Answer
$x=0, x=-\ln2$
Work Step by Step
Let $u=e^{-x}$, then $u^{2}=e^{-2x}$.
Then $e^{-2x}-3e^{-x}=-2$ will be $$u^{2}-3u=-2.$$ Simplify $$ u^{2}-3u+2=0.$$ Factor it $$ (u-1)(u-2)=0.$$ Solve $$u-1=0 \text{ or }u-2 = 0$$ $$u= 1 \text{ or }u=2.$$ Since $u = e^{-x}$, then $$e^{-x}=1 \text{ or }e^{-x}=2.$$ We find $x$: $$\begin{aligned}
\ln e^{-x}&=\ln1\text{ or } \ln e^{-x}=\ln2\\
-x&=0\text{ or } -x=\ln2\\
x&=0 \text{ or } x=-\ln2.
\end{aligned}$$
