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Solve these equations $$\log x+\frac{\log x+8\log y}{\log ^2x+\log^2y}=2$$ $$\log y+\frac{8\log x-\log y}{\log^2x+\log^2y}=0$$ Does an elegant solution exist? If not, how do I solve two cubic equations. See here How to solve this system of equations of degree 3?

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    $\begingroup$ The logs just complicate stuff. Take $X= \log x$ and $Y = \log y$ to get a much simpler looking system. Try to multiply the first equation by $X$ and the second by $Y$ and subtract the two equations. $\endgroup$ – Winther Apr 24 '15 at 15:15
  • $\begingroup$ That is what someone said on the previous question(see link). After that I get b in terms of a, it will he quite tedious to substitute back... $\endgroup$ – user167045 Apr 24 '15 at 15:20
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Let $\log x = u$ and $\log y = v$. Your system becomes

$$ \eqalign{u + \dfrac{u+8v}{u^2+v^2} &= 2\cr v + \dfrac{8u-v}{u^2+v^2} &= 0\cr}$$ Put the $2$ on the left side, and multiply by $u^2+v^2$: $$ \eqalign{{u}^{3}+u{v}^{2}-2\,{u}^{2}-2\,{v}^{2}+u+8\,v &= 0\cr {u}^{2}v+{v}^{3}+8\,u-v &= 0\cr}$$ The resultant of the two left sides with respect to $v$ is $$260\,u \left( u-3 \right) \left( u+1 \right) \left( {u}^{2}-2\,u+5 \right) $$ which must be $0$ for a solution. With $u=0$ we get $v=0$ (not a solution to the original system). With $u=3$ we get $v=-2$. With $u=-1$ we get $v=2$. With $u = 1 \pm 2 i$ (the roots of $u^2 - 2 u + 5$) we get $v = \pm 2 i$.

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