Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 112992 by Khanacademy last updated on 10/Sep/20

Commented by Khanacademy last updated on 10/Sep/20

$$(\mathit{x}:\mathit{y})=?$$

Answered by bobhans last updated on 11/Sep/20

$$\sqrt[4]{\mathrm{x}}+\sqrt{\mathrm{y}}=\frac{8}{3}\frac{\ln \left(\mathrm{y}\right)}{\ln \left(\mathrm{x}\right)}$$$$\sqrt[4]{\mathrm{x}}+\sqrt{\mathrm{y}}=\frac{2}{3}\frac{\ln \left(\mathrm{x}\right)}{\ln \left(\mathrm{y}\right)}$$$$\iff \frac{8\ln \left(\mathrm{y}\right)}{3\ln \left(\mathrm{x}\right)}=\frac{2\ln \left(\mathrm{x}\right)}{3\ln \left(\mathrm{y}\right)};{4\ln }^2\left(\mathrm{y}\right)-\ln {}^2\left(\mathrm{x}\right)=0$$$$\{2\ln \left(\mathrm{y}\right)-\ln \left(\mathrm{x}\right)\}\{2\ln \left(\mathrm{y}\right)+\ln \left(\mathrm{x}\right)\}=0$$$$\mathrm{case}\left(1\right)\Rightarrow {\mathrm{y}}^2=\mathrm{x}$$$${\mathrm{x}}^{\sqrt[4]{\mathrm{x}}+\mathrm{x}}={\left(\sqrt{\mathrm{x}}\right)}^{\frac{8}{3}}\Rightarrow {\mathrm{x}}^{\sqrt[4]{\mathrm{x}}+\mathrm{x}}={\mathrm{x}}^{\frac{4}{3}}$$$$\Rightarrow (\mathrm{x}-1)(\mathrm{x}+\sqrt[4]{\mathrm{x}}-\frac{4}{3})=0$$$$\{\begin{array}{l}\mathrm{x}=1\mathrm{or}\\ \mathrm{x}+\sqrt[4]{\mathrm{x}}-\frac{4}{3}=0;{3\mathrm{p}}^4+3\mathrm{p}-4=0,\mathrm{where}\mathrm{p}=\sqrt[4]{\mathrm{x}}\end{array}$$$$\mathrm{case}\left(2\right){\mathrm{y}}^2=\frac{1}{\mathrm{x}}$$$${\mathrm{x}}^{\sqrt[4]{\mathrm{x}}+\mathrm{x}}={\left(\sqrt{\frac{1}{\mathrm{x}}}\right)}^{\frac{8}{3}};{\mathrm{x}}^{\sqrt[4]{\mathrm{x}}+\mathrm{x}}={\mathrm{x}}^{-\frac{4}{3}}$$$$(\mathrm{x}-1)(\mathrm{x}+\sqrt[4]{\mathrm{x}}+\frac{4}{3})=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com