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, ...#