What's wrong with this logarithm calculation?

$$\displaystyle\log_a(xy)=\log_ax+\log_ay$$

The above identity holds only for $x,y>0$ and $a\not =1$, $a>0$.

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.