What's wrong with this logarithm calculation? We know that $\displaystyle\log_a(xy)=\log_ax+\log_ay$. 
Consider the following:

$$\displaystyle\ln(1)=\displaystyle\ln((-1)\times(-1))=\displaystyle\ln(-1)+\displaystyle\ln(-1)
$$

$\displaystyle\ln(1) $ is a completely valid statement, but I'm not sure if $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ is.
$\displaystyle\ln(-1) $ doesn't exist, but $\displaystyle\ln((-1)\times(-1)) $ does, and if I plug $\displaystyle\ln((-1)\times(-1)) $ into my calculator it gives a 0 as the answer (which is correct), but if I plug in $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ it says that there is a domain error (which there is). So my question is, what is wrong with the highlighted equation? 
 A: 
$$\displaystyle\log_a(xy)=\log_ax+\log_ay$$

The above identity holds only for $x,y>0$ and $a\not =1$, $a>0$. 
A: It is sort of true if we allow complex numbers to come in:
$$\log_a(1)=\frac{\ln(1)}{\ln(a)}=\frac{\pm2\pi ik}{\ln(a)}$$
$$\log_a(-1)=\frac{\ln(-1)}{\ln(a)}=\frac{\pm\pi i(2k-1)}{\ln(a)}$$
$$k=0,1,2,3,\dots$$
So it is sort of true that $\log(1)=2\log(-1)$, but this just depends on what is allowed.
$$2\log_a(-1)=\frac{\pm\pi i(4k-2)}{\ln(a)}$$
We are trying to make the two equal, so we must have:
$$2k=4n-2,n=0,1,2,3,\dots$$
We use $n$ at this stage because $\log_a(1)$ and $\log_a(-1)$ don't have to rely on the same constant, it could just as easily be $k=1$ and $n=1$ or $k=3$ and $n=2$.
In fact, now there are an infinite amount of solutions, many of which match up together.
However, all of the solutions to $2\log_a(-1)$ do not match up with all of the solutions in $\log_a(1)$, actually, it only matches up with half of the solutions.
So, like I said, it just depends on how you look at it.
A: Basically, once you go to taking the log of negative numbers, and therefore getting complex results, you are into the realm of multi-valued functions. Cf. $arcsin(\theta)$ etc.
Euler's identity says $exp(i\pi) = -1$ But also, for example, $exp(-i\pi) = -1$. So by taking logs of both sides, we get $log(-1) = (2n - 1)i\pi$ for an arbitrary integer value of n, and it is effectively by convention that we take $n = 0$ to get the principal value of the log function..
Then the effect of writing $log(1) = log(-1) + log(-1)$ simply amounts to:
$$0 = (2m - 1)i\pi + (2n - 1)i\pi$$which reduces to choosing m, n such that $m + n = 1$.
This is the same order of paradox as saying $\sqrt4 = 2$ and $\sqrt4 = -2$. By convention we take the principal (positive) value throughout; but there are circumstances where we have to break the convention to get a sensible result.
