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?