Question

Solve the following inequality:
$\sqrt{x^{{\log }_2\sqrt{x}}}⩾2$

Answer 1

$\sqrt{x^{{\log }_2\sqrt{x}}}\ge 2x>0$

Squaring on both sides we get

$x^{{\log }_2\sqrt{x}}\ge 4$

For $xϵ(0,1)$ For $x>1$

${\log }_2\sqrt{x}\le \frac{2\log 2}{\log x}{\log }_2\sqrt{x}\ge \frac{2\log 2}{\left(\log x\right)}$

$\frac{1}{2}\frac{\log 2}{\log x}\le \frac{2\log 2}{\log x}\frac{\log x}{2\log 2}\ge \frac{2\log 2\left(\log x\right)}{{\left(\log x\right)}^2}$

$\left(\log x\right)\le \frac{4{\left(\log 2\right)}^2\left(\log x\right)}{\left(\log x\right)}\left(\log x\right)(\log x-2\log 2)(\log x+2\log 2)\ge 0$

${\left(\log x\right)}^3-\left(\log x\right){\left(4\log 2\right)}^2\le 0\log xϵ[-\log 4,0]\bigcup [\log 4,\infty ]$

$\left(\log x\right)(\log x-2\log 2)(\log x+2\log 2)\le 0$

$\log xϵ(-\infty ,-\log 4]\bigcup [0,\log 4]xϵ[\frac{1}{4},1]\bigcup [4,\infty )$

$xϵ[0,\frac{1}{4})\bigcup [1,4]$

Hence $xϵ[0,\frac{1}{4}]xϵ[4,\infty )$

Taking union we get $xϵ(0,\frac{1}{4}]\bigcup [4,\infty )$