Question

Solve the following inequality:
${\log }_{(x-\sqrt{2})}\frac{x+7}{x-2}⩽{\log }_{(x-\sqrt{2})}2x$

Answer 1

For $\log \frac{x+7}{x-2}\Rightarrow$ $x+7>0x>-7$ and $x-2>0x>2$

For $\log 2x\Rightarrow$ $2x>0x>0$

For $\log (x-\sqrt{2})\Rightarrow$ $x-\sqrt{2}>0x>\sqrt{2}$

Taking intersection, $x\in (2,\infty )$
For $x>1+\sqrt{2}$

$\log \frac{x+7}{x-2}\le \log 2x\frac{x+7}{x-2}\le 2xx+7\le 2x(x-2)2x^2-5x-7\ge 0x\in (-\infty ,-1]\cup [\frac{7}{2},\infty )$

For $x\in (2,1+\sqrt{2})$

$\log \frac{x+7}{x-2}\ge \log 2x\frac{x+7}{x-2}\ge 2xx+7\ge 2x(x-2)2x^2-5x-7\le 0x\in [-1,\frac{7}{2}]$

Hence, $x\in (2,1+\sqrt{2})$

Taking union, we get $x\in (2,1+\sqrt{2})\cup [\frac{7}{2},\infty )$