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## 4.7: Exponential and Logarithmic Equations

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## 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.

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 functions.

## Using Like Bases to Solve Exponential Equations

The first technique 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 \(b>0\), \(b≠1\), \(b^S=b^T\) 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 \(3^{4x−7}=\dfrac{3^{2x}}{3}\). 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 \(x\):

\[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\]

## 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 \(b≠1\),

\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]

## How to: Given an exponential equation with 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.
- Solve the resulting equation, \(S=T\), for the unknown.

## Example \(\PageIndex{1}\): Solving an Exponential Equation with a Common Base

Solve \(2^{x−1}=2^{2x−4}\).

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

## Exercise \(\PageIndex{1}\)

Solve \(5^{2x}=5^{3x+2}\).

\(x=−2\)

## Rewrite 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 \(x\):

\[\begin{align*} 256&= 4^{x-5}\\ 2^8&= {(2^2)}^{x-5} \qquad &&\text{Rewrite each side as a power with base 2}\\ 2^8&= 2^{2x-10} \qquad &&\text{Use the one-to-one property of exponents}\\ 8&= 2x-10 \qquad &&\text{Apply the one-to-one property of exponents}\\ 18&= 2x \qquad &&\text{Add 10 to both sides}\\ x&= 9 \qquad &&\text{Divide by 2} \end{align*}\]

## 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.

## Example \(\PageIndex{2}\): Solving Equations by Rewriting Them to Have a Common Base

Solve \(8^{x+2}={16}^{x+1}\).

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

## Exercise \(\PageIndex{2}\)

Solve \(5^{2x}={25}^{3x+2}\).

\(x=−1\)

## Example \(\PageIndex{3}\): Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve \(2^{5x}=\sqrt{2}\).

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

## Exercise \(\PageIndex{3}\)

Solve \(5^x=\sqrt{5}\).

\(x=\dfrac{1}{2}\)

## 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 \(\PageIndex{4}\): Solving an Equation with Positive and Negative Powers

Solve \(3^{x+1}=−2\).

This equation 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.

## Exercise \(\PageIndex{4}\)

Solve \(2^x=−100\).

The equation has no solution.

## Solving Exponential Equations Using Logarithms

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, since \(\log(a)=\log(b)\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.

## How to: Given an exponential equation in which a common base cannot be found, solve for the unknown

- 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 \(\PageIndex{5}\): Solving an Equation Containing Powers of Different Bases

Solve \(5^{x+2}=4^x\).

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

## Exercise \(\PageIndex{5}\)

Solve \(2^x=3^{x+1}\).

\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)

## Q&A: Is there any way to solve \(2^x=3^x\)?

Yes. The solution is \(0\).

## Equations Containing \(e\)

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.

## How to: Given an equation of the form \(y=Ae^{kt}\), solve for \(t\).

- Divide both sides of the equation by \(A\).
- Apply the natural logarithm of both sides of the equation.
- Divide both sides of the equation by \(k\).

## Example \(\PageIndex{6}\): Solve an Equation of the Form \(y = Ae^{kt}\)

Solve \(100=20e^{2t}\).

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

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.

## Exercise \(\PageIndex{6}\)

Solve \(3e^{0.5t}=11\).

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

## Q&A: Does every equation of the form \(y=Ae^{kt}\) have a solution?

No. There is a solution when \(k≠0\),and when \(y\) and \(A\) 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 \(2=−3e^t\).

## Example \(\PageIndex{7}\): Solving an Equation That Can Be Simplified to the Form \(y=Ae^{kt}\)

Solve \(4e^{2x}+5=12\).

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

## Exercise \(\PageIndex{7}\)

Solve \(3+e^{2t}=7e^{2t}\).

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

## 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 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 \(\PageIndex{8}\): Solving Exponential Functions in Quadratic Form

Solve \(e^{2x}−e^x=56\).

\[\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\\ e^x+7&= 0 \qquad &&\text{or} \\ e^x-8&= 0 \qquad &&\text{If a product is zero, then one factor must be zero}\\ e^x&= -7 \qquad &&\text{or} \\ e^x&= 8 \qquad &&\text{Isolate the exponentials}\\ e^x&= 8 \qquad &&\text{Reject the equation in which the power equals a negative number}\\ x&= \ln8 \qquad &&\text{Solve the equation in which the power equals a positive number} \end{align*}\]

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 \(\ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

## Exercise \(\PageIndex{8}\)

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

## 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.

## 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 \(x\):

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

## USING THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expression \(S\) and real numbers \(b\) and \(c\),where \(b>0\), \(b≠1\),

\[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\]

## Example \(\PageIndex{9}\): Using Algebra to Solve a Logarithmic Equation

Solve \(2\ln x+3=7\).

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

## Exercise \(\PageIndex{9}\)

Solve \(6+\ln x=10\).

## Example \(\PageIndex{10}\): Using Algebra Before and After Using the Definition of the Natural Logarithm

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

\[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad \text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad \text{Use the definition of }\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad \text{Divide by 6} \end{align*}\]

## Exercise \(\PageIndex{10}\)

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

\(x=e^5−1\)

## Example \(\PageIndex{11}\): Using a Graph to Understand the Solution to a Logarithmic Equation

Solve \(\ln x=3\).

\[\begin{align*} \ln x&= 3\\ x&= e^3 \qquad \text{Use the definition of the natural logarithm} \end{align*}\]

Figure \(\PageIndex{3}\) 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: \(e^3≈20.0855\).

## Exercise \(\PageIndex{11}\)

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

\(x≈9.97\)

## 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 \(b≠1\),

\({\log}_bS={\log}_bT\) 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 \(x\):

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

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

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

## 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 \(b≠1\),

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 has the form \({\log}_bS={\log}_bT\).
- Use the one-to-one property to set the arguments equal.

## Example \(\PageIndex{12}\): Solving an Equation Using the One-to-One Property of Logarithms

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

\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad \text{Use the one-to-one property of the logarithm}\\ x^2-2x-3&= 0 \qquad \text{Get zero on one side before factoring}\\ (x-3)(x+1)&= 0 \qquad \text{Factor using FOIL}\\ x-3&= 0 \qquad \text{or } x+1=0 \text{ If a product is zero, one of the factors must be zero}\\ x=3 \qquad \text{or} \\ x&= -11 \qquad \text{Solve for x} \end{align*}\]

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.

## Exercise \(\PageIndex{12}\)

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

\(x=1\) or \(x=−1\)

## Solving Applied Problems Using Exponential and Logarithmic Equations

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.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . Table \(\PageIndex{1}\) lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

\[\begin{align} A(t)&= A_0e^{\tfrac{\ln(0.5)}{T}t}\\ A(t)&= A_0e^{\tfrac{\ln(0.5)t}{T}}\\ A(t)&= A_0{(e^{\ln(0.5)})}^{\tfrac{t}{T}}\\ A(t)&= A_0{\left (\dfrac{1}{2}\right )}^{\tfrac{t}{T}}\\ \end{align}\]

- \(A_0\) is the amount initially present
- \(T\) is the half-life of the substance
- \(t\) is the time period over which the substance is studied
- \(y\) is the amount of the substance present after time \(t\)

## Example \(\PageIndex{13}\): Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a \(1000\)-gram sample of uranium-235 to decay?

\[\begin{align*} y&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t}\\ 900&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{After } 10\% \text{ decays, 900 grams are left}\\ 0.9&= e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{Divide by 1000}\\ \ln(0.9)&= \ln \left (e^{\tfrac{\ln(0.5)}{703,800,000}t} \right ) \qquad \text{Take ln of both sides}\\ \ln(0.9)&= \dfrac{\ln(0.5)}{703,800,000}t \qquad \ln(e^M)=M\\ t&= 703,800,000\times \dfrac{\ln(0.9)}{\ln(0.5)} \qquad \text{years Solve for t}\\ t&\approx 106,979,777 \qquad \text{years} \end {align*} \]

Ten percent of \(1000\) grams is \(100\) grams. If \(100\) grams decay, the amount of uranium-235 remaining is \(900\) grams.

## Exercise \(\PageIndex{13}\)

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

\(t=703,800,000×\dfrac{\ln(0.8)}{\ln(0.5)}\)years ≈ 226,572,993 years.

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

- Solving Logarithmic Equations
- Solving Exponential Equations with Logarithms

## Key Equations

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. See Example \(\PageIndex{1}\).
- 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. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
- When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example \(\PageIndex{5}\).
- We can solve exponential equations with base \(e\),by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\).
- After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example \(\PageIndex{8}\).
- When given an equation of the form \({\log}_b(S)=c\), 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. See Example \(\PageIndex{9}\) and Example \(\PageIndex{10}\).
- We can also use graphing to solve equations with the form \({\log}_b(S)=c\). We graph both equations \(y={\log}_b(S)\) and \(y=c\) on the same coordinate plane and identify the solution as the x- value of the intersecting point. See Example \(\PageIndex{11}\).
- When given an equation of the form \({\log}_bS={\log}_bT\), 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. See Example \(\PageIndex{12}\).
- 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. See Example \(\PageIndex{13}\).

## 6.6 Exponential and Logarithmic Equations

Learning objectives.

In this section, you will:

- 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.

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 functions.

## Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b , b , S , S , and T , T , where b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , b S = b T b S = b T if and only if S = T . 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 4 x − 7 = 3 2 x 3 . 3 4 x − 7 = 3 2 x 3 . To solve for x , x , we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x x :

## Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions S and T , S and T , and any positive real number b ≠ 1 , b ≠ 1 ,

Given an exponential equation with the form b S = b T , b S = b T , where S S and T 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 . b S = b T .
- Use the one-to-one property to set the exponents equal.
- Solve the resulting equation, S = T , S = T , for the unknown.

## Solving an Exponential Equation with a Common Base

Solve 2 x − 1 = 2 2 x − 4 . 2 x − 1 = 2 2 x − 4 .

Solve 5 2 x = 5 3 x + 2 . 5 2 x = 5 3 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 . 256 = 4 x − 5 . We can rewrite both sides of this equation as a power of 2. 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x : x :

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.

## Solving Equations by Rewriting Them to Have a Common Base

Solve 8 x + 2 = 16 x + 1 . 8 x + 2 = 16 x + 1 .

Solve 5 2 x = 25 3 x + 2 . 5 2 x = 25 3 x + 2 .

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

Solve 2 5 x = 2 . 2 5 x = 2 .

Solve 5 x = 5 . 5 x = 5 .

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.

## Solving an Equation with Positive and Negative Powers

Solve 3 x + 1 = −2. 3 x + 1 = −2.

This equation has no solution. There is no real value of x x that will make the equation a true statement because any power of a positive number is positive.

Figure 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.

Solve 2 x = −100. 2 x = −100.

## Solving Exponential Equations Using Logarithms

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, since log ( a ) = log ( b ) log ( a ) = log ( b ) is equivalent to a = b , a = b , we may apply logarithms with the same base on both sides of an exponential equation.

Given an exponential equation in which a common base cannot be found, solve for the unknown.

- 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.

## Solving an Equation Containing Powers of Different Bases

Solve 5 x + 2 = 4 x . 5 x + 2 = 4 x .

Solve 2 x = 3 x + 1 . 2 x = 3 x + 1 .

Is there any way to solve 2 x = 3 x ? 2 x = 3 x ?

Yes. The solution is 0. 0.

## Equations Containing e

One common type of exponential equations are those with base e . 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 e on either side, we can use the natural logarithm to solve it.

Given an equation of the form y = A e k t , y = A e k t , solve for t . t .

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

## Solve an Equation of the Form y = Ae kt

Solve 100 = 20 e 2 t . 100 = 20 e 2 t .

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

Solve 3 e 0.5 t = 11. 3 e 0.5 t = 11.

Does every equation of the form y = A e k t y = A e k t have a solution?

No. There is a solution when k ≠ 0 , k ≠ 0 , and when y y and A A 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 . 2 = −3 e t .

## Solving an Equation That Can Be Simplified to the Form y = Ae kt

Solve 4 e 2 x + 5 = 12. 4 e 2 x + 5 = 12.

Solve 3 + e 2 t = 7 e 2 t . 3 + e 2 t = 7 e 2 t .

## 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 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.

## Solving Exponential Functions in Quadratic Form

Solve e 2 x − e x = 56. e 2 x − e x = 56.

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 e x = −7 because a positive number never equals a negative number. The solution ln ( −7 ) ln ( −7 ) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Solve e 2 x = e x + 2. e 2 x = e x + 2.

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.

## Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation log b ( x ) = y log b ( x ) = y is equivalent to the exponential equation b y = x . 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 ( 3 x − 5 ) = 3. log 2 ( 2 ) + log 2 ( 3 x − 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 : x :

For any algebraic expression S S and real numbers b b and c , c , where b > 0 , b ≠ 1 , b > 0 , b ≠ 1 ,

## Using Algebra to Solve a Logarithmic Equation

Solve 2 ln x + 3 = 7. 2 ln x + 3 = 7.

Solve 6 + ln x = 10. 6 + ln x = 10.

## Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve 2 ln ( 6 x ) = 7. 2 ln ( 6 x ) = 7.

Solve 2 ln ( x + 1 ) = 10. 2 ln ( x + 1 ) = 10.

## Using a Graph to Understand the Solution to a Logarithmic Equation

Solve ln x = 3. ln x = 3.

Figure 3 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. e 3 ≈ 20. A calculator gives a better approximation: e 3 ≈ 20.0855. e 3 ≈ 20.0855.

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

## 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 , x > 0 , S > 0 , S > 0 , T > 0 T > 0 and any positive real number b , b , where b ≠ 1 , b ≠ 1 ,

For example,

So, if x − 1 = 8 , x − 1 = 8 , then we can solve for x , x , and we get x = 9. x = 9. To check, we can substitute x = 9 x = 9 into the original equation: log 2 ( 9 − 1 ) = log 2 ( 8 ) = 3. 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 ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) . log ( 3 x − 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 : x :

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

For any algebraic expressions S S and T T and any positive real number b , b , where b ≠ 1 , b ≠ 1 ,

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

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 has the form log b S = log b T . log b S = log b T .
- Use the one-to-one property to set the arguments equal.

## Solving an Equation Using the One-to-One Property of Logarithms

Solve ln ( x 2 ) = ln ( 2 x + 3 ) . ln ( x 2 ) = ln ( 2 x + 3 ) .

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

Solve ln ( x 2 ) = ln 1. ln ( x 2 ) = ln 1.

## Solving Applied Problems Using Exponential and Logarithmic Equations

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.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . Table 1 lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

- A 0 A 0 is the amount initially present
- T T is the half-life of the substance
- t t is the time period over which the substance is studied
- A ( t ) A ( t ) is the amount of the substance present after time t t

## Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

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

- Solving Logarithmic Equations
- Solving Exponential Equations with Logarithms

## 6.6 Section Exercises

How can an exponential equation be solved?

When does an extraneous solution occur? How can an extraneous solution be recognized?

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

For the following exercises, use like bases to solve the exponential equation.

4 − 3 v − 2 = 4 − v 4 − 3 v − 2 = 4 − v

64 ⋅ 4 3 x = 16 64 ⋅ 4 3 x = 16

3 2 x + 1 ⋅ 3 x = 243 3 2 x + 1 ⋅ 3 x = 243

2 − 3 n ⋅ 1 4 = 2 n + 2 2 − 3 n ⋅ 1 4 = 2 n + 2

625 ⋅ 5 3 x + 3 = 125 625 ⋅ 5 3 x + 3 = 125

36 3 b 36 2 b = 216 2 − b 36 3 b 36 2 b = 216 2 − b

( 1 64 ) 3 n ⋅ 8 = 2 6 ( 1 64 ) 3 n ⋅ 8 = 2 6

For the following exercises, use logarithms to solve.

9 x − 10 = 1 9 x − 10 = 1

2 e 6 x = 13 2 e 6 x = 13

e r + 10 − 10 = −42 e r + 10 − 10 = −42

2 ⋅ 10 9 a = 29 2 ⋅ 10 9 a = 29

− 8 ⋅ 10 p + 7 − 7 = −24 − 8 ⋅ 10 p + 7 − 7 = −24

7 e 3 n − 5 + 5 = −89 7 e 3 n − 5 + 5 = −89

e − 3 k + 6 = 44 e − 3 k + 6 = 44

− 5 e 9 x − 8 − 8 = −62 − 5 e 9 x − 8 − 8 = −62

− 6 e 9 x + 8 + 2 = −74 − 6 e 9 x + 8 + 2 = −74

2 x + 1 = 5 2 x − 1 2 x + 1 = 5 2 x − 1

e 2 x − e x − 132 = 0 e 2 x − e x − 132 = 0

7 e 8 x + 8 − 5 = −95 7 e 8 x + 8 − 5 = −95

10 e 8 x + 3 + 2 = 8 10 e 8 x + 3 + 2 = 8

4 e 3 x + 3 − 7 = 53 4 e 3 x + 3 − 7 = 53

8 e − 5 x − 2 − 4 = −90 8 e − 5 x − 2 − 4 = −90

3 2 x + 1 = 7 x − 2 3 2 x + 1 = 7 x − 2

e 2 x − e x − 6 = 0 e 2 x − e x − 6 = 0

3 e 3 − 3 x + 6 = −31 3 e 3 − 3 x + 6 = −31

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

log ( 1 100 ) = −2 log ( 1 100 ) = −2

log 324 ( 18 ) = 1 2 log 324 ( 18 ) = 1 2

For the following exercises, use the definition of a logarithm to solve the equation.

5 log 7 n = 10 5 log 7 n = 10

− 8 log 9 x = 16 − 8 log 9 x = 16

4 + log 2 ( 9 k ) = 2 4 + log 2 ( 9 k ) = 2

2 log ( 8 n + 4 ) + 6 = 10 2 log ( 8 n + 4 ) + 6 = 10

10 − 4 ln ( 9 − 8 x ) = 6 10 − 4 ln ( 9 − 8 x ) = 6

For the following exercises, use the one-to-one property of logarithms to solve.

ln ( 10 − 3 x ) = ln ( − 4 x ) ln ( 10 − 3 x ) = ln ( − 4 x )

log 13 ( 5 n − 2 ) = log 13 ( 8 − 5 n ) log 13 ( 5 n − 2 ) = log 13 ( 8 − 5 n )

log ( x + 3 ) − log ( x ) = log ( 74 ) log ( x + 3 ) − log ( x ) = log ( 74 )

ln ( − 3 x ) = ln ( x 2 − 6 x ) ln ( − 3 x ) = ln ( x 2 − 6 x )

log 4 ( 6 − m ) = log 4 3 m log 4 ( 6 − m ) = log 4 3 m

ln ( x − 2 ) − ln ( x ) = ln ( 54 ) ln ( x − 2 ) − ln ( x ) = ln ( 54 )

log 9 ( 2 n 2 − 14 n ) = log 9 ( − 45 + n 2 ) log 9 ( 2 n 2 − 14 n ) = log 9 ( − 45 + n 2 )

ln ( x 2 − 10 ) + ln ( 9 ) = ln ( 10 ) ln ( x 2 − 10 ) + ln ( 9 ) = ln ( 10 )

For the following exercises, solve each equation for x . x .

log ( x + 12 ) = log ( x ) + log ( 12 ) log ( x + 12 ) = log ( x ) + log ( 12 )

ln ( x ) + ln ( x − 3 ) = ln ( 7 x ) ln ( x ) + ln ( x − 3 ) = ln ( 7 x )

log 2 ( 7 x + 6 ) = 3 log 2 ( 7 x + 6 ) = 3

ln ( 7 ) + ln ( 2 − 4 x 2 ) = ln ( 14 ) ln ( 7 ) + ln ( 2 − 4 x 2 ) = ln ( 14 )

log 8 ( x + 6 ) − log 8 ( x ) = log 8 ( 58 ) log 8 ( x + 6 ) − log 8 ( x ) = log 8 ( 58 )

ln ( 3 ) − ln ( 3 − 3 x ) = ln ( 4 ) ln ( 3 ) − ln ( 3 − 3 x ) = ln ( 4 )

log 3 ( 3 x ) − log 3 ( 6 ) = log 3 ( 77 ) log 3 ( 3 x ) − log 3 ( 6 ) = log 3 ( 77 )

For the following exercises, solve the equation for x , x , if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

log 9 ( x ) − 5 = −4 log 9 ( x ) − 5 = −4

log 3 ( x ) + 3 = 2 log 3 ( x ) + 3 = 2

ln ( 3 x ) = 2 ln ( 3 x ) = 2

ln ( x − 5 ) = 1 ln ( x − 5 ) = 1

log ( 4 ) + log ( − 5 x ) = 2 log ( 4 ) + log ( − 5 x ) = 2

− 7 + log 3 ( 4 − x ) = −6 − 7 + log 3 ( 4 − x ) = −6

ln ( 4 x − 10 ) − 6 = − 5 ln ( 4 x − 10 ) − 6 = − 5

log ( 4 − 2 x ) = log ( − 4 x ) log ( 4 − 2 x ) = log ( − 4 x )

log 11 ( − 2 x 2 − 7 x ) = log 11 ( x − 2 ) log 11 ( − 2 x 2 − 7 x ) = log 11 ( x − 2 )

ln ( 2 x + 9 ) = ln ( − 5 x ) ln ( 2 x + 9 ) = ln ( − 5 x )

log 9 ( 3 − x ) = log 9 ( 4 x − 8 ) log 9 ( 3 − x ) = log 9 ( 4 x − 8 )

log ( x 2 + 13 ) = log ( 7 x + 3 ) log ( x 2 + 13 ) = log ( 7 x + 3 )

3 log 2 ( 10 ) − log ( x − 9 ) = log ( 44 ) 3 log 2 ( 10 ) − log ( x − 9 ) = log ( 44 )

ln ( x ) − ln ( x + 3 ) = ln ( 6 ) ln ( x ) − ln ( x + 3 ) = ln ( 6 )

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of $6,500 $6,500 earns 7.25 % 7.25 % annual interest, compounded continuously. How much will the account be worth after 20 years?

The formula for measuring sound intensity in decibels D D is defined by the equation D = 10 log ( I I 0 ) , D = 10 log ( I I 0 ) , where I I is the intensity of the sound in watts per square meter and I 0 = 10 − 12 I 0 = 10 − 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.3 ⋅ 10 2 8.3 ⋅ 10 2 watts per square meter?

The population of a small town is modeled by the equation P = 1650 e 0.5 t P = 1650 e 0.5 t where t t is measured in years. In approximately how many years will the town’s population reach 20,000? 20,000?

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places.

1000 ( 1.03 ) t = 5000 1000 ( 1.03 ) t = 5000 using the common log.

e 5 x = 17 e 5 x = 17 using the natural log

3 ( 1.04 ) 3 t = 8 3 ( 1.04 ) 3 t = 8 using the common log

3 4 x − 5 = 38 3 4 x − 5 = 38 using the common log

50 e − 0.12 t = 10 50 e − 0.12 t = 10 using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

7 e 3 x − 5 + 7.9 = 47 7 e 3 x − 5 + 7.9 = 47

ln ( 3 ) + ln ( 4.4 x + 6.8 ) = 2 ln ( 3 ) + ln ( 4.4 x + 6.8 ) = 2

log ( − 0.7 x − 9 ) = 1 + 5 log ( 5 ) log ( − 0.7 x − 9 ) = 1 + 5 log ( 5 )

Atmospheric pressure P P in pounds per square inch is represented by the formula P = 14.7 e − 0.21 x , P = 14.7 e − 0.21 x , where x x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 8.369 pounds per square inch? ( Hint : there are 5280 feet in a mile)

The magnitude M of an earthquake is represented by the equation M = 2 3 log ( E E 0 ) M = 2 3 log ( E E 0 ) where E E is the amount of energy released by the earthquake in joules and E 0 = 10 4.4 E 0 = 10 4.4 is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing 1.4 ⋅ 10 13 1.4 ⋅ 10 13 joules of energy?

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that b log b x = x . b log b x = x .

Recall the formula for continually compounding interest, y = A e k t . y = A e k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

Recall the compound interest formula A = a ( 1 + r k ) k t . A = a ( 1 + r k ) k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t . t .

Newton’s Law of Cooling states that the temperature T T of an object at any time t can be described by the equation T = T s + ( T 0 − T s ) e − k t , T = T s + ( T 0 − T s ) e − k t , where T s T s is the temperature of the surrounding environment, T 0 T 0 is the initial temperature of the object, and k k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

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## Exponential and Logarithmic Functions

Exponential and logarithmic equations, learning objectives.

In this section, you will:

- 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.

Figure 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)

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 functions.

## Using Like Bases to Solve Exponential Equations

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

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

## Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions[latex]\,S\text{ and }T,[/latex] and any positive real number[latex]\,b\ne 1,[/latex]

Given an exponential equation with the form[latex]\,{b}^{S}={b}^{T},[/latex] where [latex]\,S\,[/latex] and [latex]\,T\,[/latex] 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[latex]\,{b}^{S}={b}^{T}.[/latex]
- Use the one-to-one property to set the exponents equal.
- Solve the resulting equation,[latex]\,S=T,[/latex] for the unknown.

## Solving an Exponential Equation with a Common Base

Solve[latex]\,{2}^{x-1}={2}^{2x-4}.[/latex]

Solve[latex]\,{5}^{2x}={5}^{3x+2}.[/latex]

[latex]x=-2[/latex]

## 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[latex]\,256={4}^{x-5}.\,[/latex]We can rewrite both sides of this equation as a power of[latex]\,2.\,[/latex]Then we apply the rules of exponents, along with the one-to-one property, to solve for[latex]\,x:[/latex]

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.

## Solving Equations by Rewriting Them to Have a Common Base

Solve[latex]\,{8}^{x+2}={16}^{x+1}.[/latex]

Solve[latex]\,{5}^{2x}={25}^{3x+2}.[/latex]

[latex]x=-1[/latex]

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

Solve[latex]\,{2}^{5x}=\sqrt{2}.[/latex]

Solve[latex]\,{5}^{x}=\sqrt{5}.[/latex]

[latex]x=\frac{1}{2}[/latex]

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.

## Solving an Equation with Positive and Negative Powers

Solve[latex]\,{3}^{x+1}=-2.[/latex]

This equation has no solution. There is no real value of[latex]\,x\,[/latex]that will make the equation a true statement because any power of a positive number is positive.

(Figure) 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.

Solve[latex]\,{2}^{x}=-100.[/latex]

The equation has no solution.

## Solving Exponential Equations Using Logarithms

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, since[latex]\,\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)\,[/latex]is equivalent to[latex]\,a=b,[/latex] we may apply logarithms with the same base on both sides of an exponential equation.

Given an exponential equation in which a common base cannot be found, solve for the unknown.

- 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.

## Solving an Equation Containing Powers of Different Bases

Solve[latex]\,{5}^{x+2}={4}^{x}.[/latex]

Solve[latex]\,{2}^{x}={3}^{x+1}.[/latex]

Is there any way to solve[latex]\,{2}^{x}={3}^{x}?[/latex]

Yes. The solution is [latex]0.[/latex]

## Equations Containing e

One common type of exponential equations are those with base[latex]\,e.\,[/latex]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[latex]\,e\,[/latex]on either side, we can use the natural logarithm to solve it.

Given an equation of the form[latex]\,y=A{e}^{kt}\text{,}[/latex] solve for[latex]\,t.[/latex]

- Divide both sides of the equation by[latex]\,A.[/latex]
- Apply the natural logarithm of both sides of the equation.
- Divide both sides of the equation by[latex]\,k.[/latex]

## Solve an Equation of the Form y = Ae kt

Solve[latex]\,100=20{e}^{2t}.[/latex]

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

Solve[latex]\,3{e}^{0.5t}=11.[/latex]

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

Does every equation of the form [latex]\,y=A{e}^{kt}\,[/latex] have a solution?

No. There is a solution when[latex]\,k\ne 0,[/latex]and when[latex]\,y\,[/latex]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[latex]\,2=-3{e}^{t}.[/latex]

## Solving an Equation That Can Be Simplified to the Form y = Ae kt

Solve[latex]\,4{e}^{2x}+5=12.[/latex]

Solve[latex]\,3+{e}^{2t}=7{e}^{2t}.[/latex]

[latex]t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)[/latex]

## 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 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.

## Solving Exponential Functions in Quadratic Form

Solve[latex]\,{e}^{2x}-{e}^{x}=56.[/latex]

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[latex]\,{e}^{x}=-7\,[/latex]because a positive number never equals a negative number. The solution [latex]\,\mathrm{ln}\left(-7\right)\,[/latex]is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Solve[latex]\,{e}^{2x}={e}^{x}+2.[/latex]

[latex]x=\mathrm{ln}2[/latex]

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.

## Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation [latex]\,{\mathrm{log}}_{b}\left(x\right)=y\,[/latex]is equivalent to the exponential equation[latex]\,{b}^{y}=x.\,[/latex]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[latex]\,{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x-5\right)=3.\,[/latex]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[latex]\,x:[/latex]

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

## Using Algebra to Solve a Logarithmic Equation

Solve[latex]\,2\mathrm{ln}x+3=7.[/latex]

Solve[latex]\,6+\mathrm{ln}x=10.[/latex]

[latex]x={e}^{4}[/latex]

## Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve[latex]\,2\mathrm{ln}\left(6x\right)=7.[/latex]

Solve[latex]\,2\mathrm{ln}\left(x+1\right)=10.[/latex]

[latex]x={e}^{5}-1[/latex]

## Using a Graph to Understand the Solution to a Logarithmic Equation

Solve[latex]\,\mathrm{ln}x=3.[/latex]

(Figure) 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[latex]\,{e}^{3}\approx 20.\,[/latex]A calculator gives a better approximation:[latex]\,{e}^{3}\approx 20.0855.[/latex]

Figure 3. The graphs of[latex]\,y=\mathrm{ln}x\,[/latex]and[latex]\,y=3\,[/latex]cross at the point[latex]\,{\text{(e}}^{3},3\right),[/latex]which is approximately (20.0855, 3).

Use a graphing calculator to estimate the approximate solution to the logarithmic equation[latex]\,{2}^{x}=1000\,[/latex]to 2 decimal places.

[latex]x\approx 9.97[/latex]

## 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[latex]\,x>0,[/latex] [latex]S>0,[/latex] [latex]T>0\,[/latex]and any positive real number[latex]\,b,[/latex] where[latex]\,b\ne 1,[/latex]

For example,

So, if[latex]\,x-1=8,[/latex]then we can solve for[latex]\,x,[/latex]and we get[latex]\,x=9.\,[/latex]To check, we can substitute[latex]\,x=9\,[/latex]into the original equation:[latex]\,{\mathrm{log}}_{2}\left(9-1\right)={\mathrm{log}}_{2}\left(8\right)=3.\,[/latex]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[latex]\,\mathrm{log}\left(3x-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right).\,[/latex]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[latex]\,x:[/latex]

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

For any algebraic expressions[latex]\,S\,[/latex]and[latex]\,T\,[/latex]and any positive real number[latex]\,b,[/latex] where[latex]\,b\ne 1,[/latex]

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

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 has the form[latex]\,{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T.[/latex]
- Use the one-to-one property to set the arguments equal.

## Solving an Equation Using the One-to-One Property of Logarithms

Solve[latex]\,\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}\left(2x+3\right).[/latex]

There are two solutions: [latex]\,3\,[/latex] or [latex]\,-1.\,[/latex]The solution [latex]\,-1\,[/latex]is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Solve[latex]\,\mathrm{ln}\left({x}^{2}\right)=\mathrm{ln}1.[/latex]

[latex]x=1\,[/latex]or[latex]\,x=-1[/latex]

## Solving Applied Problems Using Exponential and Logarithmic Equations

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.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . (Figure) lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

- [latex]{A}_{0}\,[/latex]is the amount initially present
- [latex]T\,[/latex]is the half-life of the substance
- [latex]t\,[/latex]is the time period over which the substance is studied
- [latex]y\,[/latex]is the amount of the substance present after time[latex]\,t[/latex]

## Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

[latex]t=703,800,000×\frac{\mathrm{ln}\left(0.8\right)}{\mathrm{ln}\left(0.5\right)}\text{ years }\approx \text{ }226,572,993\text{ years}.[/latex]

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

- Solving Logarithmic Equations
- Solving Exponential Equations with Logarithms

## Key Equations

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. See (Figure) .
- 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. See (Figure) , (Figure) , and (Figure) .
- When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See (Figure) .
- We can solve exponential equations with base[latex]\,e,[/latex]by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See (Figure) and (Figure) .
- After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See (Figure) .
- When given an equation of the form[latex]\,{\mathrm{log}}_{b}\left(S\right)=c,\text{}[/latex]where[latex]\,S\,[/latex]is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation[latex]\,{b}^{c}=S,\text{}[/latex]and solve for the unknown. See (Figure) and (Figure) .
- We can also use graphing to solve equations with the form[latex]\,{\mathrm{log}}_{b}\left(S\right)=c.\,[/latex]We graph both equations[latex]\,y={\mathrm{log}}_{b}\left(S\right)\,[/latex]and[latex]\,y=c\,[/latex]on the same coordinate plane and identify the solution as the x- value of the intersecting point. See (Figure) .
- When given an equation of the form[latex]\,{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T,\text{}[/latex]where[latex]\,S\,[/latex]and[latex]\,T\,[/latex]are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation[latex]\,S=T\,[/latex]for the unknown. See (Figure) .
- 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. See (Figure) .

## Section Exercises

How can an exponential equation be solved?

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

When does an extraneous solution occur? How can an extraneous solution be recognized?

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

For the following exercises, use like bases to solve the exponential equation.

[latex]{4}^{-3v-2}={4}^{-v}[/latex]

[latex]64\cdot {4}^{3x}=16[/latex]

[latex]x=-\frac{1}{3}[/latex]

[latex]{3}^{2x+1}\cdot {3}^{x}=243[/latex]

[latex]{2}^{-3n}\cdot \frac{1}{4}={2}^{n+2}[/latex]

[latex]n=-1[/latex]

[latex]625\cdot {5}^{3x+3}=125[/latex]

[latex]\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}[/latex]

[latex]b=\frac{6}{5}[/latex]

[latex]{\left(\frac{1}{64}\right)}^{3n}\cdot 8={2}^{6}[/latex]

For the following exercises, use logarithms to solve.

[latex]{9}^{x-10}=1[/latex]

[latex]x=10[/latex]

[latex]2{e}^{6x}=13[/latex]

[latex]{e}^{r+10}-10=-42[/latex]

No solution

[latex]2\cdot {10}^{9a}=29[/latex]

[latex]-8\cdot {10}^{p+7}-7=-24[/latex]

[latex]p=\mathrm{log}\left(\frac{17}{8}\right)-7[/latex]

[latex]7{e}^{3n-5}+5=-89[/latex]

[latex]{e}^{-3k}+6=44[/latex]

[latex]k=-\frac{\mathrm{ln}\left(38\right)}{3}[/latex]

[latex]-5{e}^{9x-8}-8=-62[/latex]

[latex]-6{e}^{9x+8}+2=-74[/latex]

[latex]x=\frac{\mathrm{ln}\left(\frac{38}{3}\right)-8}{9}[/latex]

[latex]{2}^{x+1}={5}^{2x-1}[/latex]

[latex]{e}^{2x}-{e}^{x}-132=0[/latex]

[latex]x=\mathrm{ln}12\text{ }[/latex]

[latex]7{e}^{8x+8}-5=-95[/latex]

[latex]10{e}^{8x+3}+2=8[/latex]

[latex]x=\frac{\mathrm{ln}\left(\frac{3}{5}\right)-3}{8}[/latex]

[latex]4{e}^{3x+3}-7=53[/latex]

[latex]8{e}^{-5x-2}-4=-90[/latex]

no solution

[latex]{3}^{2x+1}={7}^{x-2}[/latex]

[latex]{e}^{2x}-{e}^{x}-6=0[/latex]

[latex]x=\mathrm{ln}\left(3\right)[/latex]

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

[latex]\mathrm{log}\left(\frac{1}{100}\right)=-2[/latex]

[latex]{10}^{-2}=\frac{1}{100}[/latex]

[latex]{\mathrm{log}}_{324}\left(18\right)=\frac{1}{2}[/latex]

For the following exercises, use the definition of a logarithm to solve the equation.

[latex]5{\mathrm{log}}_{7}n=10[/latex]

[latex]n=49[/latex]

[latex]-8{\mathrm{log}}_{9}x=16[/latex]

[latex]4+{\mathrm{log}}_{2}\left(9k\right)=2[/latex]

[latex]k=\frac{1}{36}[/latex]

[latex]2\mathrm{log}\left(8n+4\right)+6=10[/latex]

[latex]10-4\mathrm{ln}\left(9-8x\right)=6[/latex]

[latex]x=\frac{9-e}{8}[/latex]

For the following exercises, use the one-to-one property of logarithms to solve.

[latex]\mathrm{ln}\left(10-3x\right)=\mathrm{ln}\left(-4x\right)[/latex]

[latex]{\mathrm{log}}_{13}\left(5n-2\right)={\mathrm{log}}_{13}\left(8-5n\right)[/latex]

[latex]n=1[/latex]

[latex]\mathrm{log}\left(x+3\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(74\right)[/latex]

[latex]\mathrm{ln}\left(-3x\right)=\mathrm{ln}\left({x}^{2}-6x\right)[/latex]

[latex]{\mathrm{log}}_{4}\left(6-m\right)={\mathrm{log}}_{4}3m[/latex]

[latex]\mathrm{ln}\left(x-2\right)-\mathrm{ln}\left(x\right)=\mathrm{ln}\left(54\right)[/latex]

[latex]{\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)[/latex]

[latex]\mathrm{ln}\left({x}^{2}-10\right)+\mathrm{ln}\left(9\right)=\mathrm{ln}\left(10\right)[/latex]

[latex]x=±\frac{10}{3}[/latex]

For the following exercises, solve each equation for[latex]\,x.[/latex]

[latex]\mathrm{log}\left(x+12\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(12\right)[/latex]

[latex]\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x-3\right)=\mathrm{ln}\left(7x\right)[/latex]

[latex]{\mathrm{log}}_{2}\left(7x+6\right)=3[/latex]

[latex]\mathrm{ln}\left(7\right)+\mathrm{ln}\left(2-4{x}^{2}\right)=\mathrm{ln}\left(14\right)[/latex]

[latex]{\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)[/latex]

[latex]\mathrm{ln}\left(3\right)-\mathrm{ln}\left(3-3x\right)=\mathrm{ln}\left(4\right)[/latex]

[latex]x=\frac{3}{4}[/latex]

[latex]{\mathrm{log}}_{3}\left(3x\right)-{\mathrm{log}}_{3}\left(6\right)={\mathrm{log}}_{3}\left(77\right)[/latex]

For the following exercises, solve the equation for[latex]\,x,[/latex]if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

[latex]{\mathrm{log}}_{9}\left(x\right)-5=-4[/latex]

[latex]x=9[/latex]

[latex]\mathrm{ln}\left(3x\right)=2[/latex]

[latex]\mathrm{ln}\left(x-5\right)=1[/latex]

[latex]\mathrm{log}\left(4\right)+\mathrm{log}\left(-5x\right)=2[/latex]

[latex]x=-5[/latex]

[latex]-7+{\mathrm{log}}_{3}\left(4-x\right)=-6[/latex]

[latex]\mathrm{ln}\left(4x-10\right)-6=-5[/latex]

[latex]x=\frac{e+10}{4}\approx 3.2[/latex]

[latex]\mathrm{log}\left(4-2x\right)=\mathrm{log}\left(-4x\right)[/latex]

[latex]\mathrm{ln}\left(2x+9\right)=\mathrm{ln}\left(-5x\right)[/latex]

[latex]{\mathrm{log}}_{9}\left(3-x\right)={\mathrm{log}}_{9}\left(4x-8\right)[/latex]

[latex]x=\frac{11}{5}\approx 2.2[/latex]

[latex]\mathrm{log}\left({x}^{2}+13\right)=\mathrm{log}\left(7x+3\right)[/latex]

[latex]\frac{3}{{\mathrm{log}}_{2}\left(10\right)}-\mathrm{log}\left(x-9\right)=\mathrm{log}\left(44\right)[/latex]

[latex]x=\frac{101}{11}\approx 9.2[/latex]

[latex]\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+3\right)=\mathrm{ln}\left(6\right)[/latex]

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of[latex]\,\text{\$6,500}\,[/latex]earns[latex]\,7.25%\,[/latex]annual interest, compounded continuously. How much will the account be worth after 20 years?

about[latex]\,$27,710.24[/latex]

The formula for measuring sound intensity in decibels[latex]\,D\,[/latex]is defined by the equation[latex]\,D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right),\text{}[/latex]where[latex]\,I\,[/latex]is the intensity of the sound in watts per square meter and[latex]\,{I}_{0}={10}^{-12}\,[/latex]is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of[latex]\,8.3\cdot {10}^{2}\,[/latex]watts per square meter?

The population of a small town is modeled by the equation[latex]\,P=1650{e}^{0.5t}\,[/latex]where[latex]\,t\,[/latex]is measured in years. In approximately how many years will the town’s population reach[latex]\,\text{20,000?}[/latex]

about 5 years

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate[latex]\,x\,[/latex]to 3 decimal places .

[latex]1000{\left(1.03\right)}^{t}=5000\,[/latex]using the common log.

[latex]{e}^{5x}=17\,[/latex]using the natural log

[latex]3{\left(1.04\right)}^{3t}=8\,[/latex]using the common log

[latex]{3}^{4x-5}=38\,[/latex]using the common log

[latex]x=\frac{\mathrm{log}\left(38\right)+5\mathrm{log}\left(3\right)\text{ }}{4\mathrm{log}\left(3\right)}\approx 2.078[/latex]

[latex]50{e}^{-0.12t}=10\,[/latex]using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

[latex]7{e}^{3x-5}+7.9=47[/latex]

[latex]x\approx 2.2401[/latex]

[latex]\mathrm{log}\left(-0.7x-9\right)=1+5\mathrm{log}\left(5\right)[/latex]

[latex]x\approx -\text{44655}.\text{7143}[/latex]

Atmospheric pressure[latex]\,P\,[/latex]in pounds per square inch is represented by the formula[latex]\,P=14.7{e}^{-0.21x},[/latex] where [latex]x[/latex] is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of[latex]\,8.369\,[/latex]pounds per square inch? ( Hint : there are 5280 feet in a mile)

The magnitude M of an earthquake is represented by the equation[latex]\,M=\frac{2}{3}\mathrm{log}\left(\frac{E}{{E}_{0}}\right)\,[/latex]where[latex]\,E\,[/latex]is the amount of energy released by the earthquake in joules and[latex]\,{E}_{0}={10}^{4.4}\,[/latex]is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing[latex]\,1.4\cdot {10}^{13}\,[/latex]joules of energy?

about[latex]\,5.83[/latex]

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that [latex]\,{b}^{{\mathrm{log}}_{b}x}=x.[/latex]

Recall the formula for continually compounding interest,[latex]\,y=A{e}^{kt}.\,[/latex]Use the definition of a logarithm along with properties of logarithms to solve the formula for time[latex]\,t\,[/latex]such that[latex]\,t\,[/latex]is equal to a single logarithm.

[latex]t=\mathrm{ln}\left({\left(\frac{y}{A}\right)}^{\frac{1}{k}}\right)[/latex]

Recall the compound interest formula[latex]\,A=a{\left(1+\frac{r}{k}\right)}^{kt}.\,[/latex]Use the definition of a logarithm along with properties of logarithms to solve the formula for time[latex]\,t.[/latex]

Newton’s Law of Cooling states that the temperature[latex]\,T\,[/latex]of an object at any time t can be described by the equation[latex]\,T={T}_{s}+\left({T}_{0}-{T}_{s}\right){e}^{-kt},[/latex] where[latex]\,{T}_{s}\,[/latex]is the temperature of the surrounding environment,[latex]\,{T}_{0}\,[/latex]is the initial temperature of the object, and[latex]\,k\text{}[/latex]is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time[latex]\,t\,[/latex]such that[latex]\,t\,[/latex]is equal to a single logarithm.

[latex]t=\mathrm{ln}\left({\left(\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\right)}^{-\,\frac{1}{k}}\right)[/latex]

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## Most Used Actions

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- radical\:to\:exponent\:\sqrt[4]{x}
- radical\:to\:exponent\:\sqrt[5]{8x}
- radical\:to\:exponent\:\sqrt[6]{x^2}

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- High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,...

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## Course: Algebra 2 > Unit 6

- Rewriting exponential expressions as A⋅Bᵗ

## Rewrite exponential expressions

- Equivalent forms of exponential expressions
- Your answer should be
- an integer, like 6
- a simplified proper fraction, like 3 / 5
- a simplified improper fraction, like 7 / 4
- a mixed number, like 1 3 / 4
- an exact decimal, like 0.75
- a multiple of pi, like 12 pi or 2 / 3 pi

## Exponential Form

The exponential form is an easier way of writing repeated multiplication involving base and exponents. For example, we can write 5 × 5 × 5 × 5 as 5 4 in the exponential form, where 5 is the base and 4 is the power. In this form, the power represents the number of times we are multiplying the base by itself.

## How to Write in Exponential Form?

To write numbers in exponential form, we need to express them raised to certain powers of their prime factors as shown in the following examples:

- 8 = 2 × 2 × 2 = 2 3 . Therefore, the exponential form of 8 can be expressed as 2 3
- 72 = 2 × 2 × 2 × 3 × 3 = 2 3 × 3 2 . Therefore, the exponential form of 72 can be expressed as 2 3 × 3 2
- 121 = 11 2 . Therefore, the exponential form of 121 can be expressed as 11 2

These are the exponential forms of the corresponding numbers . When it comes to representing numbers, there are three forms in which we can do that. Those are exponential form, factor form, and standard form. Any number can be represented in all three forms. Let us understand it in detail through the table given below:

## Exponential Form to Logarithmic Form

An expression written in the exponential form can be easily converted to logarithmic form by using a simple formula: If e a = b, then \(log_{e}b\) = a. Let us understand this conversion with the help of an example. Convert 5 3 =125 to log form. By equating it with the formula given above, we can say that, here, b = 125, a = 3, and e = 5. So, the logarithmic form is \(log_{5}125\) = 3.

## Logarithmic to Exponential Form

Now, let us understand how to convert logarithmic to exponential form. The same formula will be applicable to this also. If \(log_{e}b\) = a is given, this implies, e a = b. Let us take an example. Convert \(log_{10}100\) = 2 to exponential form. If we equate this to the above formula, we get b=100, a=2, and e=10. So, the exponential form will be 10 2 = 100.

## Exponential Form to Radical Form

The conversion of an expression from exponential form to radical form is done by using the formula: x m/n = n √x m . Radical denotes the √ symbol which is used to represent nth roots. To convert an exponential form to a radical form, the denominator of the exponent will shift to the left of the radical sign, and the numerator will be the power of the radicand. It is only possible for fractional exponents . If the exponent is a whole number, then the radical sign is not used as it is only used to represent roots. For example, 2 3/5 can be represented in radical form as 5 √2 3 .

Observe the figure given below to understand the exponential form to radical form conversion formula along with an example.

Similarly, we can also convert a radical form to an exponential form. For example, if 3 √3 is given, it means in the exponent form, the numerator of the power will be 1 and the denominator will be 3. So, 3 √3 = 3 1/3 .

## Standard Exponential Form

If very large numbers or very small numbers are given, then it is better to use the standard exponential form to represent them. For example, it is difficult to make sense of the number 2030000000000000, but it will be easier if we write it in standard form as 2.03 × 10 15 . It is also known as a scientific notation to write numbers. To write a number in standard exponential form, we follow the steps given below:

- Step 1: Count the number of trailing zeros in the number.
- Step 2: Use the beginning of the number and write the digits from the left till the last non-zero digit followed by a 10 raised to power which is equal to the number of trailing zeros.
- Step 3: Place a decimal point after the first digit from the left side and add the number of decimal places created to the power of 10. For example, in the example written above, if we place a decimal point after 2, then two decimal places will be created in the number 2.03. So, we will add 2 to the existing power of 10. The power of 10 was 13 as there were 13 trailing zeros, but now it will become 15.
- Step 4: Therefore, the standard exponential form of the number 2030000000000000 is 2.03 × 10 15 .

Here, it is important to note that the decimal number written in the standard exponential form will always be greater than 0 and less than 10.

☛ Related Topics

- Exponential Functions
- Exponential Equations
- Exponent Rules

## Exponential Form Examples

Example 1: Write 2000 in exponential form.

Solution: To write 2000 in the exponential form, we have to find the prime factorization of 2000. The prime factorization of 2000 is 2 × 2 × 2 × 2 × 5 × 5 × 5. It means 2000 = 2 4 × 5 3 .

Example 2: Convert \(log_{4}16\) = 2 in exponential form.

Solution: The formula to convert logarithmic to exponential form is, if \(log_{e}a\) = b, this implies, e b = a. It means \(log_{4}16\) = 2 can be written as 4 2 = 16. Therefore, 4 2 = 16 is the required answer.

Example 3: Convert the following numbers written in standard form to exponential form.

Solution: To write these numbers in exponential form, we will use the prime factorization method.

a) 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 10

b) 350 = 2 × 5 × 5 × 7 = 2 1 × 5 2 × 7 1

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## Practice Questions on Exponential Form

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## FAQs on Exponential Form

What is exponential form.

The exponential form is a way of writing numbers using bases and powers. It tells us how many times are we actually multiplying a number to get the result. For example, if we observe the number 125, it appears to be a usual 3-digit number, but if we write it as 5 3 , we know that we are multiplying 5 three times to get 125, or 125 is the third power of 5.

## How to Write an Expression in Exponential Form?

In order to write an expression in exponential form, we need to find the prime factorization of its terms. For example, when we have numbers like 100, we need to find its prime factors and we get 5 × 5 × 2 × 2. Now, these prime factors can be written in the exponential form as, 5 2 × 2 2

## How to Write Log in Exponential Form?

To write an equation given in log in exponential form, we use the following conversion formula: If \(log_{e}a\) = b, this implies, e b = a.

## How to Convert Radical to Exponential Form?

To convert radical form to exponential form, we use the following formula: x m/n = n √x m . It is used only with fractional exponents.

## What is the Exponential Form of 64?

The exponential form of 64 is: 64 = 4 × 4 × 4 = 4 3 .

## What is the Difference Between Standard Form and Exponential Form?

The standard form of a number is writing it in the form of an integer like 121. However, exponential form means to write a number as the power of another number. For example, 121 is the standard form but 11 2 is its exponential form.

## What is the Exponential Form of 343?

The exponential form of 343 is 343 = 7 × 7 × 7 = 7 3 .

## IMAGES

## VIDEO

## COMMENTS

Then, take the logarithm of both sides of the equation to convert the exponential equation into a logarithmic equation. The logarithm must have the same base as the exponential expression in the equation. Use logarithmic properties to simplify the logarithmic equation, and solve for the variable by isolating it on one side of the equation.

To solve for x , we must first isolate the exponential part. To do this, divide both sides by 5 as shown below. We do not multiply the 5 and the 2 as this goes against the order of operations! 5 ⋅ 2 x = 240 2 x = 48. Now, we can solve for x by converting the equation to logarithmic form. 2 x = 48 is equivalent to log 2.

Meaning of Logarithms. Rewrite each equation in logarithmic form. 5) 7) 11) log. 13) log. Rewrite each equation in logarithmic form. 15) ©P U2P0Q1K27 nKHuOt7ap cSTosfEtYwyahree3 wLPLnCk.i F uAmlRl9 6rIiEgGhutvsJ 3r9e2sqemrTvGehdh.8 B KMbaHdHed 0wLiDtrhn eIvn6fTi3nviJtNeC zAal0gaebbLrcaY A2T.v.

Algebra 2 (FL B.E.S.T.) 11 units · 156 skills. Unit 1. Properties of functions. Unit 2. ... Rewrite exponential expressions Get 3 of 4 questions to level up! ... Solve exponential equations using exponent properties (advanced) Get 3 of 4 questions to level up! Compound interest.

5) . 7) . Rewrite each equation in exponential form. 9) log y. x. 11) log. x . Rewrite each equation in logarithmic form. 13) xy .

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.

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. b S = b T. Use the one-to-one property to set the exponents equal.

Solving exponential equations using properties of exponents. Solving exponential equations using exponent properties (advanced) Level up on the above skills and collect up to 400 Mastery points. Level up on all the skills in this unit and collect up to 1,000 Mastery points! We previously learned about integer powers—first positive and then ...

If You Experience Display Problems with Your Math Worksheet. Click here for More Algebra 2 - Exponential and Logarithmic Worksheets. This Algebra 2 Meaning of Logarithms worksheet will give you equations into rewrite to either exponential or logarithmic form. You can choose the form of the problems.

Find step-by-step Algebra 2 solutions and your answer to the following textbook question: Rewrite each equation in exponential form. $$ \log _{\mathrm{a}}(b)=c $$.

Skill plans. IXL plans. Virginia state standards. Textbooks. Test prep. Awards. Improve your math knowledge with free questions in "Solve exponential equations by rewriting the base" and thousands of other math skills.

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[latex]\,{b}^{S}={b}^{T}.[/latex] Use the one-to-one property to set the exponents equal.

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. {b}^ {S}= {b}^ {T} bS = bT. Use the one-to-one property to set the exponents equal to each other. Solve the resulting equation, S = T, for the unknown.

In this video, Sal was giving examples of using some exponent properties to help show how to rewrite exponential expressions. Exponential Propertes. Here are some exponential properties that you should be familiar with. a^ (bc) = (a^b)^c // or vice versa. a^b * a^c = a^ (b + c) a^b / a^c = a^ (b - c) a^c * b^c = (a*b)^c.

Rewrite each equation in exponential form. 1) log 6 216 = 3 2) log u v = 16 3) log 12 144 = 2 4) log n 149 = m 5) log 7 y = x 6) log 8 64 = 2 7) log 361 19 = 1 2 8) log 20 400 = 2 9) log 144 1 12 = - 1 2 ... Algebra 2 Practice- Converting from Logarithm to Exponential Name_____ ID: 1 ©G r2K0i1U5U kKHust^aR eS_ovfntCwaafrfev zLJLgCr.X D sAelplp ...

For logarithmic equations, logb(x) = y log b ( x) = y is equivalent to by = x b y = x such that x > 0 x > 0, b > 0 b > 0, and b ≠ 1 b ≠ 1. In this case, b = 2 b = 2, x = x x = x, and y = y y = y. Substitute the values of b b, x x, and y y into the equation by = x b y = x. Free math problem solver answers your algebra, geometry, trigonometry ...

a. exponents b. equalx = -1. What are the three steps to solving an exponential equation when the bases are not the same? 1. Rewrite each side to have the same base 2. Set exponents equal to each other 3. Solve the equation and check solutions. Solve: 6^-3x = 1,296. x = -4/3.

Free Exponential Form calculator - convert radicals to exponents step-by-step.

Example: 2. 5 = 2 2 2 2 2 = 32 2 . is the base, and . 5 . is the exponent. Equations that have a variable as an . exponent . are called exponential equations. exponential growth •an . increase. in a quantity at a constant percentage rate per unit interval •examples: a quantity that . doubles. every day, compounded interest,

Rewrite exponential expressions. What is the value of A when we rewrite 2 x − 6 + 2 x as A ⋅ 2 x ? Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

Example 1: Write 2000 in exponential form. Solution: To write 2000 in the exponential form, we have to find the prime factorization of 2000. The prime factorization of 2000 is 2 × 2 × 2 × 2 × 5 × 5 × 5. It means 2000 = 2 4 × 5 3. Example 2: Convert log416 l o g 4 16 = 2 in exponential form. Solution: The formula to convert logarithmic to ...

Algebra 2: Week 31 6.4 Rewriting Logarithmic & Exponential Equations ... ©I \2P0A2B1a ]Ktuttlaz NStoMfytrwea[rget LLELDCT.i k pAxlTlk srSitgnh_tbsH grJe`sreAruvBeNdT.-1-Rewrite each equation in exponential form. Let's do these together. 1) log 17 289 = 2 2) log 6 216 = 3 3) log 169 13 = 1 2 4) log 4 1 64 = -3 YOUR TURN: Rewrite each equation ...

Rewriting An Exponential Expression As A X B T Algebra Study Com. Solved Please Help Me With These Problems Rewrite Each Equation In Course Hero. Section 6 3 Exponential And Logarithmic Functions. Rewriting Using The Laws Of Exponents Worksheet. Solved 1 8 Exercises Rewrite Each Equation In Exponential Form Log V T 2 R 3 N 4 X Y Logarithmic 5