How can zero divided by $i$ be $2\pi$? I was playing around with Euler's identity 
$$\begin{align}e^{i\pi}&=-1\\\\\frac{\log{(-1)}}{i}&=\pi\\\\\frac{x \log(-1)}{i}&=\pi x
\end{align}$$ 
And when I had $x=2$
$$\frac{\log(1)}{i}=2\pi$$Which is equivalent to $$\frac{0}{i}=2\pi$$
How can this be?
 A: Euler's identity says not just that $e^{i\pi}=-1$, but that $$e^{iz}= \cos(z)+i\sin(z)$$ which is not bijective, so just defining $\ln$ as the inverse of exponentiaton is no longer well defined, it only works like that for real numbers.
Thus to define a logarithm, to make it the inverse of exponentiation you must remove a ray from the origin, which is called choosing a branch of your logarithm. When you do this you will get a result which is consistent.
A: When considering complex numbers, the usual rules about logarithms and exponentials do not apply. For example, $e^{2\pi i} = 1 = e^0$, even though $2\pi i \neq 0$. Most notably in this case, the rule $a\ln{b} = \ln(b^a)$ is not true when $b$ is not a positive real or when $a$ is not real. So it's not correct to say that $2\ln(-1) = \ln{1}$.
The logarithm and exponential rules you were taught in intro algebra (e.g., $b\log(a) = \log(a^b)$ and that $a^b = a^c$ if and only if $b = c$) are true only for real numbers. Introducing complex numbers significantly complicates the issue: since $e^x$ is no longer a one-to-one function, it's questionable what exactly $\ln$ means. This is solved by something called branch cuts, but there isn't a way to define $\ln$ so that all of the usual log rules apply.
A: 
In THIS ANSWER, I showed that $\log(z^2)\ne 2\log(z)$ in general.

This might seem paradoxical given the relationship 
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag1$$
But $(1)$ is interpreted as a set equivalence.  It means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$.  And conversely, it means that the sum of any value of $\log(z_1)$ and any value of $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.

But, for the problem of interest, simply note that
$$2\log(-1)=2i(2k+1)\pi \ne \log((-1)^2)=0$$
for $k\in \mathbb{Z}$.
