Formal Definition of Exponentiation (Rational Exponent) This question has bothered me for a long time...
I know some people define $(-8)^\frac{1}{3}\approx 1 +1.73i$ referring to root with the minimum argument by De Moivre's Theorem. For this question, I want to stick with the Calculus definition where $(-8)^\frac{1}{3}=-2$. [Note I do not know what the Calculus definition actually is here.]

$\textbf{Question:}$ What is the "well-defined" definition for $a^c$ in "Calculus" where $c$ is a rational number? I'm not trying to sound harsh, but the "middle school" answer that $\sqrt[e]{a^d}:=a^\frac{d}{e}$ is not what I'm looking for the following reason.

Mathematically speaking, defining $\sqrt[e]{a^d}:=a^\frac{d}{e}$ where $d\in \mathbb{Z}$ and $e\in \mathbb{Z}^{\neq 0}$ would require the following theorem below to state such a definition actually makes sense. I provide a counterexample for the Thm below which means the definition does not work. So, what is the actual definition for $a^c$?


Thm: Let $c,d,e,f$ be integers where  $d,f\neq 0$. If $\frac{c}{d}=\frac{e}{f}$, then $\sqrt[d]{a^c}=\sqrt[f]{a^e}$.
Counterexample: Clearly, $\sqrt[2]{(-1)^2}\neq \sqrt[3]{(-1)^3}$.
 A: For a coherent answer, you really have to go via complex numbers.
The answer is that $a^c = \exp(c \log(a))$, where $\log(a)$ is a number $b$ such that $\exp(b) = a$.  The problem is that this is multi-valued, because $\exp(2\pi i)=1$.
Now $\exp(x+iy) = \exp(x) (\cos(y) + i \sin(y))$ (where $x$ and $y$ are real) is real if and only if $\sin(y)=0$, i.e. $y = \pi k$ for some integer $k$.
If $a < 0$, the values of $\log(a)$ are $\ln(|a|) + (2n+1)i \pi$ for integer $n$.
The corresponding value of $a^c$, where $c$ is real, is if and only if $(2n+1)c$ is an integer.  Thus $a^c$ has a real value if and only if $c$ can be written as a rational number with odd denominator.
A: For real numbers the equation $x^n = b$ will, if $b\ne 0; b\ne 1$
1) If $b >0$ and $n$ is even have exactly two solutions $x = c$ for some $c > 0$ and $-c$.
1b) If $b > 0$ and $n$ is odd then there will have exactly one solution $x = c$ for som $c > 0$.
2) If $b < 0$ and $n$ is odd then there will be exactly one solution $x = c$ for some $c < 0$.  Furthmore $|c|$ will be the solution to $x^n = |b|$.
2b) If $b < 0$ and $n$ is even there aren't any solutions.
And its easy to see that:  $c > 0$ is solution to $x^n = b^m; b>0;b\ne 1$ and $c$ is also a solution to $c^p = b^q$ if and only if $\frac mn =\frac qp$ and $c$ is a solution to $x^{pn}= b^{mp} = b^{nq}$.
Therefore a formal definition of $b^{\frac ab}$ can be and usually is the one you consider niave and informal.

If $q = \frac mn$ and $m$ and $n$ are in lowest terms then $b^q = \sqrt[n]{b^m}$ assuming either $b \ge 0$ or if $b < 0$ then $n$ is odd.  Otherwise if $b < 0$ and $n$ is odd this is not defined.

Note your theorem for positive numbers allows us to make the specification that we put $q$ in lowest terms.
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Note: we completely overhaual this definition when we develop real exponents or complex numbers.
We only actuall need these for analyzing polynomials and rational functions.  And even then we can probably get away without using it most of the time.
