Question

How do you evaluate the expression `log_2 (1/32)`?

Answer 1

`-5`

Explanation:

`\log_2(1/32)`

`=\log_2(1/2^{5})`

`=\log_2(2^{-5})`

`=-5\log_2(2)`

`=-5\cdot 1`

`=-5`