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We know that $\displaystyle\log_a(xy)=\log_ax+\log_ay$.

Consider the following:

$$\displaystyle\ln(1)=\displaystyle\ln((-1)\times(-1))=\displaystyle\ln(-1)+\displaystyle\ln(-1) $$

$\displaystyle\ln(1) $ is a completely valid statement, but I'm not sure if $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ is.

$\displaystyle\ln(-1) $ doesn't exist, but $\displaystyle\ln((-1)\times(-1)) $ does, and if I plug $\displaystyle\ln((-1)\times(-1)) $ into my calculator it gives a 0 as the answer (which is correct), but if I plug in $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ it says that there is a domain error (which there is). So my question is, what is wrong with the highlighted equation?

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    $\begingroup$ $\log_a(xy)=\log_a(x)+\log_a(y)$ is only true when $x,y>0$, I think. $\endgroup$ – S.C.B. Feb 27 '16 at 11:27
  • $\begingroup$ I also think that $\ln (-1)= \pi i$, but I don't know enough math to know if this is true. $\endgroup$ – S.C.B. Feb 27 '16 at 11:31
  • $\begingroup$ If you google "ln(-1)" it says that it's equal to 3.14159265i, so I'd say you're right. I didn't know about this, cool! EDIT: math.stackexchange.com/questions/1019177/how-is-ln-1-i-pi $\endgroup$ – user265554 Feb 27 '16 at 11:35
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    $\begingroup$ @MXYMXY It mostly depends on how you introduce the complex exponential. If you define $e^z=\sum_{n\ge0}z^n/n!$, the relation $e^{\pi i}+1=0$ can be deduced. If you define $e^{x+yi}=e^x(\cos y+i\sin y)$, then $e^{\pi i}+1=0$ by definition. $\endgroup$ – egreg Feb 27 '16 at 13:15
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    $\begingroup$ @SimpleArt: true, but "identity" just means a statement that two things are equal (identical, hence identity). The identity can still be either a definition or a theorem (or neither), so calling it an identity doesn't get us off the hook of deciding whether or not it's a theorem :-) $\endgroup$ – Steve Jessop Feb 27 '16 at 15:31
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$$\displaystyle\log_a(xy)=\log_ax+\log_ay$$

The above identity holds only for $x,y>0$ and $a\not =1$, $a>0$.

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    $\begingroup$ Alright, well that settles it! $\endgroup$ – user265554 Feb 27 '16 at 11:29
  • $\begingroup$ @user265554 Yup you are right. $\endgroup$ – SchrodingersCat Feb 27 '16 at 11:30
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It is sort of true if we allow complex numbers to come in:

$$\log_a(1)=\frac{\ln(1)}{\ln(a)}=\frac{\pm2\pi ik}{\ln(a)}$$

$$\log_a(-1)=\frac{\ln(-1)}{\ln(a)}=\frac{\pm\pi i(2k-1)}{\ln(a)}$$

$$k=0,1,2,3,\dots$$

So it is sort of true that $\log(1)=2\log(-1)$, but this just depends on what is allowed.

$$2\log_a(-1)=\frac{\pm\pi i(4k-2)}{\ln(a)}$$

We are trying to make the two equal, so we must have:

$$2k=4n-2,n=0,1,2,3,\dots$$

We use $n$ at this stage because $\log_a(1)$ and $\log_a(-1)$ don't have to rely on the same constant, it could just as easily be $k=1$ and $n=1$ or $k=3$ and $n=2$.

In fact, now there are an infinite amount of solutions, many of which match up together.

However, all of the solutions to $2\log_a(-1)$ do not match up with all of the solutions in $\log_a(1)$, actually, it only matches up with half of the solutions.

So, like I said, it just depends on how you look at it.

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  • $\begingroup$ @RaymondManzoni Oh, yes, my bad. $\endgroup$ – Simply Beautiful Art Feb 27 '16 at 23:47
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Basically, once you go to taking the log of negative numbers, and therefore getting complex results, you are into the realm of multi-valued functions. Cf. $arcsin(\theta)$ etc. Euler's identity says $exp(i\pi) = -1$ But also, for example, $exp(-i\pi) = -1$. So by taking logs of both sides, we get $log(-1) = (2n - 1)i\pi$ for an arbitrary integer value of n, and it is effectively by convention that we take $n = 0$ to get the principal value of the log function..

Then the effect of writing $log(1) = log(-1) + log(-1)$ simply amounts to: $$0 = (2m - 1)i\pi + (2n - 1)i\pi$$which reduces to choosing m, n such that $m + n = 1$.

This is the same order of paradox as saying $\sqrt4 = 2$ and $\sqrt4 = -2$. By convention we take the principal (positive) value throughout; but there are circumstances where we have to break the convention to get a sensible result.

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