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Higher Order Derivatives

Solve the fol...

Question

Solve the following inequality:
${log}_{2}^{2} \frac{4 x - 3}{4 - 3 x} > - \frac{1}{2}$

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Solution

Since there is square, ${log}_{2}^{2} (\frac{4 x - 3}{4 - 3 x})$ is always $> \frac{- 1}{2}$ in its domain

$\frac{4 x - 3}{4 - 3 x} > 0 (4 x - 3) (4 - 3 x) > 0 (4 x - 3) (3 x - 4) < 0 x \in (\frac{3}{4}, \frac{4}{3})$ $4 x - 3 > 0 4 x > 3 x > \frac{3}{4}$ $4 - 3 x > 0 4 > 3 x \frac{4}{3} > x$
Taking intersection, we get $x \in (\frac{3}{4}, \frac{4}{3})$

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