# Change of variables notation clarification

In D. Mackay's book on Information Theory he states that we can make a change of variables for a probability density $P(x)$ by for example $l=\ln(x)$. Then, the density of $x$ is transformed to $$P(l)=P(x(l))\left\vert \frac{\delta x}{\delta l}\right\vert = P(x(l))x(l)$$

I'm confused about the notation of $x$ as a function of $l$: $x(l)$. What exactly is $x(l)$? If I got this right, then we have $l=\ln(x) \rightarrow x = \exp(l)$, so $$P(x(l))\left\vert \frac{\delta x}{\delta l}\right\vert = P(x(l))\left\vert \frac{\delta \exp(l)}{\delta l}\right\vert = P(x(l))\exp(l)=P(x(l))x$$ So are we essentially saying $x(l)=\exp(l)$? Is it that trivial and just a notational trick?

When doing change of variables one typically uses a smooth invertible transformation. So if $l = f(x)$ we can also think of $x = f^{-1}(l)$. For notational convenience, many authors simply write $x(l)$ instead of $f^{-1}(l)$. That is, yes it is trivial and just a notational trick.
$x(l)$ is "$x$ as a function of $l$".   In this case, $x(l)=\exp(l)$
$P(x(l))\lvert \dfrac{\partial x(l)}{\partial l}\rvert = P(\exp(l))\left\lvert\dfrac{\partial \exp(l)}{\partial l}\right\lvert = P(\exp(l)) \cdot\exp(l)$