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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?

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    $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Dec 18, 2018 at 22:32
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    $\begingroup$ 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 $x$. $\endgroup$
    – JMoravitz
    Dec 18, 2018 at 22:34
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    $\begingroup$ $\log(-x)+\log(-x)=\log(x^2)$ 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. $\endgroup$
    – egreg
    Dec 18, 2018 at 23:02
  • $\begingroup$ Related $\endgroup$
    – A.Γ.
    Dec 18, 2018 at 23:09

2 Answers 2

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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.

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  • $\begingroup$ 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 $-x<0.$ $\endgroup$
    – David K
    Dec 18, 2018 at 23:32
  • $\begingroup$ @DavidK I suppose one can't spell that out too explicitly. I was hoping I was making that completely clear. $\endgroup$
    – fleablood
    Dec 19, 2018 at 0:38
  • $\begingroup$ I suppose my comment was a bit redundant. Hopefully nobody will miss the point after the edit. $\endgroup$
    – David K
    Dec 19, 2018 at 1:40
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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.

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