Question Number 4614 by love math last updated on 14/Feb/16
$$\sqrt{{x}^{{log}_{\mathrm{2}} \sqrt{{x}}} }>\mathrm{2} \\ $$
Commented by Yozzii last updated on 14/Feb/16
$${Squaring}\:{both}\:{sides}\:{of}\:{the}\:{given}\: \\ $$$${inequality}\:{yields}\:{the}\:{following}. \\ $$$${x}^{{log}_{\mathrm{2}} \sqrt{{x}}} >\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\ast\right) \\ $$$${Taking}\:{logs}\:{to}\:{base}\:\mathrm{2}\:{on}\:{both}\:{sides} \\ $$$${of}\:\left(\ast\right)\:{we}\:{obtain}\: \\ $$$${log}_{\mathrm{2}} {x}^{{log}_{\mathrm{2}} \sqrt{{x}}} >{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{2}} \\ $$$${By}\:{the}\:{power}\:{rule}\:{we}\:{have}\: \\ $$$$\left({log}_{\mathrm{2}} \sqrt{{x}}\right)\left({log}_{\mathrm{2}} {x}\right)>\mathrm{2}{log}_{\mathrm{2}} \mathrm{2}=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({log}_{\mathrm{2}} {x}\right)\left({log}_{\mathrm{2}} {x}\right)>\mathrm{2} \\ $$$$\left({log}_{\mathrm{2}} {x}\right)^{\mathrm{2}} >\mathrm{4} \\ $$$$\left({log}_{\mathrm{2}} {x}−\mathrm{2}\right)\left({log}_{\mathrm{2}} {x}+\mathrm{2}\right)>\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right){log}_{\mathrm{2}} {x}−\mathrm{2}>\mathrm{0}\:{and}\:{log}_{\mathrm{2}} {x}+\mathrm{2}>\mathrm{0} \\ $$$$\Rightarrow\:{x}>\mathrm{4}\:{and}\:{x}>\mathrm{2}^{−\mathrm{2}} \Rightarrow\:{x}>\mathrm{4} \\ $$$$ \\ $$$${or}\:\Rightarrow\left(\mathrm{2}\right)\:{log}_{\mathrm{2}} {x}−\mathrm{2}<\mathrm{0}\:{and}\:{log}_{\mathrm{2}} {x}+\mathrm{2}<\mathrm{0} \\ $$$$\Rightarrow{x}<\mathrm{4}\:{and}\:{x}<\mathrm{2}^{−\mathrm{2}} \Rightarrow\:{x}<\mathrm{1}/\mathrm{4} \\ $$$${x}\:{must}\:{be}\:{strictly}\:{positive}\:{however} \\ $$$${for}\:\:\sqrt{{x}}\in\mathbb{R}.\:{So},\:{truly}\:\mathrm{0}<{x}<\mathrm{1}/\mathrm{4}. \\ $$$$ \\ $$$${Hence},\:{the}\:{solution}\:{set}\:\Upsilon\:{for}\:{the}\: \\ $$$${inequality}\:{is}\:\Upsilon=\left\{{x}\in\mathbb{R}\mid\mathrm{0}<{x}<\mathrm{1}/\mathrm{4}\:{or}\:{x}>\mathrm{4}\right\}. \\ $$$$ \\ $$