Solve for $x$: $\frac1e = e^{2x}$ I tried  making it to $e^{-1} = e^{2x}$

and had the exponents equal each other $-1=2x$

and the I solved for $x$, making it $x=-1/2$, but that answer is wrong.

please help

I don't know why that answer is wrong.
Multiply both sides by $e$; then $e^{2x+1}=1$. The solutions to $e^z=1$ are given by $z=2\pi ni$ for $n\in\mathbb{Z}$, thus the solutions for $x$ are given by
$$
x=\frac{2\pi ni -1}{2},\qquad n\in\mathbb{Z}.
$$
If you only want real solutions, take $n=0$ to get $x=-1/2$ as you mentioned.
$\frac1e = e^{2x}$ is in fact $e^{-1} = e^{2x}$  which applying lans is $lane^{-1} = lane^{2x}$ but this is simply ${-1} = {2x}$ which can only mean $x=-0.5$, so you are , in fact, correct.

Peace and Chicken grease.