Uniform prior distribution on log scale

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Can anyone please suggest me a distribution whose $log$ transformation is uniform and it should be a well known distribution? I am not sure if it exists. I know that if we consider a uniform distribution $x \sim U(a,b)$ and and take $y = e^x$. Then, $log(y)$ follows uniform distribution. However, $y$ is not a well known distribution. Can you suggest me a well known distribution which is close to this?

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The distribution of your $y$ has two names:

  1. log-uniform distribution (also spelled as log uniform) which follows properties of the inverse transform
  2. Reciprocal distribution because it has a probability density function $\it f(y) \propto \frac 1y$

$y$ has PDF $f(y,c,d) = \frac 1{y \ \left(\it ln(d)-\it ln(c)\right)} \ \text for \ y \in [c,d]$ where $c=e^a, d=e^b$

I don't know if it's well known but Google is giving tens of thousands answers.

https://en.wikipedia.org/wiki/Reciprocal_distribution

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Let me provide you with a full derivation so that you do not feel you are missing anything.

Let random variable $X$ have pdf $f_X(x)$. Let $Y=g(X)$ where $g$ is monotone. Let $h = g^{-1}$ be the be inverse function of $g$. Then the pdf of $Y$ is given by $$ f_Y(x)=f_X(h(x))h^\prime(x). $$

You can read here for the above statement.

In your case, you have $g(X)=ln(X)$. Hence $h(x) = e^x$. You want the distribution of $Y$ to be uniform, i.e. $$ f_Y(x)= \frac{1_{[a, b]}}{b-a}$$ for some real value $a < b$. Plug in h(x), you get $$ f_X(e^x)e^x= \frac{1_{[a, b]}}{b-a}.$$

Let $y=e^x$. you get $$ f_X(y) = \frac{1_{[e^a, e^b]}}{y(b-a)}, y > 0. $$ In the last formula, $y$ is just a variable, you can use $x$ if you prefer. Then you get $$ f_X(x) = \frac{1_{[e^a, e^b]}}{x(b-a)}, x > 0. $$ That is your distribution. You have NO other options.