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.
property of logarithmic equality
For any [latex]M \gt 0[/latex], [latex]N \gt 0[/latex], [latex]b \gt 0[/latex], and [latex]b \ne 1[/latex],
If [latex]\mathrm{log}_b(M) = \mathrm{log}_b(N)[/latex], then [latex]M = N[/latex]
Because of the property of logarithmic equality, we may apply logarithms with the same base to both sides of an exponential equation.
How To: Given an exponential equation Where a common base cannot be found, solve for the unknown
Apply the logarithm to 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.
Note in the paragraph above the reiteration of the one-to-one property of logarithms. When applying logarithms with the same base to both sides of an exponential equation, we often use the common logarithm, [latex]\log[/latex] or the natural logarithm, [latex]\ln[/latex]. The choice is yours which to use in most situations, but if either base in a given exponential equation is [latex]10[/latex], use [latex]\log[/latex] or if the base is [latex]e[/latex], use [latex]\ln[/latex] to take advantage of the identity property of logarithms.Solve [latex]{5}^{x+2}={4}^{x}[/latex].
[latex]\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 the power rule for logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive property}.\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 }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the properties 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]
Solve [latex]4^{x-5} = 35[/latex].
[latex]\begin{array}{l} 4^{x-5} = 35 \hfill \\ x-5 = \ln(35) / \ln(4) \hfill & \text{Take } \ln \text{ of both sides and use the change of base formula.} \hfill \\ x = \frac{\ln(35)}{\ln(4)} + 5 \hfill & \text{Solve for } x. \hfill \\ \end{array}[/latex]
The approximate value of [latex]x[/latex] is [latex]x \approx 7.565[/latex].
Equations Containing [latex]e[/latex]
One common type of exponential equations are those with base [latex]e[/latex]. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base [latex]e[/latex] 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 [latex]t[/latex]
Divide both sides of the equation by [latex]A[/latex].
Apply the natural logarithm to both sides of the equation.
Divide both sides of the equation by [latex]k[/latex].
Solve [latex]100=20{e}^{2t}[/latex].
[latex]\begin{array}{l}100\hfill & =20{e}^{2t}\hfill & \hfill \\ 5\hfill & ={e}^{2t}\hfill & \text{Divide by the coefficient 20}\text{.}\hfill \\ \mathrm{ln}5\hfill & =\mathrm{ln}{{e}^{2t}}\hfill & \text{Take ln of both sides.}\hfill \\ \mathrm{ln}5\hfill & =2t\hfill & \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, then we use a calculator.
Just as you have done when solving various types of equations, isolate the term containing the variable for which you are solving before applying any properties of equality or inverse operations. That’s why, in the example above, you must divide away the [latex]A[/latex] first. Remember that the functions [latex]y=e^{x}[/latex] and [latex]y=\mathrm{ln}\left(x\right)[/latex] are inverse functions. Therefore, [latex]\mathrm{ln}\left({e}^{x}\right)=x[/latex] for all [latex]x[/latex], and [latex]e^{\mathrm{ln}\left(x\right)}=x[/latex] for [latex]x>0[/latex].Solve [latex]4{e}^{2x}+5=12[/latex].
[latex]\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Subtract 5 from both sides}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide both sides by 4}.\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 taking the logarithm of 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.
Solve [latex]{e}^{2x}-{e}^{x}=56[/latex].
[latex]\begin{array}{l}{e}^{2x}-{e}^{x}=56\hfill \\ {e}^{2x}-{e}^{x}-56=0\hfill & \text{Get one side of the equation equal to zero}.\hfill \\ \left({e}^{x}+7\right)\left({e}^{x}-8\right)=0\hfill & \text{Factor by the FOIL method}.\hfill \\ {e}^{x}+7=0\text{ or }{e}^{x}-8=0 & \text{If a product is zero, then one factor must be zero}.\hfill \\ {e}^{x}=-7{\text{ or e}}^{x}=8\hfill & \text{Isolate the exponentials}.\hfill \\ {e}^{x}=8\hfill & \text{Reject the equation in which the power equals a negative number}.\hfill \\ x=\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.