Question

How do you calculate `log _2 (1/8)`?

Answer 1

-3

Explanation:

Let `log_2(1/8)=n` and we want to calculate the value of n.

`color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(log_b x=nhArrx=b^n)color(white)(a/a)|)))`

here b = 2 and x`=1/8rArr1/8=2^nrArrn=-3`

Since `2^-3=1/2^3=1/8`

`rArrlog_2(1/8)=-3`