Let $\ln X \sim \operatorname{Exp}(1)$ and $X,Y$ i.i.d. What is the distribution of $Z=XY$? Let $\ln X \sim \operatorname{Exp}(1)$ and $X,Y$ i.i.d. 
What is the distribution of $Z=XY$?
My attempt: $$P(Z\le z)=P(XY\le z)=P(\ln(XY)\le \ln(z))=P(\ln(X)+\ln(Y)\le \ln(z))$$
We need the convolution, since $\ln X $ and $\ln Y$ are independent we get, $$\int_0^{\ln(z)} f_{\ln(X)}\circledast f_{\ln(Y)}=\int_0^{\ln(z)}e^{-y}e^{-x+y}\,dy=e^{-x}ln(z),$$ i.e. $e^{-x}\ln(z)$ is the density of $XY$
 A: Apparantly $X$ is a positive rv (if not then $\ln X$ would not be well defined).
So you can start with
$$\ln Z=\ln XY=\ln X+\ln Y=R+S$$ 
Now let's find the distribution of $R+S$ where $R,S$ are iid with $\mathsf{Exp}(1)$-distribution.
For $x>0$ we find:
\begin{align}
P(R+S>x)
& =\int_0^{\infty} P(R+S>x\mid S=s)e^{-s}ds
=\int_0^{\infty} P(R>x-s)e^{-s}ds \\
& =\int_0^xe^{s-x}e^{-s}ds+\int_x^{\infty}e^{-s}ds
= e^{-x}(1+x)
\end{align}
Then for $z>1$ $$P(Z>z)=P(\ln Z>\ln z)=P(R+S>\ln z)=e^{-\ln z}(1+\ln z)=\frac{1+\ln z}{z}$$
This leads to $F_Z(z)=1-\frac{1+\ln z}{z}$ for $z>1$ and $F_Z(z)=0$ for $z\leq1$.

It seems that you think that $R+S$ has exponential distribution with rate $2$.
That however is the case for $\min(R,S)$ as you can see here.
A: Here's the solution using convolution:
Let $\overline{X} := \ln(X)$ and $\overline{Y}$ analogously.
We can now calculate the PDF of $\ln(XY) = \overline{X} + \overline{Y}$ using convolution:
For $x \ge 0$ we have
$$
f_{\ln(XY)}(x)
= f_{\overline{X}} \ast f_{\overline{Y}}
= \int_{-\infty}^{\infty} f_{\overline{X}}(y) f_{\overline{Y}}(x - y) \ \text{d}y
= \int_{0}^{x} e^{-y} e^{y - x}  \ \text{d}y
= \int_{0}^{x} e^{-x}  \ \text{d}y
= x e^{x}
$$ 
and $f_{\ln(XY)}(x) = 0$ otherwise.
Therefore, for $x \ge 0$ we have
$$
P(\ln(XY) \le x)
= \int_{-\infty}^{x}f_{\ln(XY)}(y)\ \text{d}y
= \int_{0}^{x} y e^{-y} \ \text{d}y
= 1 - e^{-x}(x + 1).
$$
For $z \ge 1$ this implies
$$
P(XY \le z)
= 1 - \frac{1 + \ln(z)}{z}
.
$$
and $P(XY \le z) = 0$ for $z \le 1$.
A: Here's another way without using the functional equation of the logarithm and instead deriving the PDF of $X$ directly.
Since $\ln(X) \sim$Exp$(1)$, for all $k \ge 0$ we have
\begin{equation*}
 \mathbb{P}(X \le e^k)
 = \mathbb{P}(\ln(X) \le k)
 = 1 - e^{-k}
\end{equation*}
By substitution one obtains
\begin{equation*}
 \mathbb{P}(X \le a)
 = 1 - \frac{1}{a}.
\end{equation*}
for all $a \ge 1$. 
We can now obtain the density function by calculating the derivative: $f_X(x) = \frac{1}{x^2}$.
Now, for $z \ge 1$ we have 
\begin{align*}
\mathbb{P}(XY \le z)
& = \int_{1}^{\infty} 1_{(-\infty,z]}(x,y) \ \text{d}\mathbb{P}(x,y)
= \int_{1}^{\infty} \int_{\mathbb{R}^+} 1_{\left(1, \frac{z}{y}\right)}(x) \ \text{d}\mathbb{P}_{X} \ \text{d}\mathbb{P}_{Y} \\
& = \int_{1}^{z} \int_{\mathbb{R}^+} 1_{\left(1, \frac{z}{y}\right)}(x) \frac{1}{x^2} \frac{1}{y^2} \ \text{d}x \ \text{d}y \\ 
& = \int_{1}^{z} \frac{1}{y^2} \left[ \int_{1}^{\frac{z}{y}} \frac{1}{x^2} \ \text{d}x \right] \text{d}y \\
& = \int_{1}^{z} \frac{1}{y^2} \left(1 - \frac{y}{z}\right) \text{d}y
= 1 - \frac{1 + \ln(z)}{z}.
\end{align*}
Note: Yet another way to this by realising that the sum of exponential distributed random variables is gamma distributed, can be found on this question.
