$\log_{-2}(-8) = \frac{\log8+i\pi}{\log2+i\pi}$ (which is definitely not 3)
But what if we allowed all values (not just the principal value) of $\log(-1)$?
i.e, $\log(-1) = i(2n+1)\pi$ (n is an integer)
$\Rightarrow \log_{-2}(-8) = \frac{\log8+i(2n+1)\pi}{\log2+i(2m+1)\pi}$ (for integers n and m)
But the right hand side of the above expression is not 3 for any value of n and m..
Shouldn't 3 be one solution to $\log_{-2}(-8)$?
$$n = 1$$ $$m = 0$$ $$\frac{\log8+ i(2n+1)\pi}{\log2+ i(2m+1)\pi} = \frac{3\log 2 + 3\pi i}{\log 2 + \pi i} = 3$$