Решите неравенствоlog3(x2+7x-5)>1

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$D=49+32=81; \sqrt{D}=9$

$x_{1/2}= \frac{-7 \pm9}{2}=1; -8$

Ответ: $x\in(-\infty; -8) (1;+ \infty)$

ОДЗ:

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$D=49+20=69; \sqrt{D}= \sqrt{69} \approx8,3$

$x_{1/2} \frac{-7\pm8.3}{2} \approx0.65; -7,65$