PDF of e raised to an exponential random variable 
Let $X$ be an exponential random variable with parameter $\lambda > 0$. For $\lambda > 1$, compute the PDF of $\mathrm e^X$.

Is $P(\mathrm e^X \le \mathrm e^x)$ the same as $P(X \le x)$? Does this mean that the PDF of $\mathrm e^X$ is just the PDF of $X$? 

(1) Yes, as $\exp$ is monotone $\exp(X) \le \exp (x)$ is equivalent to $X\le x$.
(2) No, that does not imply that $\exp(X)$ has the same pdf as $X$. We have 
$$ F_{\exp X}(x) = P\bigl(\exp(X) \le x\bigr) = P(X \le \log x) = F_X(\log x) $$
where $F_X$ denotes the cdf of $X$. Taking derivatives, we have for the pdf
$$ f_{\exp X}(x) = f_X(\log x) \cdot \frac 1x. $$
