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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

Any workings sir?

$$\mathrm{Any}\:\mathrm{workings}\:\mathrm{sir}? \\ $$

Answered by MM42 last updated on 18/Jun/24

((11))^(1/3) =((14641))^(1/(12))    &  (√7)=((117649))^(1/(12))    &  ((45))^(1/4) =((91125))^(1/(12))   ⇒(III) ✓

$$\sqrt[{\mathrm{3}}]{\mathrm{11}}=\sqrt[{\mathrm{12}}]{\mathrm{14641}}\:\:\:\&\:\:\sqrt{\mathrm{7}}=\sqrt[{\mathrm{12}}]{\mathrm{117649}}\:\:\:\&\:\:\sqrt[{\mathrm{4}}]{\mathrm{45}}=\sqrt[{\mathrm{12}}]{\mathrm{91125}} \\ $$$$\Rightarrow\left({III}\right)\:\checkmark \\ $$$$ \\ $$

Commented by Tawa11 last updated on 18/Jun/24

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by A5T last updated on 18/Jun/24

(√7)=(√(√(49)))=((49))^(1/4) >((45))^(1/4)   ((45))^(1/4) >^? ((11))^(1/3) ⇔(((45))^(1/4) )^(12) =45^3 >(((11))^(1/3) )^(12) =11^4   45^3 >(12×3)^3 =12^3 ×3^3 >11×11×11×11=11^4   ⇔((45))^(1/4) >((11))^(1/3) ⇒(√7)>((45))^(1/(45)) >(1)^(1/3) 1

$$\sqrt{\mathrm{7}}=\sqrt{\sqrt{\mathrm{49}}}=\sqrt[{\mathrm{4}}]{\mathrm{49}}>\sqrt[{\mathrm{4}}]{\mathrm{45}} \\ $$$$\sqrt[{\mathrm{4}}]{\mathrm{45}}\overset{?} {>}\sqrt[{\mathrm{3}}]{\mathrm{11}}\Leftrightarrow\left(\sqrt[{\mathrm{4}}]{\mathrm{45}}\right)^{\mathrm{12}} =\mathrm{45}^{\mathrm{3}} >\left(\sqrt[{\mathrm{3}}]{\mathrm{11}}\right)^{\mathrm{12}} =\mathrm{11}^{\mathrm{4}} \\ $$$$\mathrm{45}^{\mathrm{3}} >\left(\mathrm{12}×\mathrm{3}\right)^{\mathrm{3}} =\mathrm{12}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} >\mathrm{11}×\mathrm{11}×\mathrm{11}×\mathrm{11}=\mathrm{11}^{\mathrm{4}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{45}}>\sqrt[{\mathrm{3}}]{\mathrm{11}}\Rightarrow\sqrt{\mathrm{7}}>\sqrt[{\mathrm{45}}]{\mathrm{45}}>\sqrt[{\mathrm{3}}]{\mathrm{1}}\mathrm{1} \\ $$

Commented by Tawa11 last updated on 18/Jun/24

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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