Showing that $-\ln{X} \sim \exp{\alpha}$ for $X \sim Beta(\alpha, 1)$ The CDF for $X \sim Beta(\alpha,1)$ is given by:
$$F(x) = \frac{\int_{0}^{x}t^{\alpha-1}dt}{\int_{0}^{1} t^{\alpha-1}dt}$$
I am given to understand that $-\ln{X} \sim \exp(\alpha)$ if $\alpha > 0$.
How do I go about showing this? In all my attempts, I really don't know what to do with the presence of the natural logarithm term. 
EDIT: If this is not so trivial, it might help to look at the Kumaraswamy Distribution.
 A: Yes, but I think you first need to explicitly figure out the cdf of $X$, $F_X$, which turns out to be:
$$ F_X(x):=F_X(x; \alpha,1) = x^\alpha. $$
Then, the usual approach:
$$P(Y:=-\ln X \leq y) = P\left(X\geq {\rm e}^{-y}\right) = 1-F_X\left({\rm e}^{- y}\right) =1- {\rm e}^{-\alpha y}$$
A: Let $g(x) = -\log x$, hence $Y = g(X) = -\log X$, so that $X = g^{-1}(Y) = e^{-Y}$.  Then $$F_Y(y) = \Pr[Y \le y] = \Pr[-\log X \le y] = \Pr[X \ge e^{-y}] = 1 - F_X(e^{-y}).$$  Then differentiation of the CDF gives $$f_Y(y) = f_X(e^{-y})e^{-y}$$ by the chain rule.  Now if $X \sim \mathrm{Beta}(\alpha,1)$, then $f_X(x) \propto x^{\alpha - 1}$, hence we have $$f_Y(y) \propto e^{-y(\alpha-1)} e^{-y} = e^{-\alpha y}$$ which is clearly exponential with rate parameter $\alpha$.  There is no need to explicitly compute $F_X$; it suffices to know the PDF of $X$.  Indeed, this approach can be used even when $X \sim \mathrm{Beta}(\alpha,\beta)$ in general, although the resulting distribution for $Y$ is not generally exponential.  I leave it to the reader as an exercise to determine whether such a distribution has a name.
