Find $x,y \in \mathbb{R}$ such that \begin{align*} 2^x+3^y=7, \\ 2^y+3^x=11 \end{align*} Obviously, $(x,y)=(2,1)$ is a solution, but I can't prove that it's the only one. I tried using contradiction arguments, monotony, convexity and $\log$s, but didn't succed.

  • $\begingroup$ eliminate one variable from the given system $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '17 at 15:15
  • $\begingroup$ This is similar to the Ramanujan school problem. In case you are wondering what the problem was, it concerned to the solutions of : $$\sqrt {x}+y =7$$ $$\sqrt {y }+x =11$$ $\endgroup$ – Rohan Jan 8 '17 at 15:15
  • $\begingroup$ @Dr.SonnhardGraubner when I eliminate one variable, I get $\log_{2}3=\log_{7-3^y}(11-2^y)$ $\endgroup$ – Shroud Jan 8 '17 at 15:17
  • $\begingroup$ i got this here $${2}^{{\frac {\ln \left( -{2}^{x}+7 \right) }{\ln \left( 3 \right) }} }+{3}^{x}-11 =0$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '17 at 15:20

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