Question

Solve the following inequality:
$\frac{\sqrt{x-5}}{{\log }_{\sqrt{2}}(x-4)-1}⩾0.$

Answer 1

$\frac{\sqrt{x-5}}{{log}_{\sqrt{2}}(x-4)-1}\ge 0\Rightarrow \frac{\sqrt{x-5}}{{log}_2({x-4)}^2-1}\ge 0$

$\sqrt{x-5}$ always $\ge 0$ if $x\ge 5\Rightarrow x\in (5,\infty )$

Therefore, ${log}_2({x-4)}^2-1<0\Rightarrow {(x-4)}^2<2\Rightarrow x^2-8x+16<2\Rightarrow x^2-8x+14<0$

Roots $=\frac{8\pm \sqrt{64-56}}{2}=4\pm \sqrt{2}x\in (4-\sqrt{2},4+\sqrt{2})$

But $x\ge 5$

Therefore, $x\in (5,4+\sqrt{2}).$