Question

Solve the following inequality:
${\log }_{4/3}(\sqrt{x+3}-\sqrt{x})+{\log }_{4/9}\left(\frac{2}{3}\right)⩾0$

Answer 1

$\frac{\log (\sqrt{x+3}-\sqrt{x})}{2\log 2-\log 3}+\frac{\log 2-\log 3}{2(\log 2-\log 3)}\ge 0\frac{\log (\sqrt{x+3}-\sqrt{x})}{2\log 2-\log 3}+\frac{1}{2}\ge 0\log (\sqrt{x+3}-\sqrt{x})\ge -\left(\frac{2\log 2-\log 3}{2}\right)\log (\sqrt{x+3}-\sqrt{x})\ge -\log \left(\frac{2}{\sqrt{3}}\right)\sqrt{x+3}-\sqrt{x}\ge \frac{\sqrt{3}}{2}\sqrt{x+3}\ge \sqrt{x}+\frac{\sqrt{3}}{2}x+3\ge x+\frac{3}{4}+\sqrt{3x}\frac{9}{4}\ge \sqrt{3x}{\left(\frac{9}{4}\right)}^2\ge 3x\frac{27}{16}\ge x$

Also,

$\sqrt{x+3}-\sqrt{x}>0\sqrt{x+3}>\sqrt{x}x+3>x3>0x\in R$

And

$x+3>0x>-3x\in (-3,\infty )$ $x>0x\in (0,\infty )$

Taking intersection, we get $x\in (0,\frac{27}{16})$