Question

How do you solve `log_x 9 = -2`?

Answer 1

You must change to exponential form: `log_an = x -> a^x = n`

Explanation:

`x^-2 = 9`

Make the exponent positive by using the rule `x^-n = 1/x^n`

`1/x^2 = 9`

`1 = 9x^2`

`1/9 = x^2`

`sqrt(1/9) = x`

`1/3 = x` (a negative answer is not possible since the log of a negative number is undefined)

Practice exercises:

1. Solve for x in `log_(2x)4 = -4`

Challenge problem

What is the value of x in `log_x4 = x`?