How do you solve ${log (x - 1)}^{2} = log (- 5 x - 1)$ $log (x-1)^2 = log (-5x-1)$ ?

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How do you solve ${log (x - 1)}^{2} = log (- 5 x - 1)$ $log (x-1)^2 = log (-5x-1)$ ?

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Answer 1 · 2015-09-02T16:20:14

$x = - 1$ $x = -1$ or $x = - 2$ $x = -2$

Explanation:

${log (x - 1)}^{2} = log (- 5 x - 1)$ $log (x-1)^2 = log (-5x-1)$
$x^{2} - 2 x + 1 = - 5 x - 1$ $x^2-2x+1 = -5x-1$
$x^{2} + 3 x + 2 = 0$ $x^2+3x+2 = 0$
$(x + 1) (x + 2) = 0$ $(x+1)(x+2) = 0$
$x = - 1$ $x = -1$ or $x = - 2$ $x = -2$

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How do you solve ${log (x - 1)}^{2} = log (- 5 x - 1)$ $log (x-1)^2 = log (-5x-1)$ ?

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How do you solve ${log (x - 1)}^{2} = log (- 5 x - 1)$ $log (x-1)^2 = log (-5x-1)$ ?