algebra precalculus - Why isn't $-2$ solution for $x$?

I came across an logarithm problem recently. I don't know why solution to this problem cannot be $-2$. Now, don't downvote now because you don't know why I'm asking this. I know that logarithms' domains must be greater than $0$, can't take a negative number. Just read the whole post and wait until I get to the point to see why I'm asking this "ridiculous question".

This is the problem:
$$\log_{10}x=1-\log_{10}(x-3)$$
You can solve it like this:
$$\log_{10}x=\log_{10}10-\log_{10}(x-3)$$
$$\log_{10}x=\log_{10}\frac{10}{x-3}$$

$$x=\frac{10}{x-3}$$
$$x(x-3)=10$$
$$x^2-3x-10=0$$
$$(x+2)(x-5)=0$$
$$x\stackrel{?}{=}-2,\,\,x\stackrel{?}{=}5$$
If you plug in $5$, it will work just fine. You will get when simplified $\log_{10}5=\log_{10}5$. But if you plug in $-2$, you get something like this:
$$\log_{10}-2=\log_{10}\frac{10}{-2-3}$$
$$\log_{10}-2=\log_{10}-2$$
I know that negative logarithms with any base don't exist, but I thought of something. There's a logarithm property which states:
$$\log_by=\log_bx \Rightarrow y=x$$

Using this property, we can say:
$$\log_{10}-2=\log_{10}-2 \Rightarrow -2=-2$$
And, now (I hope) everyone will agree that $-2$ is indeed equal to $-2$ (itself).

So why besides $5$, isn't $-2$ solution to this problem? Even if the property that I stated doesn't apply to this case, could I theoretically invent a new imaginary number unit of $v_b=\log_b(-1)$ like someone back in the days did with $i=\sqrt{-1}$ and state that $\log_{10}(-2)=v_{10}+\log_{10}2$ ?

Answer

Okay we've talked a lot about complex logarithms, let's try and solve the problem over the complex numbers. (I think that is what the OP is really interested in; it has been reiterated that the only solution over the reals is 5)

We have:

$\log_{10} (-2)$

The change of base formula is consistent for all complex numbers, of which 10 is one. ($0i+10$). But we must be careful about branch selection. In this exercise we will denote $L(x)$ as the principal branch of the natural complex logarithm.

Therefore:

$\log_{10} (-2)$ =$\frac{L(-2)}{L(10)}$

By the definition of the principle branch:

$\log_{10} (-2)$ =$\frac{\ln(2)+iπ}{\ln(10)}$

As @uqtredd1 has noted, this solution is not one that should be submitted in the context of the presented problem, but is completely extraneous. The complex logarithm is a multivalued function, which is why branch selection was so important. There are many more possible "answers" to what $\log_{10} (-2)$ is in the complex sense.