Question Number 174884 by daus last updated on 13/Aug/22
Answered by aleks041103 last updated on 13/Aug/22
$$\left({log}_{{b}} {a}\right)\left({log}_{{a}} {b}\right)={log}_{{a}} \left({b}^{{log}_{{b}} {a}} \right)={log}_{{a}} {a}=\mathrm{1} \\ $$$$\Rightarrow{log}_{{x}} \mathrm{2}=\frac{\mathrm{1}}{{log}_{\mathrm{2}} {x}} \\ $$$${let}\:{log}_{\mathrm{2}} {x}={t} \\ $$$$\mathrm{1}+{t}−\frac{\mathrm{6}}{{t}}>\mathrm{0} \\ $$$$\frac{{t}^{\mathrm{2}} +{t}−\mathrm{6}}{{t}}>\mathrm{0} \\ $$$$\frac{\left({t}+\mathrm{3}\right)\left({t}−\mathrm{2}\right)}{{t}}>\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2},\mathrm{3}} =−\mathrm{3},\mathrm{0},\mathrm{2} \\ $$$$\Rightarrow{t}={log}_{\mathrm{2}} {x}\in\left(−\mathrm{3},\mathrm{0}\right)\cup\left(\mathrm{2},\infty\right) \\ $$$$\Rightarrow{x}\in\left(\frac{\mathrm{1}}{\mathrm{8}},\mathrm{1}\right)\cup\left(\mathrm{4},\infty\right) \\ $$