Question Number 79263 by jagoll last updated on 23/Jan/20
$$\mathrm{4}^{\mathrm{2x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\:^{\mathrm{2}} \mathrm{log}^{\mathrm{2}} \left(\mathrm{2x}\right)>\:^{\mathrm{2}} \mathrm{log}\left(\mathrm{x}\right) \\ $$$$\left\{^{\mathrm{2}} \mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{2}^{\mathrm{2x}} \right\} \\ $$
Commented by john santu last updated on 24/Jan/20
$$\left({i}\right)\:{x}>\mathrm{0} \\ $$$$\left({ii}\right)\frac{\mathrm{4}^{\mathrm{2}{x}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}\right)\right\}^{\mathrm{2}} \\ $$$$>\mathrm{log}_{\mathrm{2}} \left({x}\right)\left\{−\mathrm{log}_{\mathrm{2}} \left({x}\right)−\mathrm{4}^{{x}} \right\} \\ $$$${let}\:\mathrm{4}^{{x}} =\:{u}\:,\:\mathrm{log}_{\mathrm{2}} \left({x}\right)=\:{v} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >−\mathrm{4}{v}^{\mathrm{2}} −\mathrm{4}{uv} \\ $$$${u}^{\mathrm{2}} +\mathrm{4}{uv}+\mathrm{4}{v}^{\mathrm{2}} +\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$\left({u}+\mathrm{2}{v}\right)^{\mathrm{2}} +\left(\mathrm{1}+{v}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{u}+\mathrm{2}{v}=\mathrm{0}\Rightarrow\mathrm{4}^{{x}} +\mathrm{2}.\mathrm{log}_{\mathrm{2}} \left({x}\right)=\mathrm{0}}\\{\mathrm{1}+{v}=\mathrm{0}\Rightarrow\mathrm{1}+\mathrm{log}_{\mathrm{2}} \left({x}\right)=\mathrm{0}}\end{cases}\: \\ $$$${x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\therefore\:{x}\in\mathbb{R}\:\wedge{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by john santu last updated on 24/Jan/20
$${check}\:{for}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}^{\mathrm{2}} > \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left\{\mathrm{log}_{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \right\} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}.\:\mathrm{0}\:>\:\left(−\mathrm{1}\right).\left(−\mathrm{1}\right) \\ $$
Commented by john santu last updated on 24/Jan/20
$${i}\:{think}\:{about}\:{this}\:{typing}\:{error}. \\ $$$${should}\:\mathrm{4}^{\mathrm{2}{x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{log}_{\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{2}{x}\right)\geqslant \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}\right)\left\{\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{2}^{\mathrm{2}{x}} \right\} \\ $$
Commented by jagoll last updated on 24/Jan/20
$$\mathrm{thanks} \\ $$