# How do you solve log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1?

Jun 11, 2016

$x = \frac{1}{6} \left(1 + \sqrt{85}\right)$

#### Explanation:

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

$\frac{{x}^{2} + 4 x}{{x}^{3} - x} = 3$ or $\frac{x + 4}{{x}^{2} - 1} = 3$

or

$3 {x}^{2} - 3 = x + 4$ or finally $3 {x}^{2} - x - 7 = 0$

solving for $x$ we get $\left\{x = \frac{1}{6} \left(1 - \sqrt{85}\right)\right\} , \left\{x = \frac{1}{6} \left(1 + \sqrt{85}\right)\right\}$ but we choose $x = \frac{1}{6} \left(1 + \sqrt{85}\right)$ because $x = \frac{1}{6} \left(1 - \sqrt{85}\right) < 0$ and makes $\left({x}^{3} - x\right) < 0$