I saw here on math.stackexchange.com an equation which has very nice solutions (by solutions I mean a proof): $3^x+28^x=8^x+27^x$, where $x$ is a real number. However, I think there must be an elementary solution. Has someone an idea? Maybe considering a function and seeing how the first derivative and second derivative behave. I think we should divide the equation with $3^x$ to work with strictly increasing functions. I know that x=0 and x=2 are solutions! They are also the only ones! Ther exists a proof using the rules of Descartes and one simply on geneal case. Is there anothr way to prove the solutions are only 0 and 2? I don't know why but my question was erased...

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    $\begingroup$ This question is posted here yesterdey. $\endgroup$ – Adi Dani Jan 8 '14 at 16:28
  • $\begingroup$ $x=2$ is a solution. $\endgroup$ – mathlove Jan 8 '14 at 16:38
  • $\begingroup$ See also: See also: Solve $3^x + 28^x=8^x+27^x$ $\endgroup$ – Martin Sleziak May 13 '17 at 12:41

Notice that $0$ is a solution for. We can observe that the equation can be transformed like $$3^x+3^{x\log_328} = 3^{x\log_38}+3^{3x}$$ I'm trying to do something with this idea... Ok, let's go!

Observe that $3^x+3^{x\log_328} \sim 3^x+3^{x\log_327} = 3^{4x}$ and that $3^{x\log_38}+3^{3x} \sim 3^{2x}+3^{3x}=3^{5x}$ when $x$ tends to + infinity. Take the function $$f(x) =3^{-x}$$and notice that $$f(x) \sim 3^x+3^{x\log_328} - 3^{x\log_38}-3^{3x}$$ $f$ is estrictly decreasing on $(2, \infty)$. I believe that this is the reason for no more solutions that $0$ and $2$.

  • $\begingroup$ Hey! I know the solutions. I asked if there exists another way to prove that they are the only? $\endgroup$ – Tommy T. Jan 8 '14 at 17:15

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