I think I have a good answer to this one, but I'd like to make sure it's correct.

If $\displaystyle X$ has exponential distribution with parameter $\displaystyle \lambda$, find the probability density function of $\displaystyle Y=\log X$.

Here is what I did.

$\displaystyle F_Y(y) = P \{ Y \leq y \} = P \{\log X \leq y \} = P \{ X \leq e^y \} = \int_{0}^{e^y} \lambda e^{- \lambda x} dx$

Let $\displaystyle u=\lambda x$, $\displaystyle du=\lambda dx$

$\displaystyle \int_{0}^{e^y} \lambda e^{- \lambda x} dx = \int_{0}^{\lambda e^y} e^{-u} du = -e_0^{\lambda e^y} = -e^{-x} \Big |_0^{e^y} = -e^{-e^y} + 1 = 1 - e^{-e^y}$

So,

$\displaystyle \frac {d F_{|X|} (x) } {dx} = p(x) = e^x e^{-e^x}$

Does this look solid?