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Question Number 185805 by Rupesh123 last updated on 27/Jan/23

Answered by MJS_new last updated on 28/Jan/23

$${(2+\sqrt{3})}^{2023}=x$$$$\frac{1}{\sqrt{3}}\times \frac{x^2+x^{-2}+1}{x+x^{+1}}=\frac{1+n^2}{n}\Rightarrow n={\left(\frac{\sqrt{3}x}{x^2-1}\right)}^{\pm 1}=$$$$={\left(\frac{2\sinh \left(2023\ln (2+\sqrt{3})\right)}{\sqrt{3}}\right)}^{\pm 1}$$$$\mathrm{which}\mathrm{is}\mathrm{very}\mathrm{close}\mathrm{to}{(2\pm \sqrt{3})}^{2023}\approx {10}^{\pm 1157}$$

Answered by manxsol last updated on 28/Jan/23

$${(2+\sqrt{3})}^2=7+\sqrt{48}$$$$a=2+\sqrt{3}$$$$a^2=7+\sqrt{48}$$$$\frac{1}{\sqrt{3}}\left[\frac{{\left(a^{2023}\right)}^2+\frac{1}{{\left(a^{2023}\right)}^2}+1}{a^{2023}-\frac{1}{a^{2023}}}\right]=n+\frac{1}{n}$$$$p=a^{2023}$$$$\frac{1}{\sqrt{3}}\left[\frac{p^2+\frac{1}{p^2}+1}{p-\frac{1}{p}}\right]=n+\frac{1}{n}$$$$\frac{1}{\sqrt{3}}\left[\frac{{(p-\frac{1}{p})}^2+3}{p-\frac{1}{p}}\right]=n+\frac{1}{n}$$$$\frac{1}{\sqrt{3}}[(p-\frac{1}{p})+\frac{3}{(p-\frac{1}{p})}]=n+\frac{1}{n}$$$$\frac{(p-\frac{1}{p})}{\sqrt{3}}+\frac{1}{\frac{(p-\frac{1}{p})}{\sqrt{3}}}=n+\frac{1}{n}$$$$n_1=(a^{2023}-\frac{1}{a^{2023}})\frac{1}{\sqrt{3}}$$$$n_2=\frac{\sqrt{3}}{(a^{2023}-\frac{1}{a^{2023}})}$$$$z=a^{2023}$$$$z={(2+\sqrt{3})}^{2023}$$$$logz=2023log(2+\sqrt{3})$$$$z={10}^{1157.049889}$$$$\frac{1}{z}={10}^{-1157.049889}\approx 0$$$$n_1=\frac{1}{\sqrt{3}}{10}^{1157.049}$$$$n_2=\sqrt{3}{10}^{-1157.048}$$

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