Does $2 \log(-1) = 0$? It occurred to me, following the rule of logarithms, that $a\log(b)=\log(b^a)$
Thus $2 \log(-1) = \log((-1)^2) = \log(1) = 0$
 A: In complex analysis the answer is no.  You do not have $\ln(a^b)=b\ln(a)$, it simply isn't an identity when we move beyond positive real arguments.
Things are different, however, in $p$-adics.  Select base 2 and consider the Maclaurin series
$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$
Plug in $x=-2$ and convert all terms to $2$-adics; the series converges $2$-adically ... to $...000000 = 0$!  We indeed have $2\ln(-1)=2×0=0$ in $2$-adics.
It's an example of how $p$-adics logarithms differ from what we are used to seeing in real or complex analysis.  In the case of $2$-adics, if we want to have the one-to-one and "inverse of exponential" properties that often make logarithms useful we have to limit the domain to arguments $\equiv 1 \bmod 4$.
A: The rule of logarithms (for reals) you mention can be valid when both $\log$ arguments, $b$ and $b^a$, are positive. This is not the case here. 
If you think further, your equality would mean that, dividing by $2$, $\log -1$ is equal to $0$, which is not valid.
A: As far as I know you can assign as input to any lnx function only numbers greater than 0, meaning that the following must happen: x>0. This is very obvious, cause there cannot exist a number such that when it is raised to the e to be equal to -1. So this 2ln(-1) cannot be defined. The same happens to any function of the form logax.
