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Question Number 161868 by cortano last updated on 23/Dec/21

$${\left(\sqrt{2+\sqrt{3}}\right)}^x+\frac{1}{{\left(\sqrt{2+\sqrt{3}}\right)}^x}=2^x$$$$x=?$$

Answered by Rasheed.Sindhi last updated on 23/Dec/21

$$a=\sqrt{2+\sqrt{3}},\frac{1}{a}=\frac{1}{\sqrt{2+\sqrt{3}}}\times \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$$$$=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4-3}}=\sqrt{2-\sqrt{3}}$$$$a^2+\frac{1}{a^2}={\left(\sqrt{2+\sqrt{3}}\right)}^2+{\left(\sqrt{2-\sqrt{3}}\right)}^2$$$$=2+\sqrt{3}+2-\sqrt{3}=4=2^2$$$$a^2+\frac{1}{a^2}=2^2$$$$x=2$$

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