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Question Number 167539 by Tawa11 last updated on 19/Mar/22

$$\mathrm{Suppose}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{real},\mathrm{such}\mathrm{that},{\log }_4\mathrm{x}={\log }_6\mathrm{y}={\log }_9(\mathrm{x}+\mathrm{y}),$$$$\mathrm{find}\frac{\mathrm{y}}{\mathrm{x}}$$

Answered by Rasheed.Sindhi last updated on 19/Mar/22

$${\log }_4\mathrm{x}={\log }_6\mathrm{y}={\log }_9(\mathrm{x}+\mathrm{y})=\mathrm{k}\left(say\right)$$$$\mathrm{x}=4^{\mathrm{k}},\mathrm{y}=6^{\mathrm{k}},\mathrm{x}+\mathrm{y}=9^{\mathrm{k}}$$$$4^{\mathrm{k}}+6^{\mathrm{k}}=9^{\mathrm{k}}$$$${\left(\frac{4}{9}\right)}^{\mathrm{k}}+{\left(\frac{6}{9}\right)}^{\mathrm{k}}=1$$$${\left(\frac{2}{3}\right)}^{2\mathrm{k}}+{\left(\frac{2}{3}\right)}^{\mathrm{k}}=1$$$${\left(\frac{2}{3}\right)}^{\mathrm{k}}=\mathrm{u}$$$${\mathrm{u}}^2+\mathrm{u}-1=0$$$$\mathrm{u}=\frac{-1\pm \sqrt{5}}{2}$$$${\left(\frac{2}{3}\right)}^{\mathrm{k}}=\frac{-1\pm \sqrt{5}}{2}$$$$\frac{\mathrm{y}}{\mathrm{x}}=\frac{6^{\mathrm{k}}}{4^{\mathrm{k}}}={\left(\frac{3}{2}\right)}^{\mathrm{k}}={\left({\left(\frac{2}{3}\right)}^{\mathrm{k}}\right)}^{-1}={\left(\frac{-1\pm \sqrt{5}}{2}\right)}^{-1}$$$$=\frac{2}{-1\pm \sqrt{5}}\cdot \frac{-1\mp \sqrt{5}}{-1\mp \sqrt{5}}=\frac{-2(1\pm \sqrt{5})}{1-5}$$$$=\frac{1\pm \sqrt{5}}{2}$$

Commented by Tawa11 last updated on 19/Mar/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by Rasheed.Sindhi last updated on 19/Mar/22

$$$$$${\log }_4\mathrm{x}={\log }_6\mathrm{y}={\log }_9(\mathrm{x}+\mathrm{y})=\mathrm{k}\left(say\right)$$$$\mathrm{x}=4^{\mathrm{k}},\mathrm{y}=6^{\mathrm{k}},\mathrm{x}+\mathrm{y}=9^{\mathrm{k}}\Rightarrow 4^{\mathrm{k}}+6^{\mathrm{k}}=9^{\mathrm{k}}$$$$1+\frac{6^{\mathrm{k}}}{4^{\mathrm{k}}}=\frac{9^{\mathrm{k}}}{4^{\mathrm{k}}}$$$$1+{\left(\frac{3}{2}\right)}^{\mathrm{k}}={\left(\frac{3}{2}\right)}^{2\mathrm{k}}$$$${\left(\frac{3}{2}\right)}^{2\mathrm{k}}-{\left(\frac{3}{2}\right)}^{\mathrm{k}}-1=0$$$${\left(\frac{3}{2}\right)}^{\mathrm{k}}=\mathrm{u}$$$${\mathrm{u}}^2-\mathrm{u}-1=0$$$$\mathrm{u}=\frac{1\pm \sqrt{5}}{2}$$$${\left(\frac{3}{2}\right)}^{\mathrm{k}}=\frac{1\pm \sqrt{5}}{2}$$$$\frac{\mathrm{y}}{\mathrm{x}}=\frac{6^{\mathrm{k}}}{4^{\mathrm{k}}}={\left(\frac{3}{2}\right)}^{\mathrm{k}}=\frac{1\pm \sqrt{5}}{2}$$

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