What is the $\log_{10}$ transformation of the Rayleigh distribution?

Question

If one takes the Rayleigh distribution, $$f(x; \sigma) = \frac{x}{\sigma^{2}} e^{-x^{2}/(2\sigma^{2})}$$ and makes a transformation into $\log_{10}$ units, i.e. $\log_{10}(x)$ what is the resultant distribution? Is this a known distribution -- what is it called?

Answer 1

Consider \begin{equation} \begin{split} P(Y \leq y) &= P(\log_{10}(X) \leq y)\\ &= P(X \leq 10^y) \end{split} \end{equation} Differentiating w.r.to $y$ \begin{equation} \begin{split} f_Y(y) &= f_X(10^y) \frac{d}{dy}(10^y) \\ &= \frac{10^y}{\sigma^2} e^{-10^{2y}/2\sigma^2} 10^y (\ln 10) \\ &= \frac{10^{2y}}{\sigma^2} e^{-10^{2y}/2\sigma^2} (\ln 10) \\ \end{split} \end{equation} I don't know if it belongs to any known distribution.