# How do you solve log_3(x+6)-log_9x=log_9 2?

Nov 2, 2016

$x = \emptyset$

#### Explanation:

Use the change of base formula, ${\log}_{a} \left(n\right) = \log \frac{n}{\log} a$, to start the solving process.

$\log \frac{x + 6}{\log} 3 - \log \frac{x}{\log} 9 = \log \frac{2}{\log} 9$

$\log \frac{x + 6}{\log} 3 - \log \frac{x}{\log} {3}^{2} = \log \frac{2}{\log} {3}^{2}$

$\log \frac{x + 6}{\log} 3 - \log \frac{x}{2 \log 3} = \log \frac{2}{2 \log 3}$

Put on a common denominator.

$\frac{2 \log \left(x + 6\right)}{2 \log 3} - \log \frac{x}{2 \log 3} = \log \frac{2}{2 \log 3}$

${\log}_{9} {\left(x + 6\right)}^{2} - {\log}_{9} x = {\log}_{9} 2$

We now use the property ${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \left(\frac{n}{m}\right)$ to solve.

${\log}_{9} \left({\left(x + 6\right)}^{2} / x\right) = {\log}_{9} 2$

$\frac{{x}^{2} + 12 x + 36}{x} = 2$

${x}^{2} + 12 x + 36 = 2 x$

${x}^{2} + 10 x + 36 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 10 \pm \sqrt{{10}^{2} - \left(4 \times 1 \times 36\right)}}{2 \left(1\right)}$

$x = \frac{- 10 \pm \sqrt{- 44}}{2}$

$x = \frac{- 10 \pm 2 \sqrt{11} i}{2}$

$x = - 5 \pm \sqrt{11} i$

This does not satisfy the original equation, and hence there are no solutions.

Hopefully this helps!