I have an assignment question looks like this

$\displaystyle 2e^-^2^x) = 5$

I figured, first time around, that I could just apply natural logarithm to both, and get

$\displaystyle -4x = ln(5)$

and then x = ln(5)/-4

This does not result in the correct answer.

Why? I can get the correct answer (by dividing both sides by two initially)

If I have 2e^(-2x) and apply the natural logarithm to it, what does it become?