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 $X$? 

Yes, because the exponential function is increasing $P(e^X \le e^x) = P(X \le x)$.
This means the CDF $F_{e^X}(e^x) = F_X(x)$, or $F_{e^X}(t) = F_X(\ln (t))$ for $t > 0$.  Differentiate to get the PDF.
