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Study Guides > Intermediate Algebra

Read: Exponential Equations with Unlike Bases

Learning Objectives

  • Use logarithms to solve exponential equations whose terms cannot be rewritten with the same base
  • Solve exponential equations of the form  [latex]y=A{e}^{kt}[/latex] for t
  • Recognize when there may be extraneous solutions, or no solutions for 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. In our first example we will use the law of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Note how first, we rewrite the exponential terms as logarithms.

Example

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

Answer: There is no easy way to get the powers to have the same base for this equation.

[latex]\begin{array}{c}\text{ }{5}^{x+2}={4}^{x}. \\ \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]

In general we can solve exponential equations whose terms do not have like bases in the following way:
  1. Apply the logarithm of both sides of the equation.
    • If one of the terms in the equation has base [latex]10[/latex], use the common logarithm.
    • If none of the terms in the equation has base [latex]10[/latex], use the natural logarithm.
  2. Use the rules of logarithms to solve for the unknown.

Think About It

Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]? Use the textbox below to formulate an answer or example before you look at the solution. [practice-area rows="1"][/practice-area]

Answer: Yes. The solution is [latex]x = 0[/latex].

 Equations Containing [latex]e[/latex]

Base e is a very common base found in science, finance and engineering applications. When we have an equation with a base e on either side, we can use the natural logarithm to solve it. Earlier, we introduced a formula that models continuous growth, [latex]y=A{e}^{kt}[/latex].  This formula is found in business, finance, and many biological and physical science applications. In our next example we will show how to solve this equation for t, the elapsed time for the behavior in question.

Example

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

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

Analysis of the Solution

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.

Think About It

Does every equation of the form [latex]y=A{e}^{kt}[/latex] have a solution? Write your thoughts or an example in the textbox below before you check the answer. [practice-area rows="1"][/practice-area]

Answer: No. There is a solution when [latex]k\ne 0[/latex], and when y and A are either both [latex]0[/latex] or neither [latex]0[/latex], 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].

We will provide one more example using the continuous growth formula, but this time we have to do a couple steps of algebra to get in it a form that can be solved.

Example

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

Answer: [latex]\begin{array}{c}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]

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.

In the next example we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions, but reject one that does not satisfy the original equaiton.

Example

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

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

 Analysis of the Solution

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[/latex] because a positive number never equals a negative number. The solution [latex]x=\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.

Think About It

Does every logarithmic equation have a solution? Write your ideas, or a counter example in the box below. [practice-area rows="1"][/practice-area]

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

Summary

The inverse operation of exponentiation is the logarithm, so we can use logarithms to solve exponential equations whose terms do not have the same bases.  This is similar to using multiplication to "undo" division or addition to "undo" subtraction. It is important to check exponential equations for extraneous solutions or no solutions.

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  • Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..