Question

How do you evaluate `log_x(1/36)= -(2/3)`?

Answer 1

`x=216`

Explanation:

`x^(-2/3) =1/36`
`x^(-2/3) =6^-2`
`(x^(-2/3))^(-3/2) =(6^-2)^(-3/2)`
`x=6^3=216`