# Logs of negative numbers

$\log_{-2}(-8) = \frac{\log8+i\pi}{\log2+i\pi}$ (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 $\log_{-2}(-8)$?

• Try n = 1, m = 0. – user361424 Dec 16 '16 at 21:29
• thank you so much. totally brain dead right now. didnt see the log8 as 3log2 at all :) – Somesh Thakur Dec 16 '16 at 21:52

## 1 Answer

$$n = 1$$ $$m = 0$$ $$\frac{\log8+ i(2n+1)\pi}{\log2+ i(2m+1)\pi} = \frac{3\log 2 + 3\pi i}{\log 2 + \pi i} = 3$$

• This deserves my +1, I couldn't see this so easily. – Simply Beautiful Art Dec 16 '16 at 21:33
• 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 :) – Somesh Thakur Dec 16 '16 at 21:50