What is wrong with this reasoning?


We can obvisouly work out $\ln(4)$, but we can't with $\ln(-2)$. The reason I am asking is because I have a situation in another problem where I arrived to something similiar to $2\ln(-2)$ and I don't know if I should just write $\ln(4)$.

  • $\begingroup$ I think you could be interested in the complex logarithm function (it is a multifunction though) which should be defined for this. I think you'd have $ln(-2)=In|-2|+i\pi = In(2)+i\pi$ I think so that $2In(-2)=2In(2)+2\pi i$ $\endgroup$ – Euler_Salter Dec 30 '16 at 13:15
  • $\begingroup$ The ratio of the natural log of negative one to the square root of negative one happens to be $\pi$...not that that helps any. ;) $\endgroup$ – Wildcard Dec 30 '16 at 13:16
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    $\begingroup$ The rule $\ln(x^a)=a \ln x$ is only valid for $x>0$, so you can't apply it with $x=-2$. In that other problem where you arrived at $\ln(-2)$, you must have done something strange, but you'll have to tell us more about that before we can help you. $\endgroup$ – Hans Lundmark Dec 30 '16 at 13:17
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    $\begingroup$ The property of logs which you are using, namely $\log a^n=n\log a$ only holds for $a>0$, as you have discovered. $\endgroup$ – lulu Dec 30 '16 at 13:17
  • $\begingroup$ For complex number z in polar form: $z = r·e^{iθ}$ The complex logarithm: $\log z = \ln r + iθ$ Is defined for negative z. so we have $\ln z=\ln r+i\theta \to z=e^{r}+e^{i\theta} $ but $e^{i\theta} = \cos(\theta ) + i \sin( \theta)$ so if $z\in R$ then $e^{i\theta}=\pm1$ so for negative z (r is positive because it is Radius) we must take $e^{i\theta}=-1$ but $0<e^{r}$ so we haven't $\ln (-2)$ $\endgroup$ – W.R.P.S Dec 30 '16 at 14:04

This is actually a reason why mathematicians used to think that logarithms of negative numbers are equal to logarithms of positive numbers (it used to be somewhere on the Wikipedia, I'll try to find it). However, logarithms of negative numbers can indeed be defined, but they are not so simple.


This equality is true if we define $\ln(-2)$, and such is explained in this post. However, the inverse is not true:


This is because there just so happens to be the case that

$$e^{2\pi i}=1$$


$$2\ln(-2)=2\ln(2)+2k\pi i$$

for some whole number $k$.

Thus, we end up with

$$e^{2\ln(-2)}=e^{2\ln(2)+2k\pi i}=e^{2\ln(2)}e^{2k\pi i}=e^{2\ln(2)}$$

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    $\begingroup$ this guy's high school must be insane $\endgroup$ – Euler_Salter Dec 30 '16 at 13:18
  • $\begingroup$ @Euler_Salter Which guy? ;-) $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 13:19
  • $\begingroup$ This girl? To be honest I don't know the neutral appellative in English, so I just sticked to the past male appellative, but anyway good job $\endgroup$ – Euler_Salter Dec 30 '16 at 13:21

I am pretty sure you did something wrong with your calculation and here is the proof assume $\exists$ number $x<0$ so that $\ln(x)=y$ with $y\in\mathbb{R}$ then


$$x=e^y $$ this leads us that $e^y$ is negative which is obviously a contradiction to its definition as it is always postive .

  • $\begingroup$ You seem to miss the fact that it is $x=(e^y)^2$, so the end result is still positive. $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 13:32
  • $\begingroup$ What I meant is that you can't reach ln(-2) with any mathematical arguments . There must has been a mistake that you did while calculating. $\endgroup$ – doumham Dec 30 '16 at 13:45
  • $\begingroup$ You mean me in my answer? No, I didn't, you can reach $\ln(-2)$, but that requires analytic stuff. $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 13:46
  • $\begingroup$ why don't you show us the question if you are so sure ;) $\endgroup$ – doumham Dec 30 '16 at 13:48
  • $\begingroup$ What do you mean? Since it seems you aren't very familiar with these things, please see here $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 13:49

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