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

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Answer 1 · 2016-04-27T00:52:17

Use the log rule $log_an - log_am = log_a(n / m)$

$log_(1/3)((x^2 + 4x)/(x^3 - x)) = -1$

$(x^2 + 4x)/(x^3 - x) = (1/3)^-1$

$x^2 + 4x = 3(x^3 - x)$

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

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

$0 = x(3x^2 + x - 7)$

Solving the inner trinomial by the quadratic formula:

$x = (-1 +- sqrt(1^2 - (4 xx 3 xx -7)))/(2 xx 3)$

$x = (-1 +- sqrt(85))/6 and 0$

Hopefully this helps!

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