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Question Number 211920 by Spillover last updated on 24/Sep/24

Answered by aleks041103 last updated on 24/Sep/24

$$\sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{...}}}}=x^{1/2}{x}^{2/4}{x}^{3/8}...x^{n/2^n}...=$$$$=x\wedge \left(\underset{n=1}{\sum ^{\mathrm{\infty }}}n{2}^{-n}\right)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{nr}^n=r\underset{n=0}{\sum ^{\mathrm{\infty }}}{nr}^{n-1}=r\frac{d}{dr}\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}{r}^n\right)=$$$$=r\frac{d}{dr}\left(\frac{1}{1-r}\right)=\frac{r}{{(1-r)}^2}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}n{2}^{-n}=\frac{1/2}{{(1-1/2)}^2}=\frac{1}{2.\frac{1}{4}}=2$$$$\Rightarrow \sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{...}}}}=x^2$$$$--------------$$$$\sqrt[3]{a-b\sqrt[3]{a-b\sqrt[3]{a-b...}}}=x$$$$\Rightarrow \sqrt[3]{a-bx}=x$$$$x^3+bx-a=0$$$$x^3+18x-215=0\Rightarrow x=5$$$$--------------$$$$Ans.={log}_{\sqrt{5}}\left(\frac{{(5+5\sqrt{5})}^2}{(6+2\sqrt{5})5}\right)=$$$$=2{log}_5\left(\frac{5{(1+\sqrt{5})}^2}{6+2\sqrt{5}}\right)=2$$$$\Rightarrow Ans.=2$$

Commented by Spillover last updated on 25/Sep/24

$$great.thanks$$

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