number theory - Why aren't logarithms defined for negative $x$?

Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$)^[1], are true, why aren't logarithms defined for negative numbers?

Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this?

Answer

In complex analysis, $x$ can be negative. For example $e^{i\pi} = -1$, so $\ln{(-1)} = i\pi$.

I hadn't seen a log with a negative base, but I thought one could define it with the normal change of base formula: $\log_{b}{x} = \frac{\ln{x}}{\ln{b}}$. However, this turns out to be inconsistent Might be inconsistent, at the very least, it doesn't give 3:

$$\log_{(-2)}{(-8)} = \frac{\ln{(-8)}}{\ln{(-2)}} = \frac{\ln{8} + i\pi}{\ln{2} + i\pi} = \frac{3 \ln{2} + 3i\pi - 2i\pi}{\ln{2} + i\pi} = 3 - \frac{2i\pi}{\ln{2} + i\pi} \ne 3$$

This is because the complex log has a branch cut in it. For example: $e^{3i\pi} = -1$, but $\ln{(-1)} = i\pi$, not $3i\pi$. The cut is made so that $\mathrm{Im}(\ln{z}) \in \left(-\pi,\pi\right]$.