How do you evaluate $log_144 (-12)$ ?

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Answer 1 · 2016-11-19T05:25:44

Though unreal, the answer in complex form is

$1/2 + i((4k+1)pi/2)/ln 12$ , k = 0, +-1, +-2, +-3, ... .

#log_144 (-12) is unreal. Yet, I evaluate it for you.

Use $i^2=-1$ .

$log_144(-12)$

$=ln (-12)/ln 144$

$=ln(i^2 12)/ln 12^2$

$=(2 ln i + ln 12)/(2 ln 12)$

$ln i/ln12+1/2$

Interestingly,

i = e^(i(4k+1)pi/2), k = 0, +-1, +-2, +-3, ...#

So, $ln i = ln (e(i(4k+1)pi/2))$

$=i(4k+1)pi/2$ .

So, the answer is

$1/2 + i((4k+1)pi/2)/ln 12$ , k = 0, +-1, +-2, +-3, ...#

How do you evaluate log_144 (-12)log_144 (-12)?