Question

How do you use `log_2 3 = 1.5850` to approximate the value of `log_2 96`?

Answer 1

`log_2 96=6.5850`

Explanation:

`log_2 96`

= `log_2 (2xx2xx2xx2xx2xx3)`

= `log_2 (2^5xx3)`

= `5log_2 2+log_2 3`

= `5xx1+1.5850`

= `6.5850`