Answer
$x\approx-18.892$
Work Step by Step
Taking the natural logarithm of both sides, the given equation, $
e^{-0.103x}=7
$ is equivalent to
\begin{align*}\require{cancel}
\ln e^{-0.103x}&=\ln 7
.\end{align*}
Using the properties of logarithms, the equation above is equivalent to
\begin{align*}\require{cancel}
-0.103x(\ln e)&=\ln7
&(\text{use }\log_b x^y=y\log_b x)
\\
-0.103x(1)&=\ln7
&(\text{use }\ln e=1)
\\
-0.103x&=\ln7
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
\dfrac{\cancel{-0.103}x}{\cancel{-0.103}}&=\dfrac{\ln7}{-0.103}
\\\\
x&=\dfrac{\ln7}{-0.103}
\\\\
x&\approx-18.892
.\end{align*}
Hence, the solution to the equation $
e^{-0.103x}=7
$ is $
x\approx-18.892
$.
