# How do you solve 2log_5x = log_5 9?

Feb 24, 2016

First simplify using the rule $a \log n = \log {n}^{a}$

#### Explanation:

${\log}_{5} \left({x}^{2}\right) = {\log}_{5} \left(9\right)$

Put everything to one side of the equation.

$0 = {\log}_{5} \left(9\right) - {\log}_{5} \left({x}^{2}\right)$

Simplify using the rule ${\log}_{a} m - {\log}_{a} n = {\log}_{a} \left(\frac{m}{n}\right)$

$0 = {\log}_{5} \left(\frac{9}{x} ^ 2\right)$

Convert to exponential form.

${5}^{0} = \frac{9}{x} ^ 2$

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

${x}^{2} = 9$

$x = \pm 3$

However, only +3 works as a solution because the log of a negative number is non-defined.

The solution set is x = 3.

Hopefully this helps!