$\ln(-1) - \ln(-2)$ is it definable or have answer?
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?
$\endgroup$ 14 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$$$\ln(-1) = i\pi$$ $$\ln(-2) = \ln(2)+i\pi$$ $$\ln(-1)-\ln(-2) = -\ln(2) = \ln(1/2)$$
note Euler's equation: $0 = e^{i\pi}+1$ and take the natural logarithm to get the first line.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ $\begingroup$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 $z_1,z_2\in \mathbb R^{*-}$
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