A negative base is a point of conflict between the three commonly used meanings of exponentiation.
- For the continuous real exponentiation operator, you're not allowed to have a negative base.
- For the discrete real exponentiation operator, we allow fractional exponents with odd denominators, and
$$(-a)^{b/c} = \sqrt[c]{(-a)^b}= \left( \sqrt[c]{-a} \right)^b = (-1)^b a^{b/c} $$
(and this is allowed because every real number has a unique $c$-th root)
- For the complex exponentiation operator, exponentiation is multivalued. An exponentiation with denominator $n$ generally takes on $n$ distinct values, although one is generally chosen as the "principal" value.
For $(-5)^{2/3}$, these three exponentiation operators give
- Undefined
- $\sqrt[3]{25}$
- $\omega \sqrt[3]{25}$ is the principal value. The other two are $\sqrt[3]{25}$ and $\omega^2 \sqrt[3]{25}$, where $\omega = -\frac{1}{2} + \mathbf{i} \frac{\sqrt{3}}{2}$ is a cube root of $1$.
Unfortunately, which meaning of exponentiation is meant is rarely ever stated explicitly, and has to be guessed from context.
I'm guessing that the second one is meant.
In case you're curious, here is part of the rationale for the first and third conventions.
In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other negative. And it's not even clear that $(-5)^{\pi/5}$ should be meaningful!
For the third convention, the whole thing is like the idea of $\pm 2$ being the 'square root of 4', but for the fact the complexes cannot be cleanly separated into "negative" and "positive" to let us choose a specific one nicely.
A method is chosen for the principal value, based trying to get positive bases 'right' and trying to keep continuity as much as possible, but alas this convention gets the negative bases 'wrong'.
In some sense, this can be viewed as the principal value of $(-5)^{2/3}$ chosen to be "two-thirds of the way" from positive to negative.
For quick-and-dirty approximations, I'm fond of the musical logarithms (writeup by Sanjoy Mahajan of work due to I. J. Good, who credits his father). Essentially this boils down to a mathematical fact:
- $2^{10} \approx 10^3$, and taking 120th roots, $2^{1/12} \approx 10^{1/40}$.
and a "musical" fact:
- many rational numbers with small numerator and denominator can be approximated as powers of $2^{1/12}$.
I call this a "musical" fact because $2^{1/12}$ is the frequency ratio corresponding to an (equal-tempered) semitone.
For example: $3/2$ is the frequency ratio corresponding to the musical interval of a perfect fifth, which is seven semitones; thus $3/2 \approx 2^{7/12} \approx 10^{7/40}$ and so $\log_{10} 3/2 \approx 7/40$.
Best Answer
What D.L. gave as an answer is a good way which can be slightly improved for faster convergence of the series.
Suppose that we write $x=\lfloor x\rfloor+y$. So we could use $$a^x=a^{\lfloor x\rfloor }\sum_{n=0}^{\infty} \frac{(ln(a).y)^n}{n!}$$ This could be for $y<\frac{1}{2}$. Otherwise we should use $x=\lceil x\rceil-y$ and use $$a^x=a^{\lceil x\rceil}\sum_{n=0}^{\infty} \frac{(-ln(a).y)^n}{n!}$$ So, you have integer powers of $a$ and a very few logarithms to remember.