Question

How do you solve `-5 = log_2 x`?

Answer 1

`x=1/32`

Explanation:

`log_2 x=-5`

Using the rule `log_a M=N rArr M=a^N`, you can rewrite `log_2x=-5` as `x=2^-5`

Using the indices rule `a^-m=1/a^m`,

`x=1/2^5`

`x=1/32`