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Question Number 155510 by VIDDD last updated on 01/Oct/21

Answered by puissant last updated on 04/Oct/21

$$I=\sqrt{{log}_{\sqrt{2}}\sqrt{\sqrt{2}\times 2^{\frac{1}{4}}}+{log}_{\sqrt[4]{2}}\sqrt[4]{2^{\frac{3}{4}}}}$$$$\Rightarrow I=\sqrt{{log}_{\sqrt{2}}{2}^{\frac{3}{8}}+{log}_{\sqrt[4]{2}}{2}^{\frac{3}{16}}}$$$$\Rightarrow I=\sqrt{\frac{\frac{3}{8}ln2}{\frac{1}{2}ln2}+\frac{\frac{3}{16}ln2}{\frac{1}{4}ln2}}$$$$\Rightarrow I=\sqrt{\frac{3}{4}+\frac{3}{4}}=\sqrt{\frac{6}{4}}=\sqrt{\frac{3}{2}}$$$$$$$$\therefore \because I=\frac{1}{2}\sqrt{6}...$$

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