How do you evaluate #log_2(log_9 3)= log_x6#?

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Answer 1 · 2016-04-21T20:40:06

#x=1/6#

#log_2(log_9 3)=log_x 6#

#log_9 3=(log_3 3)/(log_3 9)=(cancel(log_3 3))/(2cancel(log_3 3))=1/2#

#log_2 (1/2)=log_x 6#

#log_2 1-log_2 2=log_x 6#

#log_2 1=0" ;" log_2 2=1#

#0-1=log_x 6#

#log_x 6=-1#

#x^-1=6#

#1/x=6#

#x=1/6#