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Integration by Partial Fractions

Solve the fol...

Question

Solve the following inequality:
${log}_{5} \sqrt{3 x + 4} \cdot {log}_{x} 5 > 1$

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Solution

$\frac{1}{2} \frac{log (3 x + 4)}{log 5} \frac{log 5}{log x} > 1 \frac{log (3 x + 4)}{log x} > 2 F o r x > 1 log (3 x + 4) > 2 log x 3 x + 4 > x^{2} x^{2} - 3 x - 4 < 0 x^{2} - 4 x + x - 4 < 0 (x + 1) (x - 4) < 0 x \in (- 1, 4) H e n c e x \in (- 1, 4) F o r x \in (0, 1) log (3 x + 4) < 2 log x 3 x + 4 < x^{2} x^{2} - 3 x - 4 > 0 x^{2} - 4 x + x - 4 > 0 (x + 1) (x - 4) > 0 x \in (- \infty, - 1) \cup (4, \infty) H e n c e x \in ϕ$
Taking union, we get $x \in (1, 4)$

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