Question

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

Answer 1

`-6`.

Explanation:

`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`.