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?

We have $$a^{x-y}=1$$ if $a$ is real, either $a=1$ or $a=-1,x-y$ is even — lab bhattacharjee, Jan 20 '14 at 10:04
The expression $a^x$ all on its own becomes probelmatic when $a < 0$! — , Jan 20 '14 at 10:05
@AlexR For $a = 1$ i know it will not, but for $a>0$ it will, correct? — Voyager, Jan 20 '14 at 10:10
$a>0$ includes $a=1$, but with this exception it's still only correct for $x,y\in\mathbb R$ — AlexR, Jan 20 '14 at 10:12

mathlove · Answer 1 · 2014-01-20T10:26:21.713

3

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

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

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

edited Jan 20 '14 at 10:26

answered Jan 20 '14 at 10:06

mathlove

139,939

Ok, but if $a<0$ and $a \neq -1$, than it works, correct? – Voyager Jan 20 '14 at 10:08
@Voyager Did you read Hurkyl's comment? – Did Jan 20 '14 at 10:08
@mathlove The same objection applies to $a\gt0$. – Did Jan 20 '14 at 10:09
1

@Voyager: Yes, it's correct. – mathlove Jan 20 '14 at 10:22

score 3 · Answer 2 · answered Jan 20 '14 at 10:11

3

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 $a\in\mathbb R_-\cup\{0\}$.

answered Jan 20 '14 at 10:11

AlexR

24,905

1

The principal branch of $\ln$ is defined for $a < 0$, it is $\ln|a| + i\pi$, but it is not continous across the branch cut. – gammatester Jan 20 '14 at 11:08
@gammatester this is a matter of convention. If you leave it undefined, it is analytic on its domain; if you do define it, it's no longer analytic, making things a lot harder. – AlexR Jan 20 '14 at 11:09

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

2 Answers2