Question Number 98885 by M±th+et+s last updated on 16/Jun/20
$${find}\:{the}\:{range} \\ $$$$ \\ $$$${f}\left({x}\right)={log}_{\mathrm{4}} {log}_{\mathrm{2}} {log}_{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right) \\ $$
Commented by MJS last updated on 17/Jun/20
$${f}\left({x}\right)\:\mathrm{defined}\:\mathrm{for}\:\mathrm{0}<{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:−\infty<{f}\left({x}\right)<+\infty \\ $$$$\mathrm{log}_{\mathrm{1}/\mathrm{2}} \:{x}\:=−\mathrm{log}_{\mathrm{2}} \:{x} \\ $$$$\Rightarrow\:{x}>\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left(−\mathrm{log}_{\mathrm{2}} \:{x}\right) \\ $$$$\Rightarrow\:−\mathrm{log}_{\mathrm{2}} \:{x}\:>\mathrm{0}\:\Rightarrow\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{4}} \:\mathrm{log}_{\mathrm{2}} \:\left(−\mathrm{log}_{\mathrm{2}} \:{x}\right) \\ $$$$\Rightarrow\:\mathrm{log}_{\mathrm{2}} \:\left(−\mathrm{log}_{\mathrm{2}} \:{x}\right)\:>\mathrm{0}\:\Rightarrow\:\mathrm{0}<{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by M±th+et+s last updated on 17/Jun/20
$${thank}\:{you}\:{sir} \\ $$