Question #37b1c

1 Answer
Dec 9, 2017

Given #log_16 27=a#

#log_16 27=a#

#=>log_16 3^3=a#

#=>3log_16 3=a#

#=>3/log _3 16=a#

#=>3/log _3 2^4=a#

#=>3/(4log _3 2)=a#

#=>log _2 3=(4a)/3#

#=>log _2 3+log_2 2=(4a)/3+log_2 2#

#=>log _2 (3* 2)=(4a)/3+1#

#=>log _2 6=(4a+3)/3#

#=>1/log _6 2=(4a+3)/3#

#=>4/(4log _6 2)=(4a+3)/3#

#=>4/(log _6 2^4)=(4a+3)/3#

#=>4/(log _6 16)=(4a+3)/3#

#=>log _6 16=12/(4a+3)#