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This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this?

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  • $\begingroup$ This problem is well solved. The self-intersection occurs when $y=x=e$. See this and this. $\endgroup$ – Ng Chung Tak Apr 25 '17 at 12:08
  • $\begingroup$ Yes I know, but both of the links display solutions using logarithmic manipulation and differentiation, which was not supposed to be used here. $\endgroup$ – Parseval Apr 25 '17 at 12:10
  • $\begingroup$ The answer is the point $(e,e)$, and the number $e$ is inseparably linked to logarithms and differentiation. It seems very unnatural to explicitly avoid those concepts. Why would you want to? $\endgroup$ – Arthur Apr 25 '17 at 12:12
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    $\begingroup$ Just a curious chap. I like math and I just like exploring different solutions to problems. $\endgroup$ – Parseval Apr 25 '17 at 12:16
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Here's a method that uses logarithms, but not differentiation.

Of course, one portion of the line is simply that of $y = x$.

Following the method here, we find that the other portion of the curve $x^y = y^x$ for which $x \neq y$ is parameterized by $$ (a^{1/(a-1)},a^{a/(a-1)}), \quad a \neq 1 $$ It stands to reason (given that the curve intersects) that the point of this intersection is $$ \lim_{a \to 1}(a^{1/(a-1)},a^{a/(a - 1)}) $$ So, it suffices to verify the limits $$ \lim_{a \to 1} a^{1/(a-1)} = \lim_{a \to 1} a^{a/(a-1)} = e $$ Note that if we rewrite these limits in terms of $b = 1/(a - 1)$ (or conversely $a = 1/b + 1$) and consider only the limit from the right, we're really evaluating $$ \lim_{b \to \infty}\left(1 + \frac 1b \right)^b \qquad \lim_{b \to \infty} \left(1 + \frac 1b\right)^{b + 1} $$ which both come out to $e$, by whichever definition of $e$ you prefer.

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Ok the function $f\colon \mathbb R^2 \to \mathbb R$ is continuous. And $f(1,2)=-1$ and $f(2,1)=1$ so it has to take the values between -1 and 1. Thus for some point we have that $f$ is zero.

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  • $\begingroup$ It's zero in the point $(e,e)$, which contradicts your conclusion? $\endgroup$ – Parseval Apr 25 '17 at 12:32
  • $\begingroup$ No - I'm saying that $f$ is zero - somewhere. It might very well be $(e,e)$ :o) $\endgroup$ – Martin Apr 25 '17 at 12:35
  • $\begingroup$ So If I ask you where I can by milk, instead of answering "Walmart 2 blocks away" you would answer: "on Earth somewhere"? :P $\endgroup$ – Parseval Apr 25 '17 at 12:37
  • $\begingroup$ No - you asked me "are there anywhere I can buy som milk?", and I answred (annoyingly) "Yes!" ;o) $\endgroup$ – Martin Apr 25 '17 at 12:40
  • $\begingroup$ Ok- read the question again, and you do seem keen to know where ;o) It's a matter of what one means by solving. $\endgroup$ – Martin Apr 25 '17 at 12:42

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