$(-32)^{\frac{2}{10}}\neq(-32)^{\frac{1}{5}}$?

Question

Is exponentiation by rational numbers defined only for simple fractions?

$(-32)^{\frac{2}{10}}=\sqrt[10]{(-32)^2}=\sqrt[10]{1024}=\pm2$ (and $8$ other complex roots)

$(-32)^{\frac{1}{5}}=\sqrt[5]{(-32)^1}=\sqrt[5]{-32}=-2$ (and $4$ other complex roots)

How do you settle this conflict?

Answer 1

Please read this. In general, $c^a=e^{a \log c}$ when $c\in\mathbb{C}$ and $a$ is a fraction will produce a set, so when you say $c_1^{a_1}=c_2^{a_2}$, you are comparing two sets. In particular,$$(-32)^\frac{1}{5}=2 (-1)^\frac{1}{5}=2 e^{i(\pi+2k\pi)\frac{1}{5}}=\{2e^{i\frac{\pi}{5}+n\alpha}|n=0,1,2,3,4,\alpha=\frac{2\pi}{5}\}$$

$$(-32)^\frac{2}{10}=2 (-1)^\frac{2}{10}=2 e^{i(\pi+2k\pi)\frac{2}{10}}=\{2e^{i\frac{\pi}{5}+n\alpha}|n=0,1,2,3,4,\alpha=\frac{2\pi}{5}\}$$

Answer 2

Of the several different but related definitions for exponentiation, the one that accepts a rational exponent should be phrased to exclude negative bases, in order to avoid exactly this problem.

In fact, the only one of the usually encountered definitions that give meaning to a negative integer to a non-integral power is the one for complex numbers, which comes with its own problems of being multi-valued.

One could choose to define $a^{1/n}$ for arbitrary real $a$ and odd $n$ as a root extraction -- but again this doesn't scale well to other rational exponents, and in general it is not really thought to be worth the trouble to maintain this special case when one can just write $\sqrt[n]{a}$ instead.

Alternatively, one would worm around the problem by defining $a^{p/q}=\sqrt[q]{a^p}$ if and only if $p/q$ is in lowest terms, but I'm not aware of any use of this somewhat tortuous interpretation except as a way to gain the warm fuzzies of having defined exponentiation with a slightly larger domain than the next guy uses.