Answer
$x\approx261.291$
Work Step by Step
Taking the natural logarithm of both sides, the given equation, $
e^{0.012x}=23
$ is equivalent to
\begin{align*}\require{cancel}
\ln e^{0.012x}&=\ln23
.\end{align*}
Using the properties of logarithms, the equation above is equivalent to
\begin{align*}\require{cancel}
0.012x(\ln e)&=\ln23
&(\text{use }\log_b x^y=y\log_b x)
\\
0.012x(1)&=\ln23
&(\text{use }\ln e=1)
\\
0.012x&=\ln23
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
\dfrac{\cancel{0.012}x}{\cancel{0.012}}&=\dfrac{\ln23}{0.012}
\\
x&=\dfrac{\ln23}{0.012}
.\end{align*}
Using a calculator, the approximate value of $
\ln23
$ is $
3.1355
$. Thus, the equation above is equivalent to
\begin{align*}
x&=\dfrac{3.1355}{0.012}
\\\\
x&=261.291
.\end{align*}
Hence, the solution to the equation $
e^{0.012x}=23
$ is $
x\approx261.291
$.
