How is $\ln(-1) = i\pi$? How do I derive: 
$\ln(-1)=i\pi$ and 
$\ln(-x)=\ln(x)+i\pi$  for $x>0$ and  $x \in\mathbb R$
Thanks for any and all help!

What you said is not correct unless you select a branch. By definition, $log_{\theta}(-1):=ln|-1|+iarg_{\theta} (-1)= 0+iarg_{\theta}(-1)$ So you need $arg_{\theta}(-1)= \pi$. This is true for only one value $\theta$ (so that the branch goes from $\theta$ to $\theta + 2\pi^{-}$). So it seems like the branch with $ -\pi <arg\leq \pi$ should do it.
EDIT
  What you ultimately need in a branch log is to have the relation $$e^{logz}=z  $$ , so that whatever argument $\theta$ you assign to z , you have $$ e^{i\theta}=cos(\theta)+isin(\theta)= i\pi$$. It then follows that $sin\theta= \pi$, so that $\theta =sin^{-1}(\pi)$
