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I've been struggling with this problem, of finding all real solutions x,y, to the following system of two nonlinear equations with logarithms. Would really appreciate help with solving the problem. Thanks in advance!

Here's the system:

$$ x^2+3x+ln(2x+1)=y $$ $$ y^2+3y+ln(2y+1)=x $$

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  • $\begingroup$ What is with $$x=y=0$$? $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '18 at 15:01
  • $\begingroup$ MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \ln x gives $\ln x$ in contrast to ln x which gives $ln x$ $\endgroup$ – Ross Millikan Sep 16 '18 at 15:10
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Define for $x\ge 0$ $$f(x)=x^2+3x+\ln(2x+1)$$ The solutions of the system are pairs of the form $(x,f(x))$ where $f(f(x))=x$. So let's define $$g(x)=f(f(x))$$ As S. Graubner has noted, $f(0)=0$, so $(0,0)$ is a solution. On the other hand, $$g'(x)=f'(x)\cdot f'(f(x))$$ Since $f(x)\ge 0$ and $f'(x)\ge 3$, we have that $g'(x)\ge 9$, so $g(x)-x$ has no more zeros in $(0,\infty)$.

Note: details are deliberately missing.

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