Suppose $X$ is a normal r.v. with mean $\mu$ and variance $\sigma^2$. Let $Y = e^X$ so that $Y$ is lognormally distributed. The pdf of a lognormal distribution in terms of the value $x$ is
$$ F_Y(x) = \frac{1}{x\sqrt{2\pi\sigma^2}}\exp\bigg\{\frac{-(\log x-\mu)^2}{2\sigma^2}\bigg\} $$
Thus, given values from $X$, one can compute the probability density of the lognormal variable $Y$. However, what if you want to express the PDF in terms of $y$, the values $Y$ can take? Basically, I want to have a pdf $F_Y(y)$ in terms of the values $y$. Would it just be
$$ F_Y(y) = \frac{1}{\log y\sqrt{2\pi\sigma^2}}\exp\bigg\{\frac{-(\log(\log y)-\mu)^2}{2\sigma^2}\bigg\} $$
I am asking because I am working on a problem related to finance where I have the pdf of a normally distributed random variable $X$, call it $Q(X)$. I want the PDF of $S/S_0$, call it $P(S)$, where $X = \log(S/S_0)$. So, would $P(S)$ be given by the second equation?