Given that $X$ is a random variable with cdf $F(x)= 1-e^{-x}$ when $x>0$ and $0$ when $x \le 0$, find $P(0 \le e^X \le 4)$.
I know the formula for $P(a \le X \le b)$ but here how I convert $e^X$ into $X$ . I assume that $y=e^X$ but in that case if I take log on both sides then it is not defined at $X=0$.
Since $\exp(x)>0$ for any $x \in \mathbb{R}$, we have $$P(0 \le e^X \le 4) = P(0<e^X \le 4).$$ Moreover, we know that $\log$ maps $(0,4]$ injective onto $(-\infty,\log(4)]$, because this function is monotonically increasing and continuous. Thus $$P(0<e^X \le 4) = P(X \le \log(4)) = F(\log(4)) = 1-e^{-\log(4)}=1-1/4=3/4.$$