#### Answer

$x\approx2.811$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
e^{0.4x}=4^{x-2}
,$ take the natural logarithm of both sides. Use the properties of logarithms and of equality to isolate the variable. Approximate the answer with $3$ decimal places.
$\bf{\text{Solution Details:}}$
Taking the natural logarithm of both sides results to
\begin{array}{l}\require{cancel}
\ln e^{0.4x}=\ln 4^{x-2}
.\end{array}
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
0.4x\ln e=(x-2)(\ln 4)
.\end{array}
Since $\ln e=1,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
0.4x(1)=(x-2)(\ln 4)
\\\\
0.4x=(x-2)(\ln 4)
.\end{array}
Using the Distributive Property and the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
0.4x=x(\ln 4)-2(\ln 4)
\\\\
0.4x=x\ln 4-2\ln 4
\\\\
0.4x-x\ln 4=-2\ln 4
\\\\
x(0.4-\ln 4)=-2\ln 4
\\\\
x=\dfrac{-2\ln 4}{0.4-\ln 4}
\\\\
x\approx2.811
.\end{array}