Answer
$x=-\frac{3}{5}$
Work Step by Step
We rewrite both sides, so that the base is $\frac{1}{e}$ on both sides.
$e^{x-1}=(\frac{1}{e}^4)^{x+1}$
By the law of exponents: $(a^x)^y=a^{xy}$ and $a^x=\frac{1}{a}^{-x}$
Left side: $e^{x-1}=(\frac{1}{e}^{-1})^{x-1}=\frac{1}{e}^{-x+1}$
Right side: $((\frac{1}{e})^4)^{x+1}=(\frac{1}{e})^{4x+4}$
The rewritten equation is:
$\frac{1}{e}^{-x+1}=\frac{1}{e}^{4x+4}$
Then, since the sides have the same base, we set the exponents equal, because the two sides are equal only if the powers are equal too: if $b^m=b^n$ and $b \ne0$ , $b \ne1$ then $m=n$
$-x+1=4x+4
\\1-4=4x+x$
$-3=5x
\\\frac{-3}{5}=\frac{5x}{5}$
$-\frac{3}{5}=x$
