Solve ${log}_{1 / 2} (x + 1) > {log}_{2} (2 - x)$ .

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Solution

${log}_{\frac{1}{2}} (x + 1) > {log}_{2} (2 - x)$
$\Rightarrow - {log}_{2} (x + 1) > {log}_{2} (2 - x)$
$\Rightarrow {log}_{2} (2 - x) + {log}_{2} (x + 1) < 0$
$\Rightarrow {log}_{2} (2 - x) (x + 1) < 0$
$\Rightarrow (2 - x) (x + 1) < 2^{\circ}$
$\Rightarrow - (x + 1) (x - 2) < 1$
$\Rightarrow (x + 1) (x - 2) > - 1$
$\Rightarrow x^{2} - x - 2 + 1 > 0$
$\Rightarrow x^{2} - x - 1 > 0$
$\Rightarrow [x - (\frac{1 + \sqrt{5}}{2})] [x - (\frac{1 - \sqrt{5}}{2})] > 0$
$\Rightarrow x \in (- \infty, \frac{1 - \sqrt{5}}{2}) & x \in (\frac{1 - \sqrt{5}}{2}, \infty)$
$\Rightarrow x \in (- \infty, \frac{1 - \sqrt{5}}{2}) \cup (\frac{1 + \sqrt{5}}{2}, \infty)$

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