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Please help solve the system of equations. $$x^{x+y} =y^n$$ $$y^{x+y} =x^{2n}\cdot y^{n} $$ Where it is given that$ n=0$. I tried to solve it by taking log both sides but couldn't arrive at any result.
My attempt :(
Taking log both sides $$ (x+y) \log x = n\cdot \log y$$ And then I divided both the equations to get $${(\frac{x} {y}) } ^{x+y} =\frac{1} {x^{2 n}} $$ Taking log both sides $$ (x+y)( \log x-\log y)= \frac{-\log x} {2n}$$ But I can't find a way to proceed from here. Please help

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  • $\begingroup$ If it is given that $n = 0$, it is trivial to solve. $y^0 = 1$ and $x^{0}y^{0} = 1$ $\endgroup$ – automaticallyGenerated Jul 6 '19 at 17:01
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Answer: $$x=\frac{-1+\sqrt{1+8n}}{2},~~n>0, ~~\mbox{and}~~ y=x^2.$$ Proof:

$$x^{x+y}=y^n~~~~(1),~~~~ y^{x+y}=x^{2n}y^n~~~~(2)$$ Here $x,y,n>0$, multiply (1) and (2) to get $$(xy)^{x+y}=(xy)^{2n} \Rightarrow x+y=2n~~~~(3).$$ Use (3) in (1) to get $$x^2=y~~~(4)$$ From (3) and (4), we get $$x^2+x-2n=0 \Rightarrow x=\frac{-1+\sqrt{1+8n}}{2},~~n>0$$

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  • $\begingroup$ Don't forget $x = y = 1$. $x = y = 1$ is a solution whether $n > 0$ or $n < 0$. In fact it is the only solution if $n < 0$. $\endgroup$ – automaticallyGenerated Jul 6 '19 at 16:59
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$(x+y)\log x = n\log y$ and $(x+y)\log y = 2n\log x + n\log y$

Dividing the first equation by the second,

$$\frac{\log x}{\log y}=\frac{\log y}{2\log x+\log y}$$

Putting $\frac{\log x}{\log y}=t$, we have $t=\frac{1}{2t+1}$ or $(2t-1)(t+1)=0$.

Substituting $x=y^t$ and $x+y=\frac nt$, it should be easily solvable from here.

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