does ln(-1)=0? What did I do wrong in my proof? I know that $\ln(-1)$, or the logarithm, is not defined for negative numbers. But I came up with a short proof that leads to a contradiction. 
$\ln(1)=0$
$\ln(-1*-1)=0$
Since everything is real, by the logarithm property, this equals to
$\ln(-1)+\ln(-1)=0$ 
$ 2\ln(-1)=0$
Use whatever method you want, but basically $\ln(-1)$ would have to be $0$. To look at it from a different perspective, use the complex definition of the logarithm so that $\ln(-1)=\pi i$.
$2\ln(-1)=0$
$2\pi i=0$, which is not true (obviously) 
 A: If you only work with real numbers, then the equality $\ln\bigl((-1)^2\bigr)=2\ln(-1)$ makes no sense, since $\ln(-1)$ is undefined. And if you are willing to work with complex numbers, then you have a problem: there is not function $\ln\colon\mathbb{C}\setminus\{0\}\longrightarrow\mathbb C$ such that$$(\forall z,w\in\mathbb{C}\setminus\{0\}):\ln(zw)=\ln(z)+\ln(w).$$
A: The principal branch of the logarithm is defined as the function $\log:\Bbb C\setminus\{0\}\to \Bbb C$ $$\log(z):=\text{the complex number }w\text{ such that }e^w=x\text{ and }\Im w\in(-\pi,\pi]$$
While from the identity $e^{a+b}=e^ae^b$ (which holds for all $a,b\in\Bbb C$) it follows that $u=\log x+\log y$ is a complex number such that $e^u=xy$, in general there is no guarantee that $-\pi<\Im u\le \pi$, in which case $\log x+\log y\ne \log(xy)$. Such is the case for $x=y=-1$, as you've observed.
For completeness, the following identity holds in terms of the usual real functions $\ln$ and $\arccos$,: $$\log(z)=\begin{cases}\ln\lvert z\rvert+i\arccos\frac{\Re z}{\lvert z\rvert}&\text{if }\Im z\ge 0\\ \ln\lvert z\rvert-i\arccos\frac{\Re z}{\lvert z\rvert}&\text{if }\text{if }\Im z< 0\end{cases}$$ which basically relates the fact that $\log(xy)\ne \log x+\log y$ to the well-know fact that $\arccos\cos a\ne  a$.
A: You put your finger into a question that caused much headache at the beginning of the development of complex anysis,
 evento the most celebrated mathematicians.
A.I. Markuševič in his "Theory of analytic functions" report that it is said that J. Bernoulli was sustaining that
$$
{\rm Ln}( - z) = {\rm Ln}(z)
$$
by the fact that 
$$
\eqalign{
  & {\rm 1)}\;{\rm Ln}\left( {\left( { - z} \right)^2 } \right) = {\rm Ln}\left( {\left( z \right)^2 } \right)\quad 2{\rm )}\;{\rm Ln}\left( { - z} \right) + {\rm Ln}\left( { - z} \right)
 = {\rm Ln}\left( z \right) + {\rm Ln}\left( z \right)  \cr 
  & 3{\rm )}\;2{\rm Ln}\left( { - z} \right) = 2{\rm Ln}\left( z \right)\quad 4{\rm )}\,{\rm Ln}\left( { - z} \right) = {\rm Ln}\left( z \right) \cr} 
$$
Then the same Markuševič  is explaining that the fault is in the passage from  2), which is right, to 3) which is wrong.
Here we are speaking of the multivalued function: 
$$
{\rm Ln}\left( z \right) = \ln \left| z \right| + i\arg \left( z \right) + i\,2k\pi \quad \left| {\;k \in Z} \right.
$$
Then while
$$
\eqalign{
  & {\rm Ln}\left( z \right) + {\rm Ln}\left( z \right) = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,2\left( {k + j} \right)\pi  =   \cr 
  &  = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,2n\pi  = 2{\rm Ln}\left( z \right)\quad \left| {\;k,j,n \in Z} \right. \cr} 
$$
for ${\rm Ln}\left( { - z} \right)$ instead it is
$$
\left\{ \matrix{
  {\rm Ln}\left( { - z} \right) + {\rm Ln}\left( { - z} \right) = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,2\left( {k + 1} \right)\pi  + i\,2\left( {j + 1} \right)\pi  =  \hfill \cr 
   = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,2\left( {k + j + 2} \right)\pi \quad \left| {\;k,j \in Z} \right. \hfill \cr 
  2{\rm Ln}\left( { - z} \right) = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,4\left( {n + 1} \right)\pi  =  \hfill \cr 
   = 2\ln \left| z \right| + 2i\arg \left( z \right) + i\,2\left( {2n + 2} \right)\pi \quad \left| {\;n \in Z} \right. \hfill \cr}  \right.
$$
and since the respective sets are different
$$
\left\{ {k + j + 2} \right\} \ne \left\{ {2n + 2} \right\}
$$
we get
$$
{\rm Ln}\left( { - z} \right) + {\rm Ln}\left( { - z} \right) \ne 2{\rm Ln}\left( { - z} \right)
$$
