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Study Guides > College Algebra: Co-requisite Course

Exponential Equations*

Learning Objectives

  • 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
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, b≠1b>0,\text{ }b\ne 1, bS=bT{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 34x−7=32x3{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:

34x−7=32x334x−7=32x31Rewrite 3 as 31.34x−7=32x−1Use the division property of exponents.4x−7=2x−1 Apply the one-to-one property of exponents.2x=6Subtract 2x and add 7 to both sides.x=3Divide by 3.\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≠1b\ne 1, bS=bT if and only if S=T{b}^{S}={b}^{T}\text{ if and only if }S=T

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

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

Example: Solving an Exponential Equation with a Common Base

Solve 2x−1=22x−4{2}^{x - 1}={2}^{2x - 4}.

Answer: 2x−1=22x−4The common base is 2. x−1=2x−4By the one-to-one property the exponents must be equal. x=3Solve for x.\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 52x=53x+2{5}^{2x}={5}^{3x+2}.

Answer: x=–2x=–2

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

Solve 25x=2{2}^{5x}=\sqrt{2}.

Answer: 25x=212Write the square root of 2 as a power of 2.5x=12Use the one-to-one property.x=110Solve for x.\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 5x=5{5}^{x}=\sqrt{5}.

Answer: x=12x=\frac{1}{2}

Use 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, since log(a)=log(b)\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right) 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.

  1. Apply the logarithm of 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.
  2. Use the rules of logarithms to solve for the unknown.

Example: Solving an Equation Containing Powers of Different Bases

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

Answer:  5x+2=4xThere is no easy way to get the powers to have the same base. ln5x+2=ln4xTake ln of both sides. (x+2)ln5=xln4Use laws of logs. xln5+2ln5=xln4Use the distributive law. xln5−xln4=−2ln5Get terms containing x on one side, terms without x on the other.x(ln5−ln4)=−2ln5On the left hand side, factor out an x. xln(54)=ln(125)Use the laws of logs. x=ln(125)ln(54)Divide by the coefficient of x.\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 laws of logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive law}.\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 an }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the laws 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 2x=3x+1{2}^{x}={3}^{x+1}.

Answer: x=ln3ln(23)x=\frac{\mathrm{ln}3}{\mathrm{ln}}\left(23\right)

Q & A

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

No. There is a solution when k≠0k\ne 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=−3et2=-3{e}^{t}.

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

Solve 4e2x+5=124{e}^{2x}+5=12.

Answer: 4e2x+5=124e2x=7Combine like terms.e2x=74Divide by the coefficient of the power.2x=ln(74)Take ln of both sides.x=12ln(74)Solve for x.\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Combine like terms}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide by the coefficient of the power}.\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+e2t=7e2t3+{e}^{2t}=7{e}^{2t}.

Answer: t=ln(12)=−12ln(2)t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)

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.

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