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

Apr 27, 2016

#### Answer:

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

#### Explanation:

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

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

${x}^{2} + 4 x = 3 \left({x}^{3} - x\right)$

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

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

$0 = x \left(3 {x}^{2} + x - 7\right)$

Solving the inner trinomial by the quadratic formula:

$x = \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \times 3 \times - 7\right)}}{2 \times 3}$

$x = \frac{- 1 \pm \sqrt{85}}{6} \mathmr{and} 0$

Hopefully this helps!