Question Number 146272 by mathdanisur last updated on 12/Jul/21
$${log}_{\boldsymbol{{x}}} \left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\leqslant\:\mathrm{1} \\ $$
Commented by iloveisrael last updated on 13/Jul/21
$$\:\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\leqslant\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{x}\right) \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{for}\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\:>\mathrm{0}\:\forall\mathrm{x}\in\mathrm{R} \\ $$$$\mathrm{then}\:\mathrm{x}<\mathrm{1}\:,\:\mathrm{x}\neq\mathrm{1}\:\cap\:\mathrm{x}>\mathrm{0} \\ $$$$\mathrm{solution}\:\mathrm{is}\:\mathrm{0}<\mathrm{x}<\mathrm{1}\: \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${thanks}\:{Ser},\:{answer}:\:\varnothing.? \\ $$
Commented by iloveisrael last updated on 13/Jul/21
$$\mathrm{no} \\ $$
Answered by gsk2684 last updated on 12/Jul/21
$${x}^{\mathrm{2}} +\mathrm{1}\:\leqslant\:{x}^{\mathrm{1}} \:{if}\:{x}>\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}\:\leqslant\:\mathrm{0}\: \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}\:\leqslant\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:\leqslant\:\mathrm{0}\:{no}\:{solution} \\ $$$${and}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:\geqslant\:{x}\:{if}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}\:\geqslant\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:\geqslant\:\mathrm{0}\: \\ $$$${true}\:{for}\:{all}\:{x}\: \\ $$$${solution}\:{set}\:{is}\:\left(\mathrm{0}\:\mathrm{1}\right) \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${cool}\:{thanks}\:{Ser} \\ $$