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As the title says I type in google and the number say -0.693...

Is it equal to ln(1/2)?

Am I misconcept anything?

Sujaan Kunalan
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IndyZa
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  • It is right to simply it as ln(1/2) according to the rules of logarithm but if you try to solve the question as it is then, ln of negative numbers is not defined. I don't know how it can be both. – Rajath Radhakrishnan Sep 13 '13 at 16:19

4 Answers4

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Although a couple of answers here contain the gist of the matter, they are not rigorous enough.

The real logarithm function is not defined for negative reals, so we have to use the complex logarithm.

However, even the complex logarithm is not defined on the whole complex plane. Nonethless, various branches of logarithm may be defined which omit a ray through origin. Usually, the domain of complex logarithm is taken to be the complex plane minus the negative real axis, but you can remove any ray (positive real axis, for example). Thus, taking a branch of logarithm that does not omit the negative real axis, we obtain (using the principal logarithm)

$$ \log(z) = \ln(|z|) + i\arg(z) $$

and since $\arg(-1) = \arg(-2) = \pi$, we have that

$$ \log(-1) - \log(-2) = \ln(1) - \ln(2) = -\ln(2).$$

which is what google gave you.

Vishal Gupta
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$$\ln(-1) = i\pi$$ $$\ln(-2) = \ln(2)+i\pi$$ $$\ln(-1)-\ln(-2) = -\ln(2) = \ln(1/2)$$

note Euler's equation: $0 = e^{i\pi}+1$ and take the natural logarithm to get the first line.

user1729
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kaine
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    Strictly speaking, you need to work with a branch of the logarithm function, as it cannot be continuously defined on the whole complex plane (punctured at $0$). I think as long as the branch cut (= complement of a maximal open domain on which the logarithm is continuously defined) does not intersect the line segment with endpoints at $-1$ and $-2$, this equation can be made rigorous. Otherwise you're in trouble. – user43208 Sep 13 '13 at 16:37
  • It is not safe to use complex logarithm. Since $e^{2i\pi}=1$, we have $e^{3i\pi }=e^{i\pi} = -1$, so logarithm of $-1$ may be $i\pi$ and $3i\pi$ as well. But with appropriate definitions everything is OK. I don't get why someone downvoted it. – savick01 Sep 13 '13 at 16:37
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    @user43208 I think it is enough to assume that $-1$ and $-2$ can be connected by a path not "going around" the origin (trivial loop in $S^1$ after projecting punctured plane onto the circle). Anyway, we can define logarithm on the plane without a halfline starting at the origin, so there's no problem. – savick01 Sep 13 '13 at 16:44
  • @svicki01 Oh yes, I think you're right; my condition is stronger than need be. Thanks! – user43208 Sep 13 '13 at 17:08
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The natural logarithmic function is only defined for $x>0$ for $x\in\mathbb{R}$. So your statement is only true if you consider complex numbers. In the case of real numbers, both $\ln(-1)$ and $\ln(-2)$ are undefined.

Shane O Rourke
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Sujaan Kunalan
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Hint:

$$\ln z=\ln \left | z \right |+i\pi:z\in \mathbb R^{*-}$$

so $$\ln z_1-\ln z_2=\ln \left | z_1 \right |+i\pi -\ln \left | z_2 \right |-i\pi=\ln \left | z_1 \right |-\ln \left | z_2 \right |$$

for $z_1,z_2\in \mathbb R^{*-}$

mnsh
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