$X$ has density $g(x)=2/x^31_{[1,\infty]}$ Find density of $e^X$.
This may be trivial, but it would be not nice if during exam I made some even more trivial mistake, so I'd like someone to look at the solution.
We know how looks like $P(X\leq t)$. We want to know how looks like $P(e^X\leq t)$. First we see that $e^X\geq0$, so for all $t\leq0$ we have $P(e^X\leq t)=0$, therefore we can now safely assume that $t$ is positive number. We have $$P(e^X\leq t)=P(X\leq \ln t)=\int_\mathbb{-\infty}^{\ln t}2/x^31_{[1,\infty]}dx=\int_\mathbb{0}^{\ln t}2/x^31_{[1,\infty]}dx$$ Here we also see that $\ln t>1$ in order for integral to be positive, so we can assume that and continue $$\int_1^{\ln t}2/x^3dx=-1/\ln^2t+1$$
Now to find density we need to differentiate our result. So we obtain for $\ln t>1$ density $$\frac{2}{t\cdot ln^3t}$$ and $0$ everywhere else. Therefore our density is $$f(t)=\frac{2}{t\cdot ln^3t}1_{[e,\infty]}$$
Is it correct?