# Question-208567

Question Number 208567 by Tawa11 last updated on 18/Jun/24
Answered by kgmxdd last updated on 18/Jun/24
Commented by Tawa11 last updated on 18/Jun/24
$$\mathrm{Any}\:\mathrm{workings}\:\mathrm{sir}? \\$$
Answered by MM42 last updated on 18/Jun/24
$$\sqrt[{\mathrm{3}}]{\mathrm{11}}=\sqrt[{\mathrm{12}}]{\mathrm{14641}}\:\:\:\&\:\:\sqrt{\mathrm{7}}=\sqrt[{\mathrm{12}}]{\mathrm{117649}}\:\:\:\&\:\:\sqrt[{\mathrm{4}}]{\mathrm{45}}=\sqrt[{\mathrm{12}}]{\mathrm{91125}} \\$$$$\Rightarrow\left({III}\right)\:\checkmark \\$$$$\\$$
Commented by Tawa11 last updated on 18/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir} \\$$
Answered by A5T last updated on 18/Jun/24
$$\sqrt{\mathrm{7}}=\sqrt{\sqrt{\mathrm{49}}}=\sqrt[{\mathrm{4}}]{\mathrm{49}}>\sqrt[{\mathrm{4}}]{\mathrm{45}} \\$$$$\sqrt[{\mathrm{4}}]{\mathrm{45}}\overset{?} {>}\sqrt[{\mathrm{3}}]{\mathrm{11}}\Leftrightarrow\left(\sqrt[{\mathrm{4}}]{\mathrm{45}}\right)^{\mathrm{12}} =\mathrm{45}^{\mathrm{3}} >\left(\sqrt[{\mathrm{3}}]{\mathrm{11}}\right)^{\mathrm{12}} =\mathrm{11}^{\mathrm{4}} \\$$$$\mathrm{45}^{\mathrm{3}} >\left(\mathrm{12}×\mathrm{3}\right)^{\mathrm{3}} =\mathrm{12}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} >\mathrm{11}×\mathrm{11}×\mathrm{11}×\mathrm{11}=\mathrm{11}^{\mathrm{4}} \\$$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{45}}>\sqrt[{\mathrm{3}}]{\mathrm{11}}\Rightarrow\sqrt{\mathrm{7}}>\sqrt[{\mathrm{45}}]{\mathrm{45}}>\sqrt[{\mathrm{3}}]{\mathrm{1}}\mathrm{1} \\$$
Commented by Tawa11 last updated on 18/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir} \\$$