How do you evaluate $log_2 (1/64)$ ?

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Answer 1 · 2016-08-21T16:01:34

$-6$ .

$log_2(1/64)=log_2 1-log_2 64=0-log_2 (2^6)=-6log_2 2$

$=-6$ .

Or,

$log_2 (1/64)=x rArr 2^x=1/64=1/2^6=2^-6$ .

Since, $log$ function is #1-1, x=-6.

$:. log_2 (1/64)=-6$ .

How do you evaluate log_2 (1/64)log_2 (1/64)?