I'm sure this is a trivial problem, but I am having trouble committing to an answer.
If $U$~$(0,1)$, then $Y=-ln U$. I know that $\mathbb P(Y \le y)$
= $\mathbb P(g(X) \le y)$
=$\mathbb P(-lnU \le y)$
=$\mathbb P(U \ge e^{-y})$
So would my CDF be $F_Y (y)=e^{-y},0\le y \le \infty$? Then of course, my PDF would be $-e^{-y}$.