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Question Number 172082 by Mikenice last updated on 23/Jun/22
solve  log_(0.5) ^2 x+log_(0.5) x−2<_− 0
$${solve} \\ $$$${log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}−\mathrm{2}\underset{−} {<}\mathrm{0} \\ $$
Commented by mokys last updated on 23/Jun/22
log_(0.5) ^2 x+log_(0.5) x ≤2  2log_(0.5) x+log_(0.5) x≤2  3log_(0.5) x ≤2   log_(0.5) x≤(2/3)  x≤(0.5)^(2/3)  ≤ ((1/4))^(1/3)      another way :    ((lnx)/(ln(0.5)))≤((2/3)) ⇒lnx ≤((2/3)) ln (0.5)    lnx ≤ ln(0.5)^(2/3)  ⇒ x ≤ (0.5)^(2/3) ≤ ((1/4))^(1/3)     Aldolaimy
$${log}_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}\:\leqslant\mathrm{2} \\ $$$$\mathrm{2}{log}_{\mathrm{0}.\mathrm{5}} {x}+{log}_{\mathrm{0}.\mathrm{5}} {x}\leqslant\mathrm{2} \\ $$$$\mathrm{3}{log}_{\mathrm{0}.\mathrm{5}} {x}\:\leqslant\mathrm{2}\: \\ $$$${log}_{\mathrm{0}.\mathrm{5}} {x}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}\leqslant\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\leqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{4}}}\: \\ $$$$ \\ $$$${another}\:{way}\:: \\ $$$$ \\ $$$$\frac{{lnx}}{{ln}\left(\mathrm{0}.\mathrm{5}\right)}\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:\Rightarrow{lnx}\:\leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:{ln}\:\left(\mathrm{0}.\mathrm{5}\right) \\ $$$$ \\ $$$${lnx}\:\leqslant\:{ln}\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\Rightarrow\:{x}\:\leqslant\:\left(\mathrm{0}.\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \leqslant\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$ \\ $$$${Aldolaimy} \\ $$

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