Distribution of $-\log X$ if $X$ is uniform. For $X$ and $Y$ random variables; $X$ follows the uniform distribution.
(1): if $Y=-\log X$
(2): then it can be shown that $-\log X$ is distributed as $\exp(1)$ {i.e. exponential with mean 1}.
Why is this so? Intuitively statement (2) make sense to me. But i'd like a mathematical proof.
-Probably wrong working:
(1) seems to imply $\exp(-Y) = X$ (which is like saying $X$ is exponentially distributed, which is a contradiction, since its actually uniform!); or is it incorrect for me to do this since $X$ and $Y$ are random variables?
Ultimately how do I prove (2)?
Thanks
 A: To begin with, $Y = -\ln(X)$ only works when $[a,b] \subset \mathbb{R}_{>0}$. So, let

$$
X \sim \mathcal{U}[a,b] \quad\text{where}\quad [a,b] \subset \mathbb{R}_{>0}.
$$

then $Y$ is log-uniformly distributed, in symbols

$$
Y \sim \mathcal{LU}[\alpha,\beta] \quad\text{with}\quad \alpha = -\exp(a) \ \text{and} \ \beta = -\exp(b).
$$

This can be derived by applying the transformation rule to the probability density function (PDF) of $X$, given by
$$
f_\mathcal{U}(x \mid a,b) = \begin{cases} \dfrac{1}{b - a}, & a \leq x \leq b \\ 0, & \text{else} \end{cases}.
$$
Suppose $\alpha = -\exp(a)$ and $\beta = -\exp(b)$. The PDF of $Y = -\ln(X)$ is then
$$
\begin{eqnarray}
 f_\mathcal{U}(-\ln(x) \mid a,b) \cdot \left| \dfrac{\mathrm{d} (-\ln(x))}{\mathrm{d}x} \right|
 &=& \begin{cases} \dfrac{1}{(b - a) \cdot x}, & a \leq -\ln(x) \leq b \\ 0, & \text{else} \end{cases} \\
 &=& \begin{cases} \dfrac{1}{(b - a) \cdot x}, & -\exp(a) \leq x \leq -\exp(b) \\ 0, & \text{else} \end{cases} \\
 &=& \begin{cases} \dfrac{1}{(\ln(-\beta) - \ln(-\alpha)) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
 &=& \begin{cases} \dfrac{1}{\ln\left( \dfrac{-\beta}{-\alpha} \right) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
 &=& \begin{cases} \dfrac{1}{(\ln(\beta) - \ln(\alpha)) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
 &=& f_\mathcal{LU}(x \mid \alpha,\beta).
\end{eqnarray}
$$
The result is the PDF of the log-uniformly distributed variable $Y \sim \mathcal{LU}[\alpha,\beta]$.
A: reiterate @S4M's answer but with $Y = -\log(X)$, rather than $Y = \log(X)$, according to OP's post:
$$ P(Y<y) = P(-\log(X)<y) = P(X>\exp(-y)).$$
Given that $X$ is uniformly distributed between $[0,1]$,
$$ P(Y<y) = P(X>\exp(-y)) = F(1 -\exp(-y)),$$
which is the CDF of exponential distribution, with $\lambda=1$.
A: The trick to resolve this kind of problems is to calculate the distribution of $Y$: $F(y) = P(Y<y)$. In this case, we have $F(y)=P(\log X<y)=P(X<e^y) =\int_0^{e^y}dt$. 
Now if you make the change of variable $t=e^u$, you are able to transform this expression into something of the form $F(y) = \int_{-\infty}^y f(u)du$, and then $f$ will be the density you are looking for.
