill & {\text{Rewrite 3 as 3}}^{1}.\hfill \\ {3}^{4x - 7}\hfill & ={3}^{2x - 1}\hfill & \text{Use the division property of exponents}\text{.}\hfill \\ 4x - 7\hfill & =2x - 1\text{ }\hfill & \text{Apply the one-to-one property of exponents}\text{.}\hfill \\ 2x\hfill & =6\hfill & \text{Subtract 2}x\text{ and add 7 to both sides}\text{.}\hfill \\ x\hfill & =3\hfill & \text{Divide by 3}\text{.}\hfill \end{array}$

A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions S and T, and any positive real number $b\ne 1$ ,

${b}^{S}={b}^{T}\text{ if and only if }S=T$

How To: Given an exponential equation Of the form ${b}^{S}={b}^{T}$ , where S and T are algebraic expressions with an unknown, solve for the unknown

Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form ${b}^{S}={b}^{T}$ .
Use the one-to-one property to set the exponents equal to each other.
Solve the resulting equation, S = T, for the unknown.

Example: Solving an Exponential Equation with a Common Base

Solve ${2}^{x - 1}={2}^{2x - 4}$ .

Show Solution

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

Try It

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

Show Solution

$x=–2$

Rewriting Equations So All Powers Have the Same Base

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:

$\begin{array}{l}256={4}^{x - 5}\hfill & \hfill \\ {2}^{8}={\left({2}^{2}\right)}^{x - 5}\hfill & \text{Rewrite each side as a power with base 2}.\hfill \\ {2}^{8}={2}^{2x - 10}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ 8=2x - 10\hfill & \text{Apply the one-to-one property of exponents}.\hfill \\ 18=2x\hfill & \text{Add 10 to both sides}.\hfill \\ x=9\hfill & \text{Divide by 2}.\hfill \end{array}$

How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it

Rewrite each side in the equation as a power with a common base.
Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form ${b}^{S}={b}^{T}$ .
Use the one-to-one property to set the exponents equal to each other.
Solve the resulting equation, S = T, for the unknown.

Example: Solving Equations by Rewriting Them to Have a Common Base

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

Show Solution

$\begin{array}{llllll}\text{ }{8}^{x+2}={16}^{x+1}\hfill & \hfill \\ {\left({2}^{3}\right)}^{x+2}={\left({2}^{4}\right)}^{x+1}\hfill & \text{Write }8\text{ and }16\text{ as powers of }2.\hfill \\ \text{ }{2}^{3x+6}={2}^{4x+4}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ \text{ }3x+6=4x+4\hfill & \text{Use the one-to-one property to set the exponents equal to each other}.\hfill \\ \text{ }x=2\hfill & \text{Solve for }x.\hfill \end{array}$

Try It

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

Show Solution

$x=–1$

Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

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

Show Solution

$\begin{array}{l}{2}^{5x}={2}^{\frac{1}{2}}\hfill & \text{Write the square root of 2 as a power of }2.\hfill \\ 5x=\frac{1}{2}\hfill & \text{Use the one-to-one property}.\hfill \\ x=\frac{1}{10}\hfill & \text{Solve for }x.\hfill \end{array}$

Try It

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

Show Solution

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

Q & A

Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation we may obtain an expression that is undefined.

Example: Determining When an Equation has No Solution

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

Show Solution

Try It

Solve ${2}^{x}=-100$ .

Show Solution

Using Logarithms to Solve Exponential Equations

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall that since $\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)$ is equal to a = b, we may apply logarithms with the same base to both sides of an exponential equation.

How To: Given an exponential equation Where a common base cannot be found, solve for the unknown

Apply the logarithm to both sides of the equation.
- If one of the terms in the equation has base 10, use the common logarithm.
- If none of the terms in the equation has base 10, use the natural logarithm.
Use the rules of logarithms to solve for the unknown.

Example: Solving an Equation Containing Powers of Different Bases

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

Show Solution

$\begin{array}{l}\text{ }{5}^{x+2}={4}^{x}\hfill & \text{There is no easy way to get the powers to have the same base}.\hfill \\ \text{ }\mathrm{ln}{5}^{x+2}=\mathrm{ln}{4}^{x}\hfill & \text{Take ln of both sides}.\hfill \\ \text{ }\left(x+2\right)\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the power rule for logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive property}.\hfill \\ \text{ }x\mathrm{ln}5-x\mathrm{ln}4=-2\mathrm{ln}5\hfill & \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}.\hfill \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5\hfill & \text{On the left hand side, factor out }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the properties of logs}.\hfill \\ \text{ }x=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}\hfill & \text{Divide by the coefficient of }x.\hfill \end{array}$

Try It

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

Show Solution

$x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(\frac{2}{3}\right)}$

Q & A

Is there any way to solve ${2}^{x}={3}^{x}$ ?

Yes. The solution is x = 0.

Equations Containing $e$

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

How To: Given an equation of the form $y=A{e}^{kt}$ , solve for $t$

Divide both sides of the equation by A.
Apply the natural logarithm to both sides of the equation.
Divide both sides of the equation by k.

Example: Solving an Equation of the Form $y=A{e}^{kt}$

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

Show Solution

$\begin{array}{l}100\hfill & =20{e}^{2t}\hfill & \hfill \\ 5\hfill & ={e}^{2t}\hfill & \text{Divide by the coefficient 20}\text{.}\hfill \\ \mathrm{ln}5\hfill & =\mathrm{ln}{{e}^{2t}}\hfill & \text{Take ln of both sides.}\hfill \\ \mathrm{ln}5\hfill & =2t\hfill & \text{Use the fact that }\mathrm{ln}\left(x\right)\text{ and }{e}^{x}\text{ are inverse functions}\text{.}\hfill \\ t\hfill & =\frac{\mathrm{ln}5}{2}\hfill & \text{Divide by the coefficient of }t\text{.}\hfill \end{array}$

Analysis of the Solution

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

Try It

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

Show Solution

t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or

l n {(\frac{11}{3})}^{2}

$\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}$

Q & A

Does every equation of the form $y=A{e}^{kt}$ have a solution?

No. There is a solution when $k\ne 0$ , and when $y$ and [latex]A[/latex] are either both 0 or neither 0 and they have the same sign. An example of an equation with this form that has no solution is $2=-3{e}^{t}$ .

Example: Solving an Equation That Can Be Simplified to the Form $y=A{e}^{kt}$

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

Show Solution

$\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Subtract 5 from both sides}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide both sides by 4}.\hfill \\ 2x=\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Take ln of both sides}.\hfill \\ x=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Solve for }x.\hfill \end{array}$

Try It

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

Show Solution

Extraneous Solutions

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 taking the logarithm of 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: Solving Exponential Functions in Quadratic Form

Solve ${e}^{2x}-{e}^{x}=56$ .

Show Solution

$\begin{array}{l}{e}^{2x}-{e}^{x}=56\hfill \\ {e}^{2x}-{e}^{x}-56=0\hfill & \text{Get one side of the equation equal to zero}.\hfill \\ \left({e}^{x}+7\right)\left({e}^{x}-8\right)=0\hfill & \text{Factor by the FOIL method}.\hfill \\ {e}^{x}+7=0\text{ or }{e}^{x}-8=0 & \text{If a product is zero, then one factor must be zero}.\hfill \\ {e}^{x}=-7{\text{ or e}}^{x}=8\hfill & \text{Isolate the exponentials}.\hfill \\ {e}^{x}=8\hfill & \text{Reject the equation in which the power equals a negative number}.\hfill \\ x=\mathrm{ln}8\hfill & \text{Solve the equation in which the power equals a positive number}.\hfill \end{array}$

Analysis of the Solution

When we plan to use factoring to solve a problem, we always get zero on one side of the equation because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation ${e}^{x}=-7$ because a positive number never equals a negative number. The solution $x=\mathrm{ln}\left(-7\right)$ is not a real number and in the real number system, this solution is rejected as an extraneous solution.

Try It

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

Show Solution

$x=\mathrm{ln}2$

Q & A

Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Logarithmic Equations

We have already seen that every logarithmic equation ${\mathrm{log}}_{b}\left(x\right)=y$ is equal 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 ${\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3$ . To solve this equation, we can use rules of logarithms to rewrite the left side as a single log and then apply the definition of logs to solve for $x$ :

$\begin{array}{l}{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3\hfill & \hfill \\ \text{ }{\mathrm{log}}_{2}\left(2\left(3x - 5\right)\right)=3\hfill & \text{Apply the product rule of logarithms}.\hfill \\ \text{ }{\mathrm{log}}_{2}\left(6x - 10\right)=3\hfill & \text{Distribute}.\hfill \\ \text{ }{2}^{3}=6x - 10\hfill & \text{Convert to exponential form}.\hfill \\ \text{ }8=6x - 10\hfill & \text{Calculate }{2}^{3}.\hfill \\ \text{ }18=6x\hfill & \text{Add 10 to both sides}.\hfill \\ \text{ }x=3\hfill & \text{Divide both sides by 6}.\hfill \end{array}$

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression S and real numbers b and c, where $b>0,\text{ }b\ne 1$ ,

${\mathrm{log}}_{b}\left(S\right)=c\text{ if and only if }{b}^{c}=S$

Example: Using Algebra to Solve a Logarithmic Equation

Solve $2\mathrm{ln}x+3=7$ .

Show Solution

$\begin{array}{l}2\mathrm{ln}x+3=7\hfill & \hfill \\ \text{}2\mathrm{ln}x=4\hfill & \text{Subtract 3 from both sides}.\hfill \\ \text{}\mathrm{ln}x=2\hfill & \text{Divide both sides by 2}.\hfill \\ \text{}x={e}^{2}\hfill & \text{Rewrite in exponential form}.\hfill \end{array}$

Try It

Solve $6+\mathrm{ln}x=10$ .

Show Solution

$x={e}^{4}$

Example: Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve $2\mathrm{ln}\left(6x\right)=7$ .

Show Solution

$\begin{array}{l}2\mathrm{ln}\left(6x\right)=7\hfill & \hfill \\ \text{}\mathrm{ln}\left(6x\right)=\frac{7}{2}\hfill & \text{Divide both sides by 2}.\hfill \\ \text{}6x={e}^{\left(\frac{7}{2}\right)}\hfill & \text{Use the definition of }\mathrm{ln}.\hfill \\ \text{}x=\frac{1}{6}{e}^{\left(\frac{7}{2}\right)}\hfill & \text{Divide both sides by 6}.\hfill \end{array}$

Try It

Solve $2\mathrm{ln}\left(x+1\right)=10$ .

Show Solution

$x={e}^{5}-1$

Example: Using a Graph to Understand the Solution to a Logarithmic Equation

Solve $\mathrm{ln}x=3$ .

Show Solution

$\begin{array}{l}\mathrm{ln}x=3\hfill & \hfill \\ x={e}^{3}\hfill & \text{Use the definition of }\mathrm{ln}\text{.}\hfill \end{array}$

Below is a graph of the equation. On the graph the x-coordinate of the point where the two graphs intersect is close to 20. In other words ${e}^{3}\approx 20$ . A calculator gives a better approximation: ${e}^{3}\approx 20.0855$ .

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

The graphs of $y=\mathrm{ln}x$ and y = 3 cross at the point $\left(e^3,3\right)$ which is approximately (20.0855, 3).

Try It

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

Show Solution

$x\approx 9.97$

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 $b\ne 1$ ,

${\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T$

For example,

$\text{If }{\mathrm{log}}_{2}\left(x - 1\right)={\mathrm{log}}_{2}\left(8\right),\text{then }x - 1=8$

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: ${\mathrm{log}}_{2}\left(9 - 1\right)={\mathrm{log}}_{2}\left(8\right)=3$ . 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 $\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)$ . 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:

$\begin{array}{l}\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(\frac{3x - 2}{2}\right)=\mathrm{log}\left(x+4\right)\hfill & \text{Apply the quotient rule of logarithms}.\hfill \\ \text{}\frac{3x - 2}{2}=x+4\hfill & \text{Apply the one-to-one property}.\hfill \\ \text{}3x - 2=2x+8\hfill & \text{Multiply both sides of the equation by }2.\hfill \\ \text{}x=10\hfill & \text{Subtract 2}x\text{ and add 2}.\hfill \end{array}$

To check the result, substitute x = 10 into $\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)$ .

$\begin{array}{l}\mathrm{log}\left(3\left(10\right)-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(\left(10\right)+4\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(28\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(14\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(\frac{28}{2}\right)=\mathrm{log}\left(14\right)\hfill & \text{The solution checks}.\hfill \end{array}$

A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

For any algebraic expressions S and T and any positive real number b, where $b\ne 1$ ,

${\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T$

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How To: Given an equation containing logarithms, solve it using the one-to-one property

Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of the form ${\mathrm{log}}_{b}S={\mathrm{log}}_{b}T$ .
Use the one-to-one property to set the arguments equal to each other.
Solve the resulting equation, S = T, for the unknown.

Example: Solving an Equation Using the One-to-One Property of Logarithms

Solve $\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right)$ .

Show Solution

$\begin{array}{l}\text{ }\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right)\hfill & \hfill \\ \text{ }{x}^{2}=2x+3\hfill & \text{Use the one-to-one property of the logarithm}.\hfill \\ \text{ }{x}^{2}-2x - 3=0\hfill & \text{Get zero on one side before factoring}.\hfill \\ \left(x - 3\right)\left(x+1\right)=0\hfill & \text{Factor using FOIL}.\hfill \\ \text{ }x - 3=0\text{ or }x+1=0\hfill & \text{If a product is zero, one of the factors must be zero}.\hfill \\ \text{ }x=3\text{ or }x=-1\hfill & \text{Solve for }x.\hfill \end{array}$

Analysis of the Solution

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

Try It

Solve $\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}1$ .

Show Solution

$x=1$ or $x=–1$

Key Equations

One-to-one property for exponential functions	For any algebraic expressions S and T and any positive real number b, where $b>0,\text{ }b\ne 1, {b}^{S}={b}^{T}$ if and only if S = T.
Definition of a logarithm	For any algebraic expression S and positive real numbers b and c, where $b\ne 1$ , ${\mathrm{log}}_{b}\left(S\right)=c$ if and only if ${b}^{c}=S$ .
One-to-one property for logarithmic functions	For any algebraic expressions S and T and any positive real number b, where $b\ne 1$ , ${\mathrm{log}}_{b}S={\mathrm{log}}_{b}T$ if and only if S = T.

Key Concepts

We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown.
When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown.
When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side.
We can solve exponential equations with base e by applying the natural logarithm to both sides because exponential and logarithmic functions are inverses of each other.
After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.
When given an equation of the form ${\mathrm{log}}_{b}\left(S\right)=c$ , where S is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation ${b}^{c}=S$ and solve for the unknown.
We can also use graphing to solve equations of the form ${\mathrm{log}}_{b}\left(S\right)=c$ . We graph both equations $y={\mathrm{log}}_{b}\left(S\right)$ and $y=c$ on the same coordinate plane and identify the solution as the x-value of the point of intersecting.
When given an equation of the form ${\mathrm{log}}_{b}S={\mathrm{log}}_{b}T$ , where S and T are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation S = T for the unknown.
Combining the skills learned in this and previous sections, we can solve equations that model real world situations whether the unknown is in an exponent or in the argument of a logarithm.

Glossary

extraneous solution: a solution introduced while solving an equation that does not satisfy the conditions of the original equation

Module 12: Exponential and Logarithmic Equations and Models