If #log_12 27 = a#, then what is #log_6 16#?

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Answer 1 · 2016-11-27T13:41:07

A good problem

Given

#a=log_12 27=log_(12)3^3=3log_(12)3#
#=>a=3/log_3 12=3/log_3(3xx2^2)#

#=>a=3/(log_3 3+2log_3 2#

#=>a=3/(1+2log_3 2)#

#=>(1+2log_3 2)=3/a#

#=>log_3 2=1/2(3/a-1)=(3-a)/(2a)#

#=>log_2 3=(2a)/(3-a)#

Now #log_6 16=log_6 2^4=4log_6 2=4/log_2 6#

#=4/log_2 (2xx3)=4/(log_2 2+log_2 3#

#=4/(1+log_2 3)=4/(1+(2a)/(3-a)#

#=(4(3-a))/(3-a+2a)=(12-4a)/(3+a)#