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Question Number 84459 by jagoll last updated on 13/Mar/20

$$\{\begin{array}{l}{\log }_{10}\left(\mathrm{x}\right)+\frac{{\log }_{10}\left(\mathrm{x}\right)+{8\log }_{10}\left(\mathrm{y}\right)}{{\log }_{10}^2\left(\mathrm{x}\right)+{\log }_{10}^2\left(\mathrm{y}\right)}=3\\ {\log }_{10}\left(\mathrm{y}\right)+\frac{{8\log }_{10}\left(\mathrm{x}\right)-{\log }_{10}\left(\mathrm{y}\right)}{{\log }_{10}^2\left(\mathrm{x}\right)+{\log }_{10}^2\left(\mathrm{y}\right)}=0\end{array}$$ $$\mathrm{find}\mathrm{x}\mathrm{\&}\mathrm{y}$$

Commented byjohn santu last updated on 13/Mar/20

$$\mathrm{let}{\log }_{10}\left(\mathrm{x}\right)=\mathrm{a}\mathrm{\&}{\log }_{10}\left(\mathrm{y}\right)=\mathrm{b}$$ $$\left(1\right)\mathrm{a}+\frac{\mathrm{a}+8\mathrm{b}}{{\mathrm{a}}^2+{\mathrm{b}}^2}=2\wedge \left(2\right)\mathrm{b}+\frac{8\mathrm{a}-\mathrm{b}}{{\mathrm{a}}^2+{\mathrm{b}}^2}=0$$ $$\mathrm{b}(\mathrm{a}+\frac{\mathrm{a}+8\mathrm{b}}{{\mathrm{a}}^2+{\mathrm{b}}^2})=2\mathrm{b}\left(1\right)$$ $$\mathrm{a}(\mathrm{b}+\frac{8\mathrm{a}-\mathrm{b}}{{\mathrm{a}}^2+{\mathrm{b}}^2})=0\left(2\right)$$ $$\Rightarrow 2\mathrm{ab}+\frac{{8\mathrm{a}}^2+{8\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}=2\mathrm{b}$$

Commented byjagoll last updated on 13/Mar/20

$$2\mathrm{ab}+8=2\mathrm{b}\Rightarrow 4=\mathrm{b}-\mathrm{ab}$$ $$\mathrm{b}=\frac{4}{1-\mathrm{a}}\wedge {\mathrm{a}}^2\mathrm{b}+{\mathrm{b}}^3+8\mathrm{a}-\mathrm{b}=0$$ $$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{have}\mathrm{idea}\mathrm{to}$$ $$\mathrm{solving}\mathrm{this}\mathrm{problem}$$

Answered by MJS last updated on 13/Mar/20

$${\log }_{10}x=u\wedge {\log }_{10}y=v$$ $$v=pu\wedge p,u\ne 0$$ $$\{\begin{array}{l}u-3+\frac{8p+1}{(p^2+1)u}=0\\ pu-\frac{p-8}{(p^2+1)u}=0\end{array}$$ $$\{\begin{array}{l}{u}^2-3u+\frac{8p+1}{p^2+1}=0\\ u^2-\frac{p-8}{p(p^2+1)}=0\end{array}$$ $$\left(2\right)-\left(1\right)\Rightarrow u=\frac{2(4p^2+p-4)}{3p(p^2+1)}$$ $$\mathrm{insert}\mathrm{in}\left(1\right)\mathrm{or}\left(2\right)\Rightarrow$$ $$p^4+\frac{104}{55}{p}^3-\frac{133}{55}{p}^2+\frac{8}{11}p+\frac{64}{55}=0$$ $$(p^2+\frac{4-(13+\sqrt{1049})}{55}p+\frac{123+\sqrt{1049}}{110})(p^2+\frac{4(13-\sqrt{1049})}{55}p+\frac{123-\sqrt{1049}}{110})=0$$ $$\mathrm{we}\mathrm{can}\mathrm{exactly}\mathrm{solve}\mathrm{this}$$

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