Why do two logs with negative values not multiply to make a positive log? I was pondering a question while exploring logarithms, and logarithmic functions.
$$ \log(-x) + \log(-x) = \log x^2 $$
Why is this considered incorrect? Seeing as:
$$ \log(3) + \log(3) = \log3^2 = \log9 $$
What makes negatives invalid?
 A: It doesn't matter if $-x$  has a minus sign in front of it or not.  
What matters is whether $-x$ is positive of not.  If $\log (-x)$ exists then $-x$ is a a positive number and $x$ is a negative number.  
Just because it has a minus sign in front of it does not make $-x$ a negative number.  If $x < 0$ then $-x > 0$ and $-x$ is a positive number.
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There's nothing wrong with that.
If $\log (-x)$ exists then $-x > 0$ and $x < 0$ and indeed:
$\log(-x) + \log(-x) = \log ((-x)^2) = \log (x^2) = 2\log(|x|) = 2\log (-x)$.
There's absolutely absolutely nothing wrong with that at all.
Note:  There IS something wrong with saying $\log x^2 = 2\log x$ if you don't know for dead given FACT that $x > 0$.  If it is possible that $x < 0$ then  we can ONLY conclude $\log x^2 = 2\log |x|$.
But either $x > 0$ and $\log x$ exists and $\log (-x)$ does NOT.
Or 
$x < 0$ and $\log -x$ exists and $\log x$ does NOT.
Or $x = 0$ and NEITHER $\log x$ nor $\log (-x)$ exist.
Because only logs of positive numbers exist, we frequently assume, are indicate it is a given fact that $x > 0$, when we take $\log x$.  We could just as easily assume and take it as a given fact that $x < 0$ but there's no point or reason to do that.  So we don't.
A: The function log$(x)$ it's only defined for positive values of $x \in \mathbb{R}$ so it if $x$ is positive then log$(-x)$ does not exist. If $x$ is negative, you have
log$(-x)$ + log$(-x)$ = log$((-x)^2)$ = log$(x^2)$ 
and then the equation holds.
