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Question Number 89124 by pete last updated on 15/Apr/20

Commented by pete last updated on 15/Apr/20

$$\mathrm{I}\mathrm{need}\mathrm{help}\mathrm{on}\mathrm{this}\mathrm{question}\mathrm{please}$$

Commented by john santu last updated on 15/Apr/20

$$2-\sqrt{3}=\frac{2-\sqrt{3}}{2+\sqrt{3}}\times 2+\sqrt{3}=\frac{1}{2+\sqrt{3}}$$$$2-\sqrt{3}={(2+\sqrt{3})}^{-1}$$$$\Rightarrow 2\log {}_{2+\sqrt{3}}(\sqrt{x^2+1}+x)-\log {}_{2+\sqrt{3}}(\sqrt{x^2+1}-x)=3$$$$\Rightarrow \log {}_{2+\sqrt{3}}{(\sqrt{x^2+1}+x)}^2=$$$$3+\log {}_{2+\sqrt{3}}(\sqrt{x^2+1}-x)$$$$\Rightarrow {(\sqrt{x^2+1}+x)}^2={(2+\sqrt{3})}^3(\sqrt{x^2+1}-x)$$$$2x^2+1+2x\sqrt{x^2+1}=(7+4\sqrt{3})(2+\sqrt{3})$$$$(\sqrt{x^2+1}-x)$$$$$$

Commented by john santu last updated on 15/Apr/20

$$mywayitstandard$$

Commented by MJS last updated on 15/Apr/20

$$2-\sqrt{3}=\frac{(2-\sqrt{3})(2+\sqrt{3})}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\Rightarrow \ln (2-\sqrt{3})=-\ln (2+\sqrt{3})$$$$-x+\sqrt{x^2+1}=\frac{(-x+\sqrt{x^2+1})(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})}=$$$$=\frac{1}{x+\sqrt{x^2+1}}\Rightarrow \ln (-x+\sqrt{x^2+1})=-\ln (x+\sqrt{x^2+1})$$$$$$$$2\frac{\ln (x+\sqrt{x^2+1})}{\ln (2+\sqrt{3})}+\frac{\ln (-x+\sqrt{x^2+1})}{\ln (2-\sqrt{3})}=3$$$$3\frac{\ln (x+\sqrt{x^2+1})}{\ln (2+\sqrt{3})}=3$$$$\ln (x+\sqrt{x^2+1})=\ln 2+\sqrt{3}$$$$x+\sqrt{x^2+1}=2+\sqrt{3}$$$$\mathrm{and}\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{solve}\mathrm{I}\mathrm{hope}$$

Commented by pete last updated on 16/Apr/20

$$thanksverymuchsir$$

Answered by MJS last updated on 15/Apr/20

$$\mathrm{let}t=x+\sqrt{x^2+1}\iff x=\frac{t^2-1}{2t}\iff -x+\sqrt{x^2+1}=\frac{1}{t}$$$$\Rightarrow$$$$\frac{3\ln t}{\ln (2+\sqrt{3})}=3\Rightarrow t=2+\sqrt{3}\Rightarrow x=\sqrt{3}$$

Commented by john santu last updated on 15/Apr/20

$$onlyx=\sqrt{3}thesolution?$$

Commented by MJS last updated on 15/Apr/20

$$\mathrm{I}\mathrm{think}\mathrm{so}$$

Commented by john santu last updated on 15/Apr/20

$$supercreativesir$$

Commented by pete last updated on 15/Apr/20

$$\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir},\mathrm{but}\mathrm{please}$$$$\mathrm{kindly}\mathrm{elaborate}\mathrm{more}\mathrm{on}\mathrm{in}\mathrm{for}\mathrm{me}\mathrm{to}$$$$\mathrm{be}\mathrm{able}\mathrm{to}\mathrm{explain}\mathrm{it}\mathrm{to}\mathrm{my}\mathrm{students}.$$

Answered by Don08q last updated on 22/Jul/20

$${2\log }_{2+\sqrt{3}}(\sqrt{x^2+1}+x)+{\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)$$$$=3$$$$$$$${2\log }_{\frac{1}{2-\sqrt{3}}}(\sqrt{x^2+1}+x)+{\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)$$$$=3$$$$$$$$\frac{{2\log }_{2-\sqrt{3}}{(\sqrt{x^2+1}-x)}^{-1}}{{\log }_{2-\sqrt{3}}{(2-\sqrt{3})}^{-1}}+{\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)$$$$=3$$$$$$$$\frac{{2\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)}{{\log }_{2-\sqrt{3}}(2-\sqrt{3})}+{\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)$$$$=3$$$$$$$${2\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)+{\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)$$$$=3$$$$$$$${3\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)=3$$$$$$$${\log }_{2-\sqrt{3}}(\sqrt{x^2+1}-x)=1$$$$$$$$\sqrt{x^2+1}-x=2-\sqrt{3}$$$$$$$$\sqrt{x^2+1}-x=\sqrt{{\left(\sqrt{3}\right)}^2+1}-\sqrt{3}$$$$$$$$\therefore x=\sqrt{3}$$$$$$$$$$$$$$$$$$

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