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!

I'm always a little surprised people answer these with Euler's equation $e^{\pi i} = -1$ and expect it to be accepted without question that $e^{\pi i} = -1$.
So...
We need to define what $e^{z}$ means if $z\in \mathbb C$.  We want the basic properties that $e^x; x \in \mathbb R$ have to apply and carry over when we extend the definition of $e^z$ to allow complex numbers.
So first of all we want $e^{z + w} = e^z e^w$ which would mean we need for $z = x + yi$ where $x,y$ are real, sothat $e^{x + yi} = e^xe^{yi}$ which means we just need to define $e^{yi}$ somehow.
So we need to define $e^{yi} =E(y)$ for some $E:\mathbb R \to \mathbb C$ with $E(0) = 1$ (so that $e^{x} = e^{x + 0i} = e^xE(0)$.)
And second of all, we need the derivative of $e^z$ to equal $e^z$.
Some comments about derivatives of complex functions:
We can rewrite a complex function $f:x _ yi \mapsto u_{x,y} + v_{x,y} i$ as $f(z= x+yi) = u(x,y) + v(x,y)i$ where $u(x,y)$ and $v(x,y)$ are two real valued functions.
The derivative of $f(z) = \lim_{h\to 0}\frac {f(z+h) - f(z)}{h}$
If $h$ is real this is:
$f'(z) = \lim_{h\to 0} \frac {u(x+h,y) -u(x,y) +v(x+h,y)i - v(x,y)i}{h} = $
$\frac{du(x,y)}{dx} + \frac {dv(x,y)}{dx}i$.
But if $h=ik$ and is purely imaginary we have
$f'(z) = \lim_{k\to 0} \frac {u(x,y+k) -u(x,y) +v(x,y+k)i - v(x,y)i}{ik} = $
$f'(z) = \lim_{k\to 0} \frac {-u(x,y+k)i +u(x,y) i-v(x,y+k) - v(x,y)}{k} = $
$\frac {dv(x,y)}{dy} - \frac {du(x,y)}{dy}i$.
(Bear in mind that $\frac 1i = -i$ and $\frac ii = 1$)
For $f(z)$ to be differentiable (which suddenly we change the name of and call it analytical) we need the very restrictive condition:
$\frac {du(x,y)}{dx} = \frac {dv(x,y)}{dy}$ and $\frac {dv(x,y)}{dx} = -\frac {du(x,y)}{dy}$.
If we define $e^{z = x+iy} = e^xE(y)= u(x,y) + iv(x,y)$ where $u(x,y) = e^xRe(E(y)) + e^xIm(E(y))$.  We can define $U(y)=Re(E(y))$ and $V(y)=Im(E(y))$ as real value functions where:
$u(x,y) = e^xU(y)$ and $v(x,y)=e^xV(y)$.
We need $\frac {du(x,y)}{dx} = \frac {de^x}{dx}U(y) + e^x\frac{dU(y)}{dx} = e^xU(y)$.
But also $\frac {du(x,y)}{dx}= \frac {dv(x,y)}{dy} = \frac {de^x}{dy}V(y) + e^x\frac {dV(y)}{dy} = e^xV'(y)$
While $\frac {dv(x,y)}{dx} = \frac {de^x}{dx}V(y) + e^x\frac{dV(y)}{dx} = e^xV(y)$.
But also $\frac {dv(x,y)}{dx} = -\frac {du(x,y)}{dy} = \frac {de^x}{dy}U(y) + e^x\frac {dU(y)}{dy} = e^xU'(y)$
So we have $U(y) = V'(y)$ and $V(y) = U'(y)$ and also (as $e^{x + 0i} =e^{x}(U(0) + V(0)i) = e^x$) we have $U(0) = 1$ and $V(0) = 0$.
$U(y) = \cos y$ and $V(y) = \sin y$ is one such pair of functions that satisfy those conditions.
I'll omit the details as to why thoes are the only pair of functions that satisfy those conditions.

So we define $e^{z = x+iy}=e^x(\cos y + \sin y i)$.

Hence: 

$e^{\pi i} = \cos \pi + \sin \pi i = -1$

Important caveat:  $e^z$ is no longer $1-1$ so the "function" $\ln z$ is multivalued.
As $e^{z + 2k\pi i} = e^z $ , $\ln w = \ln w + 2k\pi i$.  In other word $\ln w$ is a class of many complex numbers each differing from each other by a multiple of $2\pi i$ 
