Question Number 155510 by VIDDD last updated on 01/Oct/21
Answered by puissant last updated on 04/Oct/21
$$\:{I}=\sqrt{{log}_{\sqrt{\mathrm{2}}} \sqrt{\sqrt{\mathrm{2}}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} }+\:{log}_{\sqrt[{\mathrm{4}}]{\mathrm{2}}} \sqrt[{\mathrm{4}}]{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} }} \\ $$$$\Rightarrow\:{I}=\sqrt{{log}_{\sqrt{\mathrm{2}}} \mathrm{2}^{\frac{\mathrm{3}}{\mathrm{8}}} +\:{log}_{\sqrt[{\mathrm{4}}]{\mathrm{2}}} \mathrm{2}^{\frac{\mathrm{3}}{\mathrm{16}}} } \\ $$$$\Rightarrow\:{I}=\sqrt{\frac{\frac{\mathrm{3}}{\mathrm{8}}{ln}\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}}+\frac{\frac{\mathrm{3}}{\mathrm{16}}{ln}\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2}}} \\ $$$$\Rightarrow\:{I}=\sqrt{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\:=\:\sqrt{\frac{\mathrm{6}}{\mathrm{4}}}\:=\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\therefore\because\:\:{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{6}}… \\ $$