Transformation of continuous uniform distribution

Question

Let $U$ ~ $Unif(0,1)$. Define $Y=-lnU$. Find $F_Y(y)=P(Y\le y)$.

I know that $F_Y (y)=P(g(X)\le y)=\int_{x:g(x)\le y} f_X (x)dx$.

This gives the CDF; the derivative will then give the PDF. Can anyone help me with the integral? Doesn't seem too hard but I'm having trouble committing to an answer.

Answer 1

$$P(Y \le y) = P( - \ln U \le y) = P ( U \ge e^{-y} ) = \dots$$