How do you solve ${log}_{\frac{1}{3}} (x^{2} + 4 x) - {log}_{\frac{1}{3}} (x^{3} - x) = - 1$ $log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1$ ?

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How do you solve ${log}_{\frac{1}{3}} (x^{2} + 4 x) - {log}_{\frac{1}{3}} (x^{3} - x) = - 1$ $log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1$ ?

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Answer 1 · 2016-06-11T16:20:07

$x = \frac{1}{6} (1 + \sqrt{85})$ $x = 1/6 (1 + sqrt[85])$

Explanation:

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

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

or

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

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

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How do you solve ${log}_{\frac{1}{3}} (x^{2} + 4 x) - {log}_{\frac{1}{3}} (x^{3} - x) = - 1$ $log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1$ ?

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How do you solve log13(x2+4x)−log13(x3−x)=−1log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1?

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How do you solve ${log}_{\frac{1}{3}} (x^{2} + 4 x) - {log}_{\frac{1}{3}} (x^{3} - x) = - 1$ $log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1$ ?