i'm having some trouble remembering how to solve this in terms of x.

$\displaystyle y=(e^x-e^-^x)/2$

This is what i was thinking, but, i'm not sure if it is right or not.

$\displaystyle x=(e^y-e^-^x)/2$

$\displaystyle 2x=(e^y-e^y)$

$\displaystyle log(2x)=2y*log(e)$

$\displaystyle log(2x)/(2*log(e))=y$

$\displaystyle y=log(2x)/(2*log(e))$

I feel like this is not the right answer though :/