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Учебные пособия > College Algebra

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 [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] 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 [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. 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:

[latex]\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}[/latex]

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 [latex]b\ne 1[/latex], [latex-display]{b}^{S}={b}^{T}\text{ if and only if }S=T[/latex-display]

How To: Given an exponential equation with the form [latex]{b}^{S}={b}^{T}[/latex], 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 [latex]{b}^{S}={b}^{T}[/latex].
  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 [latex]{2}^{x - 1}={2}^{2x - 4}[/latex].

Answer: [latex-display]\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}[/latex-display]

Try It

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

Answer: [latex]x=–2[/latex]

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

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

Answer: [latex-display]\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}[/latex-display]

Try It

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

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

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 [latex]\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)[/latex] 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 [latex]{5}^{x+2}={4}^{x}[/latex].

Answer: [latex-display]\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}[/latex-display]

Try It

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

Answer: [latex]x=\frac{\mathrm{ln}3}{\mathrm{ln}}\left(23\right)[/latex]

Q & A

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 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 [latex]2=-3{e}^{t}[/latex].

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

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

Answer: [latex-display]\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}[/latex-display]

Try It

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

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

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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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
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