# 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?

• What do you mean by a logarithm of a negative value? What is $\log(-1)$ to you? You are accidentally treading into the realm of complex numbers when you are probably not ready to. The complex logarithm doesn't act quite the same as the real logarithm does. Dec 18, 2018 at 22:32
• If working with the strictly real logarithm which takes real numbers as inputs and gives real numbers as outputs, you should note that $\log(-1)$ doesn't exist since there is no corresponding real value $x$ such that $e^x = -1$ since $e^x$ is strictly positive for all real values of $x$. Dec 18, 2018 at 22:34
• $\log(-x)+\log(-x)=\log(x^2)$ is true, provided $x<0$. Using it for $x>0$ is a well-known fallacy that goes back to Euler and Bernoulli, who debated quite a lot about it. Dec 18, 2018 at 23:02
• Related
– A.Γ.
Dec 18, 2018 at 23:09

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.

• I suspect OP assumed that $-x$ would be a negative value. One of the lessons of this answer is that if we do not know for sure that $x$ is positive, we don't know that $-x$ is negative. Even if there were no logarithms in the question at all, we can't just assume $-x<0.$ Dec 18, 2018 at 23:32
• @DavidK I suppose one can't spell that out too explicitly. I was hoping I was making that completely clear. Dec 19, 2018 at 0:38
• I suppose my comment was a bit redundant. Hopefully nobody will miss the point after the edit. Dec 19, 2018 at 1:40

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.