Why does $\frac{dln(x_2^p/x_1^p)}{dln(x_2/x_1)}$ equal $p$?

Question

I'm working on elasticities and I found the following identity

$$\frac{dln(x_2^p/x_1^p)}{dln(x_2/x_1)}=p$$

What do the numerator and denominator even mean here? I'm used to seeing $\frac{dy}{dx}$, not something like $d(x_2^p/x_1^p)$.

And why does the identity hold?

Answer 1

First of all, it is a property of logarithms that $\log a^p = p\log a$. You can use it here getting $$\log(x_2^p/x_1^p) = \log((x_2/x_1)^p) = p\log(x_2/x_1).$$ Set $u := \log \frac{x_2}{x_1}$. Then your left hand side can be rewritten as $$\frac{d(pu)}{du} = p\frac{du}{du} = p.$$ This is done assuming that $p$ does not depend on $\log(x_2/x_1)$ and that those d's denote derivatives.

Answer 2

Hint: $\log(a^b) = b\log(a).$