Question Number 172082 by Mikenice last updated on 23/Jun/22
$${solve} \\ $$$${log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}−\mathrm{2}\underset{−} {<}\mathrm{0} \\ $$
Commented by mokys last updated on 23/Jun/22
$${log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}\:\leqslant\mathrm{2} \\ $$$$\mathrm{2}{log}_{\mathrm{0}.\mathrm{5}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}\leqslant\mathrm{2} \\ $$$$\mathrm{3}{log}_{\mathrm{0}.\mathrm{5}} {x}\:\leqslant\mathrm{2}\: \\ $$$${log}_{\mathrm{0}.\mathrm{5}} {x}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}\leqslant\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\leqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{4}}}\: \\ $$$$ \\ $$$${another}\:{way}\:: \\ $$$$ \\ $$$$\frac{{lnx}}{{ln}\left(\mathrm{0}.\mathrm{5}\right)}\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:\Rightarrow{lnx}\:\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:{ln}\:\left(\mathrm{0}.\mathrm{5}\right) \\ $$$$ \\ $$$${lnx}\:\leqslant\:{ln}\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\Rightarrow\:{x}\:\leqslant\:\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \leqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$ \\ $$$${Aldolaimy} \\ $$