Log-uniform distributions I am having some difficulty understanding what log uniform distributions are.
Suppose that $\log X$ is uniformly distributed on the interval $[1,e]$. How do I describe $P(X=x)$? It seems like there is more probability mass on the lower numbers so that X itself is not uniformly distributed, but I am having difficulty formalizing this argument.
 A: Your definition of $X$ suggests that $X$ is a continuous random variable, but your question $\Pr[X = x]$ suggests you wish to treat it as a discrete variable.  If you were asking for the probability density function of $X$, rather than the probability mass function, then we could proceed naturally using a transformation, since $\log$ is a monotone function:  if $Y = g^{-1}(X) = \log X$, then $X = g(Y) = e^Y$ and $$f_X(x) = f_Y(g^{-1}(x)) \left| \frac{dg^{-1}}{dx} \right| = \ldots$$  This of course means that $X$ is not uniformly distributed.
A: I like @heropup's answer, but am slightly bothered by the fact that he didn't finish the derivation for the OP. To enrich his answer, I'd like to add the following picture, and some comments on the above answer:

If you follow @heropup's derivation, you'll find that
$$f_{X}(x) = \frac{I_{[e, e^e]}(x)}{x(e-1)}$$
More generally, if $Y \sim Unif(a,b)$, such that $Y = log(X)$ for some random variable $X$, then
$$ f_{X}(x) = \frac{I_{[e^a, e^b]}(x)}{x(b - a)} $$
For the sake of validating your intuition, I've made the figure of $Y \sim Unif(0,1)$, so that we see the originating random variable is actually defined on $[1, e]$ and looks something like $1/x$. As you pointed out, there is, indeed, more mass at the "beginning" of $X$'s domain and from the picture of the CDF, alone, we can see that $X$ cannot also be uniformly distributed (since the CDF is not a straight line).
I hope this fleshes out a bit of the above answer...
edit: you can find the code to make this picture here.
