TITLE: Is there a problem when defining exponential with negative base?
QUESTION [9 upvotes]: Well, this question may seem silly at first, but I'll make my point clear. Suppose $n \in \Bbb N$ and suppose $a \in \Bbb R$ is any number. Then the definition of $a^n$ is clear for any $a$ we choose. Indeed we define:
$$a^n = \prod_{k=1}^na$$
And even if $a$ is negative this has a meaning. Then we extend the definition for $n \in \Bbb Z$ and for $n \in \Bbb Q$. When we are to define to $n \in\Bbb R$ we define it as:
$$a^x=e^{x\ln a}$$
That's fine, but $\ln $ is a function defined on $\Bbb R^+$ so that if we try to compute $(-5)^\pi$ we'll get into trouble because this would be:
$$(-5)^{\pi}=e^{\pi \ln(-5)}$$
But $\ln (-5)$ is undefined. In that case, the function that $f : A \subset \Bbb R^2 \to \Bbb R$ given by $f(a,x) = a^x$ would be undefined if $a < 0$, so that $A = \Bbb R^+ \times \Bbb R$. What I thought was: we can extend this function when $a$ is negative and $x$ is rational. In that case we would set it to the old definition of exponentiation, since we would have a real raised to a rational.
So, for negative base and irrational exponent the exponential remains undefined?
Thanks very much in advance!

REPLY [2 votes]: If $a < 0$ and $a \ne -1$, then it is possible to define $f(x) = a^x$, provided we restrict our domain to $a \in \{m / n : m$ is an integer and $n$ an odd positive integer$\}$. In this case, $a^{m/n} = \sqrt[n]{a^m}$. 
For example, $(-8)^{-2/3} = 1/\sqrt[3]{(-8)^2} = 1/4$.