Why do the normal and log-normal density functions differ by a factor? If a random variable $W$ is Normally distributed, then $\exp(W)$ is Log-Normally distributed. 
However, the pdfs of these two random variables differ by a factor of $\exp(W)^{-1}$.
The Normal pdf for $W$ is
$$P(w \in W) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(w - \mu)^2}{2 \sigma^2}}$$
but the Log-Normal pdf for $\exp(W)$ is
$$P(\exp(w) \in \exp(W)) = \frac{1}{\exp(w)\sigma\sqrt{2\pi}}\ e^{-\frac{\left(w-\mu\right)^2}{2\sigma^2}}$$
Why is this the case? Shouldn't they be equivalent?
 A: The factor you speak of is simply the Jacobian of the transformation, the determinant of the transformation from one variable (or set of variables) to another.
It has nothing in particular to do with normality and everything to do with the ($\exp$) transformation.
It's a straightforward calculation:
If $Y=\exp(X)$ and the cdf of $X$ is $F_X$, then
$P(Y\leq y) = P(\exp(X)\leq y) = P(X\leq \log(y)) = F_X(\log(y))$ 
Hence the pdf of $Y$ is $\frac{\partial}{\partial y} F_X(\log(y)) = f_X(\log(y))\cdot\frac{1}{y}\,,\quad$ (by the chain rule for differentiation).
More generally, if $Y$ is some different monotonic increasing function of $X$, $Y=g(X)$, then
$P(Y\leq y) = P(g(X)\leq y) = P(X\leq g^{-1}(y)) = F_X(g^{-1}(y))\,$
and so the pdf of $Y$ is $\frac{\partial}{\partial y} F_X(g^{-1}(y)) = f_X(g^{-1}(y))\cdot\frac{\partial}{\partial y}g^{-1}(y)$.
A similar relationship (up to a sign change) holds for monotonic decreasing transformations.
See here, which deals with both forms.
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You can't write your second density that way, because you're dealing with the density for the transformed variable; if you're going to substitute $w$ into that density you need the Jacobian for the inverse transform (the one back to $w$), which takes you straight back to the density for $w$ you started with.
The first post whuber links to in his comment gives some additional intuition for what's going on.
A: I'd like to provide a simple and straightforward demonstration to where this factor comes from.
Let $p(x)$ be the PDF of a normal distribution $\mathcal{N}\left(\mu,\sigma^2\right)$. Then, of course:
$$\int\limits_{-\infty}^{+\infty} p(x)dx = \frac{1}{\sqrt{2\pi}\sigma} {\rm e}^ {-\frac{\left( x-\mu \right)^2}{2\sigma^2} } {\rm d}x = 1$$
If you now make a substitution $x = \log y$, i.e. $x$ is distributed normally, and $y$ has a log-normal distribution, you'll get
$$\int\limits_{-\infty}^{+\infty} p(\log y){\rm d} (\log y) = 1 = \frac{1}{\sqrt{2\pi}\sigma} {\rm e}^ {-\frac{\left( \log y-\mu \right)^2}{2\sigma^2} } \boxed{{\rm d}(\log y)} $$
and the boxed term is $\frac{{\rm d}y}{y}$, showing where does the $\frac{1}{y}$ come from. So that the PDF of $y$ is
$$\frac{1}{\sqrt{2\pi}\sigma y} e^ {-\frac{\left( \log y-\mu \right)^2}{2\sigma^2} }$$
