How do you evaluate $log_64 2$ ?

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Answer 1 · 2018-04-06T07:07:44

$log_64 2=1/6$

Let $log_64 2=t$

then $64^t=2$

or $(2^6)^t=2^1$

or $2^(6t)=2^1$

i.e. $6t=1$

and $t=1/6$

Hence $log_64 2=1/6$

Answer 2 · 2018-04-06T07:11:44

$log_64 2 =1/6$

$log_64 2=x$

$64^x=2$

$64^(1/2)=8$ (square root)

$8^(1/3)=2$ (cube root)

$64^((1/2)*(1/3))=2$

$64^(1/6)=2$

$log_64 2 =1/6$

How do you evaluate log_64 2log_64 2?