Solve: ${log}_{3} (2 x^{2} + 6 x - 5) > 1$

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Solution

${log}_{3} (2 x^{2} + 6 x - 5) > 1$
$2 x^{2} + 6 x - 5 > 3$
$2 x^{2} + 6 x - 8 > 0$
$x^{2} + 3 x - 4 > 0$
$x^{2} + 4 x - x - 4 > 0$
$x (x + 4) - (x + 4) = 0$
$(x + 4) (x - 1) > 0$
$x \in (- \infty, - 4) \cup (1, \infty)$

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