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# The Natural Base e

## The Natural Base e 1.11 - Solution

When bases are not the same, we can solve an exponential equation by taking logarithms on each side of the equation. In each of our equations, the base is the number $e,$ so we will take the natural logarithm on each side. $\begin{gathered} m=n \quad \Leftrightarrow \quad \ln (m) = \ln (n) \end{gathered}$ Note that in order to take their logarithms, both $m$ and $n$ must be positive numbers. With this in mind, let's begin by solving the equation given in A.
$9e^{3x}=27$
$e^{3x}=3$
$\ln(\text{LHS})=\ln(\text{RHS})$
$\ln \left(e^{3x}\right)= \ln (3)$
$\ln\left(e^a\right) = a$
$3x= \ln (3)$
$x= \dfrac{\ln (3)}{3}$
$x=0.3662040962\ldots$
$x\approx 0.366$
Proceeding in the same way, we can solve the rest of the equations.
Equation Isolate Exponential Expression Take Logarithms on Each Side $x$
$\ \ \ \textbf{A. }9e^{3x}=27$ $e^{3x}=3$ $\ln \left(e^{3x}\right)= \ln \left(3\right)$ $\approx 0.366$
$\ \ \ \textbf{B. }9e^{x}=27$ $e^{x}=3$ $\ln \left(e^{x}\right)= \ln \left(3\right)$ $\approx 1.099$
$\ \ \ \textbf{C. }9e^{3x-4}=27$ $e^{3x-4}=3$ $\ln \left(e^{3x-4}\right)= \ln \left(3\right)$ $\approx 1.700$
$\ \ \ \textbf{D. }9e^{3x}+2=27$ $e^{3x}=\dfrac{25}{9}$ $\ln \left(e^{x}\right)= \ln \left(\dfrac{25}{9}\right)$ $\approx 1.022$

Now we can match each equation with its solution. \begin{aligned} \textbf{I: }& 9e^{3x}=27 \quad \quad &\textbf{iii: } x\approx 0.366 \\ \textbf{II: }& 9e^{x}=27 \quad \quad &\textbf{i: } x\approx 1.099 \\ \textbf{III: }& 9e^{3x-4}=27 \quad \quad &\textbf{iv: } x\approx 1.700 \\ \textbf{IV: }& 9e^{x}+2=27 \quad \quad &\textbf{ii: } x\approx 1.022 \end{aligned}