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$$ \left\{\begin{array}{l} x^y = y^x \\ x-y\cdot\log_xy=(x+y)\cdot\log_xy \end{array} \right. $$

Thanks for your time!

Florin M.
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  • You got one answer and now you are already expecting another. Without even trying. – UrošSlovenija May 22 '13 at 11:32
  • @MFG Flay: I edtied your question. please feel free to edit it to your liking. – Dennis Gulko May 22 '13 at 11:32
  • Dear, UrosSlovenija. Not true, this exercies I've been trying to resolve for a while . Just today i've found about this forum, and that's the reason I'm posting the problemes. I guarantee you that this exercises gave me some headache ttrying to solve them. – Florin M. May 22 '13 at 11:33
  • But you can still give us some thoughts. If you tried hard then you should have some insight – UrošSlovenija May 22 '13 at 11:34

1 Answers1

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With rules for logarithms the second equation translates to $$ x - y\log_xy = \log_x y^x + y \log_x y $$ and using the first equation this simplifies to $$ x= y+2y (1+\log_xy) $$ and then $$ \log_xy = \frac{x-y}{2y} $$ From first equation you can express $$ y = x^{\frac{y}{x}} $$ implying $$ \frac{y}{x} = \frac{x-y}{2y} $$ which translates to $$ x^2 -xy - 2y^2 = 0 $$ and factoring this you get $$ (x+y)(x-2y)=0 $$ that gives $2$ solutions which, as usual with equation with logarithms, need to be checked. Try.

UrošSlovenija
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  • Thank you... And, of course, x,y can't be 1 , and x,y > 0. Right? – Florin M. May 22 '13 at 11:43
  • I've corrected the question, someone else edit it wrong – Florin M. May 22 '13 at 15:17
  • @MFGFlay Are you sure the equations are now correct? And where are you getting these problems? Homework or something else? – UrošSlovenija May 22 '13 at 16:45
  • UrosSlovenija, yes it is correct. While I was watching your demonstration I saw that something isn't wrong, because I tried to solve it many times but no success for me ... Then I checked the equation system and It was wrong, Now it is correct. Yep , it is homework – Florin M. May 22 '13 at 17:46
  • @MFGFlay i edited. check and get back to me – UrošSlovenija May 23 '13 at 06:23
  • Well, the first case: x=-y, is impossible, because x,y > 0. The 2nd case: x=2y => I obtained : y=2, x=4. I guess these are the solutions, thanks ! – Florin M. May 23 '13 at 08:44
  • Sorry, I checked it. I'm newbie on this forum discussion ... Thanks again! – Florin M. May 23 '13 at 09:06