How do you evaluate ${log}_{2} ({log}_{9} 3) = {log}_{x} 6$ $log_2(log_9 3)= log_x6$ ?

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Answer 1 · 2016-04-21T20:40:06

$x = \frac{1}{6}$ $x=1/6$

${log}_{2} ({log}_{9} 3) = {log}_{x} 6$ $log_2(log_9 3)=log_x 6$

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

$log_2 (1/2)=log_x 6$

$log_2 1-log_2 2=log_x 6$

$log_2 1=0" ;" log_2 2=1$

$0-1=log_x 6$

$log_x 6=-1$

$x^-1=6$

$1/x=6$

$x=1/6$

How do you evaluate log_2(log_9 3)= log_x6log2(log93)=logx6log_2(log_9 3)= log_x6?