How do you evaluate #log_144 (1/12)#?

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Answer 1 · 2016-09-26T04:47:40

#-1/2#

Let #x=log_144(1/12)#

#:. 144^x=1/12#

Since #144=12^2#

#12^(2x)=1/12#

#12^(2x)=12^-1#

Equating exponents:

#2x=-1#

#x=-1/2#