Question Number 87773 by john santu last updated on 06/Apr/20
$$\:^{\mathrm{3}} \mathrm{log}\:\left(\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\left(\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\mathrm{x}^{\mathrm{4}} \right)\right)>\:\mathrm{0} \\ $$
Commented by john santu last updated on 06/Apr/20
$$\Rightarrow\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\left(\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\mathrm{x}^{\mathrm{4}} \right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{i}\right)\mathrm{x}\:\neq\:\mathrm{0} \\ $$$$\left(\mathrm{ii}\right)\:\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\left(\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\mathrm{x}^{\mathrm{4}} \right)\:>\:^{\mathrm{x}^{\mathrm{2}} } \mathrm{log}\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} \right)\:>\:\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{x}^{\mathrm{2}} \:>\mathrm{0}\: \\ $$$$\mathrm{x}\neq\:−\mathrm{1}\:,\:\mathrm{x}\:\neq\:\mathrm{1}\: \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{1}\:<\:\mathrm{x}^{\mathrm{2}} \:<\:\mathrm{2}\:\begin{cases}{−\sqrt{\mathrm{2}}\:<\:\mathrm{x}\:<\:−\mathrm{1}}\\{\mathrm{1}\:<\:\mathrm{x}\:<\:\sqrt{\mathrm{2}}}\end{cases} \\ $$
Commented by mahdi last updated on 06/Apr/20
$$\mathrm{what}\:\mathrm{meaning}\:\:\left(\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{in}\:\:\:\:\overset{\mathrm{x}^{\mathrm{2}} } {\:}\mathrm{log}\:? \\ $$$$ \\ $$
Commented by john santu last updated on 06/Apr/20
$$\mathrm{it}\:\mathrm{same}\:\mathrm{to}\:\mathrm{log}_{\mathrm{x}^{\mathrm{2}} } \:\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$
Commented by mahdi last updated on 06/Apr/20
$$\mathrm{thanks} \\ $$