# How do you solve  log_3 x = log_3(x + 2) = log_3 2 + log_3 12?

Sep 6, 2015

It is not obvious what the equation should be.

#### Explanation:

It seem likely that the equation should be

${\log}_{3} x + {\log}_{3} \left(x + 2\right) = {\log}_{3} 2 + {\log}_{3} 12$

${\log}_{3} \left(x \left(x + 2\right)\right) = {\log}_{3} \left(2 \cdot 12\right)$ $\text{ }$ (property of logarithms)

$\left(x \left(x + 2\right)\right) = \left(2 \cdot 12\right)$ $\text{ }$ (logarithms are one-to-one)

${x}^{2} + 2 x = 24$

${x}^{2} + 2 x - 24 = 0$

$\left(x + 6\right) \left(x - 4\right) = 0$

$x = - 6$ is an extraneous solution. ${\log}_{3} \left(- 6\right)$ is not defined (in the real numbers)

$x = 4$ is the only solution.