170 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.
- 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.
- Use the rules of logarithms to solve for the unknown.
Example 5: Solving an Equation Containing Powers of Different Bases
Solve [latex]{5}^{x+2}={4}^{x}\\[/latex].
Solution
[latex]\begin{cases}\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{cases}\\[/latex]
Q & A
Is there any way to solve [latex]{2}^{x}={3}^{x}\\[/latex]?
Yes. The solution is x = 0.
Equations Containing [latex]e[/latex]
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 [latex]y=A{e}^{kt}\\[/latex], 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 6: Solve an Equation of the Form [latex]y=A{e}^{kt}\\[/latex]
Solve [latex]100=20{e}^{2t}\\[/latex].
Solution
[latex]\begin{cases}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{cases}\\[/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 7: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}\\[/latex]
Solve [latex]4{e}^{2x}+5=12\\[/latex].
Solution
[latex]\begin{cases}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{cases}\\[/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.
Example 8: Solving Exponential Functions in Quadratic Form
Solve [latex]{e}^{2x}-{e}^{x}=56\\[/latex].
Solution
[latex]\begin{cases}{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{cases}\\[/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.
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.
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.