$\log(a/b) = \log(a) - \log(b)$;
Is $\log(0/x) = -\log(x)$?
I watched a video claiming $\log(1/x) = -\log(x)$, which I get because $1/x = x^{-1}$ and $\log(x^y) = y(\log(x))$ but $\log(1)$... I get it, $\log(1) = 0$.
Back to my question $\log(0/x)$?
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3It is undefined. – Henricus V. Apr 01 '16 at 04:15
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... beacause $0/x=0$ (if at least $x\ne 0$) – Hagen von Eitzen Apr 01 '16 at 04:17
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isnt 0/x undefined? – Tobi Apr 01 '16 at 04:19
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2@James No. The symbol $\dfrac0 x$ can be defined as the only real number $r$ such that $rx=0$. What number with $r$ be? – Git Gud Apr 01 '16 at 04:21
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i know x/infinity = 0 therefore x/0 = infinity therefore 0/x = 1/ infinity = r ??? – Tobi Apr 01 '16 at 04:33
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What power of a positive real number equals zero? – colormegone Apr 01 '16 at 05:00
2 Answers
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$\log\left(\frac{0}{x}\right) = \log 0$ is undefined, because $\log 0$ is itself undefined, because for any base $b\ne0$, there is no number $y$ such that $b^y=0$. (and you can't generally use $0$ as a base for the logarithm)
$\log\left(\frac{1}{x}\right) = -\log x$, because $\log 1=0$.
Dan Uznanski
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No, $\log\frac0x$ is not $-\log x$. $$\log\frac0x=\log0-\log x$$ But remember, $\log0$ doesn't exist. At least not as real solution. EDIT: Well, $\log0$ is not imaginary solution either.
The other one with $\log\frac1x$ is correct though. There are actually more than one proof for it: $$\log\frac1x=\log x^{-1}=-1\log x=-\log x$$ $$\log\frac1x=\log1-\log x=0-\log x=-\log x$$
KKZiomek
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