I arrive at the following equation:

$$\left(-1\right)^a=\left(-1\right)^{-a}$$

Intuitively, this equation could be satisfied if $a$ is either $0$ or $1$, however, if you take the log of both sides you get:

$$a\ln{\left(-1\right)}=-a\ln{\left(-1\right)}$$ $$a\left(i\pi\right)=-a\left(i\pi\right)$$ $$a=-a$$

Which can only be satisfied if $a=0$. Why can't $a=1$ since I know that $-1=\frac{1}{-1}=-1$?

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Hint : $1$/$i$ = $-i$ –  mick Oct 2 '12 at 21:43
In general, $f(x) = f(y)$ does not imply $x = y$. "I arrive the the following equation: $$\text{color}(\text{cherry}) = \text{color}(\text{strawberry}).$$ Applying the inverse of the "color" operator on both sides, you get: $$\text{cherry} = \text{strawberry}$$ which seems to say that cherries are strawberries…" –  MJD Oct 3 '12 at 3:01

Any integer satisfies the original equation. Here is your mistake:

$$(-1)^a=(-1)^{-a}$$

$$e^{ai\pi}=e^{-ai\pi\color{red}{-2k\pi}}$$

$$ai\pi=-ai\pi-2k\pi$$

$$a=-a+2k$$

$$a=k,\;\;k\in\mathbb{Z}$$

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$\ln\left((-1)^a\right) = \ln\left((-1)^{-a}\right)$
$\ln(\exp(a\cdot \ln(-1))) = \ln(\exp((-a)\cdot \ln(-1)))$
The second equation does not necessarily imply that $\;\; a\cdot \ln(-1) \: = \: (-a)\cdot \ln(-1) \;\;$,
since $\: (\pm a)\cdot \ln(-1) \:$ are not necessarily positive real numbers.