# How do you solve log_x 9 = -2?

Mar 7, 2016

You must change to exponential form: ${\log}_{a} n = x \to {a}^{x} = n$

#### Explanation:

${x}^{-} 2 = 9$

Make the exponent positive by using the rule ${x}^{-} n = \frac{1}{x} ^ n$

$\frac{1}{x} ^ 2 = 9$

$1 = 9 {x}^{2}$

$\frac{1}{9} = {x}^{2}$

$\sqrt{\frac{1}{9}} = x$

$\frac{1}{3} = x$ (a negative answer is not possible since the log of a negative number is undefined)

Practice exercises:

1. Solve for x in ${\log}_{2 x} 4 = - 4$

Challenge problem

What is the value of x in ${\log}_{x} 4 = x$?