Why the identity $ a^x = a^y \Longrightarrow x=y $ do not work for $a<0$?

Question

All books that i am reading are telling that the identiy $ a^x = a^y \Longrightarrow x=y ; \,\,\,\,\, a \in \mathbb{R} - \{0,1\} $ $ \,\,\,\,$ also do not work for $a<0$, but, for example, if $ (-2)^2 = (-2)^x $, than for sure $x=2$.

Than, the identity works for $a<0$, correct?

Answer 1

It does not work for $a\lt 0$.

If $$(-1)^x=(-1)^2,$$ then $x=4$ is another solution.

In general, $$(-1)^x=(-1)^2\Rightarrow x=2m (m\in\mathbb Z)$$

Answer 2

Note that $a^b := \exp(b \ln a)$ and even the definition of $\ln a$ is problematic for arbitrary $a\in\mathbb C$ and the complex exponential isn't bijective (unlike the real). The principal branch of the natural logarithm is not defined for $a\in\mathbb R_-\cup\{0\}$.