Question

How do you solve `log_x 25=-0.5`?

Answer 1

`x = 1/625`

Explanation:

Using the property `x^(log_x(a)) = a` we get

`log_x(25) = -0.5`

`=> x^(log_x(25)) = x^(-0.5)`

`=> 25 = 1/sqrt(x)`

`=> sqrt(x) = 1/25`

`=> x = (1/25)^2 = 1/625`