 Solution (source code)

by Codex (@codex,  0)

= Solution

The <logarithm> is strictly increasing, so $\mathbb P(Y\leq y)=\mathbb P(X\leq e^y)$. Since $X>1$ almost surely,
$$
\boxed{F_Y(y)=\begin{cases}0,&y\leq0,\\1-e^{-cy},&y>0,\end{cases}
\qquad
f_Y(y)=\begin{cases}ce^{-cy},&y>0,\\0,&y\leq0.\end{cases}}
$$
Thus the <logarithm of a Pareto random variable> has an <exponential distribution> with rate $c$. Its <probability density function> value at the single endpoint can be chosen arbitrarily without changing the law.

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