For this problem I'm only concerned with real solution, not complex ones. You can't take the square root of a negative number, but you can take the cube root of a negative.
For fraction, as long as the power is a reduced fraction and the denominator is odd, you can take the power of a negative number.
What about for an irrational power, such as $\sqrt{2}$? Is there a real answer to $(-2)^{\sqrt{2}}$. I can't tell because you can't rewrite it as a reduced fraction.
On
Suppose we consider $x^p$ with $p$ real and $x$ a negative real number. We have $x = -r$ with $r>0$, so the polar form of $x$ is $x = r e^{i \pi k}$ for $k$ any odd integer.
The (multivalued) exponential $x^p$ is defined by $r^p e^{i \pi k p}$, where $r^p$ is a well-defined real-valued exponential.
Suppose this exponential is real for some fixed $k$, that is, $e^{i \pi k p} \in \mathbb R$. This holds if and only if $\pi k p$ is an integer multiple of $\pi$, i.e., $kp \in \mathbb Z$. Since $k$ is odd, and in particular nonzero, it follows that $p$ is a rational number with odd denominator.
On
For natural $n$ $\sqrt[n]{-b}$ for negative $-b$ only has real solutions if $n$ is odd.
So $(-b)^{\frac mn}; mn \in \mathbb Z; n\ne 0;\gcd(m,n) = 1$ will only have a sensible definition is $n$ is odd.
And for that reason $b^x; x \in \mathbb R$ is only defined for real numbersin general (i.e. for irrational numbers or rationals with even denominators) if $b$ is positive.
We can only define $b^x; x$ irrational in any meaningful way as a limit. (i.e. $b^x = \lim b^q$ where $q \in \mathbb Q$ and $\lim q = x$ AND $\lim b^q $ exists. This is not the case if $b < 0$.) Or in terms of natural logarithms $x^a = e^{a\ln x}$ after somehow defining $e$ and $\ln$. Or some equivalent third method. In any event if $b < 0$ we just don't have anything that is consistant when $(-b)^q$ can jump from positive to negative to undifined infinite times within any interval of $x$.
Of course with complex roots. The definition $b^x = e^{x\ln b}$ and $e^{a + bi} = e^a*(\cos b + i \sin b)$ mean $(-b)^x$ may coincidentally have real values despite not having any meaningful definition in the reals.
A negative number has a real $n$-th root if and only if $n$ is odd.
More generally, for complex numbers $x$ and $y$, $x^y$ is a multivalued function with values $\exp(y \log x)$ for any branch of $\log x$. In particular, if $x < 0$ and $y$ is real, the values of $\log x$ are $\log |x| + \pi i n$ for odd integers $n$, and there is a real value of $x^y$ if and only if $y n$ is an integer for some odd $n$. Thus $y$ must be a rational number, $p/q$ in lowest terms, whose denominator $q$ is odd.