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Question Number 170485 by Mastermind last updated on 24/May/22

Answered by Rasheed.Sindhi last updated on 25/May/22

$${\left(\sqrt{2+\sqrt{3}}\right)}^x+{\left(\sqrt{2-\sqrt{3}}\right)}^x=2;x=?$$$$\frac{1}{{\left(\sqrt{2+\sqrt{3}}\right)}^x}=\frac{1}{{\left(\sqrt{2+\sqrt{3}}\right)}^x}\cdot \frac{{\left(\sqrt{2-\sqrt{3}}\right)}^x}{{\left(\sqrt{2-\sqrt{3}}\right)}^x}$$$$=\frac{{\left(\sqrt{2-\sqrt{3}}\right)}^x}{{\left(\sqrt{4-3}\right)}^x}={\left(\sqrt{2-\sqrt{3}}\right)}^x$$$${\left(\sqrt{2+\sqrt{3}}\right)}^x+\frac{1}{{\left(\sqrt{2+\sqrt{3}}\right)}^x}=2$$$$a+\frac{1}{a}=2$$$$a^2-2a+1=0$$$${(a-1)}^2=0$$$$a=1$$$${\left(\sqrt{2+\sqrt{3}}\right)}^x=1={\left(\sqrt{2+\sqrt{3}}\right)}^0$$$$x=0$$

Answered by Rasheed.Sindhi last updated on 25/May/22

$${\left(\sqrt{2+\sqrt{3}}\right)}^x+{\left(\sqrt{2-\sqrt{3}}\right)}^x=2;x=?$$$$Multiplyingby{\left(\sqrt{2+\sqrt{3}}\right)}^{x}tobothsides:$$$${\left(\sqrt{2+\sqrt{3}}\right)}^{2x}+{(\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}})}^x=2{\left(\sqrt{2+\sqrt{3}}\right)}^x$$$${\left(\sqrt{2+\sqrt{3}}\right)}^{2x}+{\left(\sqrt{4-3}\right)}^x=2{\left(\sqrt{2+\sqrt{3}}\right)}^x$$$$Let{\left(\sqrt{2+\sqrt{3}}\right)}^x=a$$$$a^2-2a+1=0$$$${(a-1)}^2=0$$$$a=1$$$${\left(\sqrt{2+\sqrt{3}}\right)}^x=1={\left(\sqrt{2+\sqrt{3}}\right)}^0$$$$x=0$$

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