Learning Outcomes
- 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 [latex]t[/latex]
- 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] can be rewritten as 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 laws of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Notice how we rewrite the exponential terms as logarithms first.
Example
Solve [latex]{5}^{x+2}={4}^{x}[/latex].
In general we can solve exponential equations whose terms do not have like bases in the following way:
- Apply the logarithm to 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.
- Use the rules of logarithms to solve for the unknown.
The following video provides more examples of solving exponential equations.
Think About It
Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]?
Use the text box below to formulate an answer or example before you look at the solution.
Equations Containing [latex]e[/latex]
Example
Solve [latex]100=20{e}^{2t}[/latex].
In our next example using the continuous growth formula, we have to do a couple steps of algebra to get it in a form that can be solved.
Example
Solve [latex]4{e}^{2x}+5=12[/latex].
Exponential Equations with No Solutions or Extraneous Solutions
We have seen in earlier lessons on solving equations that there are some equations where a solution does not exist and or that have extraneous solutions. We will explore such examples with exponential equations, but first, take a minute to think about when a solution to an exponential equation might not exist.
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 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 equation.
Example
Solve [latex]{e}^{2x}-{e}^{x}=56[/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.
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.
Think About It
When might an equation of the form [latex]y=A{e}^{kt}[/latex] have no solution? Write your thoughts or an example in the textbox below before you check the answer.
An equation of this form does not have a solution if y and A have different signs. An example of this is [latex]2=-3{e}^{t}[/latex].
Our next example helps to illustrate that not every equation of the form [latex]y=A{e}^{kt}[/latex] has a solution.
Example
Solve [latex]3{e}^{2x}=-6[/latex].
[latex]\begin{array}{c}3{e}^{2x}=-6\hfill & \hfill \\ {e}^{2x}=-2\hfill & \text{Divide by the coefficient of the base raised to a power}.\hfill \\ 2x=\mathrm{ln}\left(-2\right)\hfill & \text{Take ln of both sides}.\hfill \\ \end{array}[/latex]
This equation has no solution because the argument of a logarithm cannot be negative. We also could have recognized in the previous step that this equation has no solution because [latex]e[/latex] raised to any power will always produce a positive result.
Answer
No solution.