Find the values of $x$ which satisfies the inequality ${log}_{3} (2 x^{2} + 6 x - 5) > 1$

x < - 4

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x > 1

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x > - 4

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x < 1

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Solution

The correct options are
A $x < - 4$
B $x > 1$
${log}_{3} (2 x^{2} + 6 x - 5) > 1$
or $2 x^{2} + 6 x - 5 > 3^{1}$
or $2 x^{2} + 6 x - 8 > 0$ or $x^{2} + 3 x - 4 > 0$
or $(x - 1) (x + 4) > 0 \Rightarrow x < - 4$ or $x > 1$

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