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$\log_{-2}(-8) = \frac{\log8+i\pi}{\log2+i\pi}$ (which is definitely not 3)

But what if we allowed all values (not just the principal value) of $\log(-1)$?

i.e, $\log(-1) = i(2n+1)\pi$ (n is an integer)

$\Rightarrow \log_{-2}(-8) = \frac{\log8+i(2n+1)\pi}{\log2+i(2m+1)\pi}$ (for integers n and m)

But the right hand side of the above expression is not 3 for any value of n and m..

Shouldn't 3 be one solution to $\log_{-2}(-8)$?

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    $\begingroup$ Try n = 1, m = 0. $\endgroup$ – user361424 Dec 16 '16 at 21:29
  • $\begingroup$ thank you so much. totally brain dead right now. didnt see the log8 as 3log2 at all :) $\endgroup$ – Somesh Thakur Dec 16 '16 at 21:52
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$$n = 1$$ $$m = 0$$ $$\frac{\log8+ i(2n+1)\pi}{\log2+ i(2m+1)\pi} = \frac{3\log 2 + 3\pi i}{\log 2 + \pi i} = 3$$

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  • $\begingroup$ This deserves my +1, I couldn't see this so easily. $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 21:33
  • $\begingroup$ Oh thank you so much.. Im a total idiot.. didnt see i can simplify the log8 as well.. was thinking of it as immovable constant term... thanks tons :) $\endgroup$ – Somesh Thakur Dec 16 '16 at 21:50

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