Solve,
${log}_{1 / 5} (2 x^{2} + 5 x + 1) < 0$

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Solution

${log}_{5} (2 x^{2} + 5 x + 1) > 0$
$2 x^{2} + 5 x + 1 > 1$
$2 x^{2} + 5 x > 0$
$x (2 x + 5) > 0$
$x ε (- \infty, - \frac{5}{2}) \cup (0, \infty)$

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({log}_{1 / 2} x)^{2} - {log}_{1 / 2} x^{2} > ({log}_{1 / 2} 3)^{2} - 1

3 x^{3} + x^{2} - 2 = 0

2 x^{2} + 5 x + 5 = 0

2 x^{2} - 5 x = 0

- 2 x^{2} + 5 x - 2 = 0

Solve graphically.

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