Question

How do you solve for `log_x (1/64) = -3`?

Answer 1

`x=4`

Explanation:

`log_x(1/64)=-3`

`x^(log_x(1/64))=x^-3`

`1/64=x^-3`

`(1/64)^(-1/3)=(x^-3)^(-1/3)`

`(64)^(1/3)=x`

`4=x`