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Question Number 93957 by O Predador last updated on 16/May/20

$$$$$${\log }_{\sqrt{17}-\sqrt{2}}\left(\frac{15}{\sqrt{19+\sqrt{136}}}\right){\mathrm{x}}^2-{\log }_{\sqrt{19}-\sqrt{3}}\left(\frac{1}{22-\sqrt{228}}\right)\mathrm{x}=3$$$$$$$$\mathrm{x}=?$$

Commented by PRITHWISH SEN 2 last updated on 16/May/20

$$\sqrt{19+\sqrt{136}}=\frac{\sqrt{34}+2}{\sqrt{2}}\Rightarrow \frac{15\sqrt{2}}{\sqrt{34}+2}=\sqrt{17}-\sqrt{2}$$$$22-\sqrt{228}={(\sqrt{19}-\sqrt{3})}^2$$$${\log }_{\sqrt{17}-\sqrt{2}}(\sqrt{17}-\sqrt{2}){\mathrm{x}}^2-{\log }_{\sqrt{19}-\sqrt{3}}\mathrm{x}+2=3$$$${2\log }_{\sqrt{17}-\sqrt{2}}\mathrm{x}={\log }_{\sqrt{19}-\sqrt{3}}\mathrm{x}$$$$\mathrm{and}\mathrm{the}\mathrm{only}\mathrm{possible}\mathrm{solution}\mathrm{for}\mathrm{this}\mathrm{is}$$$$\mathbf{x}=1$$

Commented by O Predador last updated on 25/May/20

$$\mathrm{Simplifying}\mathrm{the}\mathrm{radicals}{\mathrm{shouldn}}^{\prime }\mathrm{t}\mathrm{we}\mathrm{consider}$$$$\mathrm{it}\mathrm{a}\mathrm{second}\mathrm{degree}\mathrm{equation}?$$

Commented by O Predador last updated on 25/May/20

$$\mathrm{Simplificando}\mathrm{os}\mathrm{radicais}\mathrm{n}\overline{\mathrm{a}o}\mathrm{deve}-\mathrm{se}\mathrm{considerar}$$$$\mathrm{uma}\mathrm{equa}\underset{}{\mathrm{c}}\overline{\mathrm{a}o}\mathrm{do}\mathrm{segundo}\mathrm{grau}?$$

Commented by PRITHWISH SEN 2 last updated on 25/May/20

$$\mathrm{English}\mathrm{please}$$

Commented by PRITHWISH SEN 2 last updated on 25/May/20

$$\mathrm{Actually}$$$$19+\sqrt{136}=\frac{38+2\sqrt{136}}{2}=\frac{{\left(\sqrt{34}\right)}^2+2^2+2.\sqrt{34}.2}{2}$$$$=\frac{{\left(\sqrt{34}\right)}^2+2^2+2.\sqrt{34}.\sqrt{4}}{2}=\frac{{(\sqrt{34}+2)}^2}{{\left(\sqrt{2}\right)}^2}$$$$\mathrm{I}\mathrm{think}\mathrm{this}\mathrm{will}\mathrm{help}\mathrm{you}.$$

Commented by O Predador last updated on 26/May/20

$$$$$$\mathrm{Grades}:\{\begin{array}{l}\sqrt{19+\sqrt{136}}=\sqrt{17}+\sqrt{2}\\ 22-\sqrt{228}={(\sqrt{19}-\sqrt{3})}^2\end{array}$$$$$$$${\log }_{\sqrt{17}-\sqrt{2}}(\sqrt{17}-\sqrt{2}){\mathrm{x}}^2-{\log }_{\sqrt{19}-\sqrt{3}}\left(\frac{1}{{(\sqrt{19}-\sqrt{3})}^2}\right)\mathrm{x}=3...$$$$$$$$...{\mathrm{x}}^2+2\mathrm{x}=3...$$$$...{\mathrm{x}}^2+2\mathrm{x}+1=3+1...$$$$...\mathrm{x}+1=\pm 2...$$$$$$$$\{\begin{array}{l}{\mathrm{x}}_1=-3\\ {\mathrm{x}}_2=1\end{array}$$$$$$$$\mathrm{Solution}:\{-3,1\}$$$$$$$${\mathrm{Wouldn}}^{\prime }\mathrm{t}\mathrm{that}\mathrm{be}\mathrm{so}?$$

Commented by PRITHWISH SEN 2 last updated on 26/May/20

$$\mathrm{no}.$$$${\log }_{\sqrt{17}-\sqrt{2}}(\sqrt{17}-\sqrt{2}){\mathrm{x}}^2={\log }_{\sqrt{17}-\sqrt{2}}(\sqrt{17}-\sqrt{2})+{\log }_{\sqrt{17}-\sqrt{2}}{\mathrm{x}}^2$$

Commented by PRITHWISH SEN 2 last updated on 26/May/20

$$\mathrm{no}.$$$$\mathrm{because}{\log }_{\mathrm{a}}\mathrm{bc}={\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{a}}\mathrm{c}$$

Commented by O Predador last updated on 26/May/20

$$\mathrm{but},\mathrm{logarithms}\mathrm{are}\mathrm{the}‘‘{\mathrm{x}}^{″}\mathrm{coefficient}.$$

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