Question Number 90434 by john santu last updated on 23/Apr/20
$$\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{8}{x}\right).\mathrm{log}_{\mathrm{3}} \left(\mathrm{27}{x}\right)}{{x}^{\mathrm{2}} −\mid{x}\mid}\:\leqslant\:\mathrm{0}\: \\ $$
Commented by jagoll last updated on 28/Apr/20
$${how}\:\frac{\mathrm{1}}{\mathrm{8}}\:<\:\frac{\mathrm{1}}{\mathrm{27}}\:?\: \\ $$
Commented by john santu last updated on 23/Apr/20
$$\left(\mathrm{1}\right)\:{x}>\mathrm{0} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{8}{x}−\mathrm{1}\right)\left(\mathrm{27}{x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}}\:\leqslant\:\mathrm{0} \\ $$$$\frac{\left({x}−\frac{\mathrm{1}}{\mathrm{8}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{27}}\right)}{{x}\left({x}−\mathrm{1}\right)}\:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{0}\:<\:{x}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{27}}\:\vee\:\frac{\mathrm{1}}{\mathrm{8}}\:\leqslant\:{x}\:<\:\mathrm{1} \\ $$
Commented by mahdi last updated on 24/Apr/20
$$\frac{\mathrm{1}}{\mathrm{8}}\leqslant\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{27}}\:\cup\:\mathrm{1}\leqslant\mathrm{x} \\ $$