Solve the following inequality:
${log}_{a} (x - 1) + {log}_{a} x > 2$

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Solution

${log}_{a} (x - 1) + {log}_{a} x > 2$
${log}_{a} (x^{2} - x) > 2 a \neq 0$
If $a > 1$
${log}_{a} (x^{2} - x) > 2$
$(x^{2} - x) > a^{2}$
$x^{2} - x - a^{2} > 0$
${(x - \frac{1}{2})}^{2} - (a^{2} + \frac{1}{4}) > 0$
$(x - \frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}}) (x - \frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}})$
$x ϵ (- \infty, \frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}}) ⋃ (\frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}}, \infty)$
Hence, $x (\frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}}, \infty)$
If $a < 1$
${log}_{a} (x^{2} - x) < 2$
$(x^{2} - x) < a^{2}$
$x^{2} - x - a^{2} < 0$
${(x - \frac{1}{2})}^{2} - {(a^{2} + \frac{1}{4})}^{2} < 0$
$(x - \frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}}) (x - \frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}}) < 0$
$x ϵ (\frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}}, \frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}})$
Hence $x (\frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}}, \infty)$
Taking union we get $x ϵ (\frac{1}{2} - \sqrt{a^{2} + \frac{1}{4}}, \infty) ⋃ (\frac{1}{2} + \sqrt{a^{2} + \frac{1}{4}}, \infty)$

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