I'm assuming the support of [tex]X[/tex] is [tex]x\ge1[/tex], judging by the usage of [tex]Y=\ln X[/tex]. The CDF of [tex]X[/tex] is
[tex]F_X(x)=\mathbb P(X\le x)=\begin{cases}1-\dfrac1{x^\alpha}&\text{for }x\ge1\\\\0&\text{otherwise}\end{cases}[/tex]
and so
[tex]F_Y(y)=\mathbb P(Y\le y)=\mathbb P(\ln X\le y)=\mathbb P(X\le e^y)=F_X(e^y)[/tex]
[tex]\implies F_Y(y)=\begin{cases}1-e^{-\alpha y}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}[/tex]
[tex]\implies f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_Y(y)=\begin{cases}\alpha e^{-\alpha y}&\text{for }y>0\\0&\text{otherwise}\end{cases}[/tex]
which is the PDF of an exponential distribution with rate parameter [tex]\alpha[/tex].
