Why is $\log_{-2}{4}$ complex? With the logarithm being the inverse of the exponential function, it follows that $ \log_{-2}{4}$ should equal $2$, since $(-2)^2=4$. The change of base law, however, implies that $\log_{-2}{4}=\frac{\log{4}}{\log{-2}}$, which is a complex number. Why does this occur when there is a real solution?

The exponential function is not invertible on the complexes. Correspondingly, the complex logarithm is not a function, it is a multi-valued function. For example, $\log(e)$ is not $1$ -- instead it is the set of all values $1 + 2 \pi \mathbf{i} n$ over all integers $n$.
How are you defining $\log_a(b)$? If you are defining it by $\log(b) / \log(a)$, then it too is a multi-valued function. The values of $\log(4)/\log(-2)$ ranges over all values $(\ln 4 + 2 \pi \mathbf{i} m)/(\ln 2 + \pi \mathbf{i} + 2 \pi \mathbf{i} n)$, where $m$ and $n$ are integers.
Do note that the set of values of this multi-valued function does include $2$; e.g. when $m =1$ and $n=0$.
