My first thought was to take the CDF of X, which we know is $1-e^{-x}$ for values of $x > 0$ and $0$ for values of $x$ less than$ 0$, and then solve the $\log(x)$ for those values, however I am not sure that this is correct. Any thoughts?
I also attempted to graph $\log(1-e^{-x})$ and of course this gave me a straight line. Not sure where to go from here. Any help would be appreciated. Thanks!
If $X$ is a random variable with pdf $f_X(x)$ and $g$ is a monotone function then the pdf of $Y=g(X)$ can be obtained by:
$f_Y(y)=f_X(g^{-1}(y))\cdot |{\frac{d}{dy}}g^{-1}(y)|$
In your case the transformation $y=g(x)=\log(x)$ and hence $x=g^{-1}(y)=e^y$, from here we also obtain $|{\frac{d}{dy}}e^y|=|e^y|=e^y$
With $\lambda=1$ we have that $f_X(x)=e^{-x}$ and then $f_Y(y)=e^{-e^y}\cdot e^y=e^{-e^y+y}$