The question is as follows:
Let a be a positive constant and $X$ a random variable, so that:
$P(X>x)=x^{-a}, x\ge1$ and 0 otherwise. Find the correct distribution model [i.e. uniform, Weibull, Normal, etc.] of ln X.
My attempt at it:
for $x\ge1$, $F_X(x)=P(X<x)=1-x^{-a}$, therefore:
$F_Y(y)=P(Y<y)=P(lnX<y)=P(X<e^y )=F_X(e^y)$
$f_y(y)=\frac {d}{dy}F_Y(y)=\frac{d}{dy}F_X(e^y)=e^y*(1-e^{y^{-a}})$, for $x \ge 1$ and 0 otherwise.
This looks to me very close to the exponential distribution model, however it is not there yet. It is as if I am missing some trick to get it there, or perhaps my method is wrong. Any hints/guidance will be extremely appreciated.
It's easier to go the other way around. So
$$\Pr(Y>y) = \Pr(\log X>y) =\Pr(X>e^y) = e^{-ay},$$
so you indeed have an exponential.