Why do two logs with negative values not multiply to make a positive log?

Question

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?

Answer 1

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.

=====

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.

Answer 2

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.