# How do you solve log_9 (3x+1) = log_3 (x) + log_3 (2)?

Feb 24, 2016

$x = 1$

#### Explanation:

Use the logarithmic identities

${\log}_{a} \left(b\right) = \frac{{\log}_{x} \left(b\right)}{{\log}_{x} \left(a\right)}$ for $x \in {\mathbb{R}}^{+}$ and

${\log}_{x} \left(a b\right) = {\log}_{x} \left(a\right) + {\log}_{x} \left(b\right)$

In this question, note that $x > 0$.

${\log}_{9} \left(3 x + 1\right) = {\log}_{3} \left(x\right) + {\log}_{3} \left(2\right)$

$= {\log}_{3} \frac{3 x + 1}{\log} _ 3 \left(9\right)$

$= {\log}_{3} \frac{3 x + 1}{2}$

${\log}_{3} \left(3 x + 1\right) = 2 {\log}_{3} \left(x\right) + 2 {\log}_{3} \left(2\right)$

$= {\log}_{3} \left(4 {x}^{2}\right)$

Since ${\log}_{3} \left(x\right)$ is one-to-one for $x \in {\mathbb{R}}^{+}$, we take inverse of ${\log}_{3} \left(x\right)$ for both sides

${3}^{{\log}_{3} \left(3 x + 1\right)} = {3}^{{\log}_{3} \left(4 {x}^{2}\right)}$

$3 x + 1 = 4 {x}^{2}$

Solve the quadratic the usual way.

$4 {x}^{2} - 3 x - 1 = 0$

$\left(4 x + 1\right) \left(x - 1\right) = 0$

$x = 1 \mathmr{and} x = - \frac{1}{4}$

Reject the negative answer as it is not in the domain.