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Why does the graph $y=(\ln y/\ln x)x (-\ln y/\ln x (x^2/y))+x$ not include all positive $y$ and $x$ for which $y = x$? When I plotted this graph on desmos , the plot did not include the line $y = x$, even though the equation is solved for all $y = x$. This equation came up in the context of me trying to prove that the equations of the solutions to $x^y = y^x$ intersect at the point $(e,e)$

Arctic Char
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Zzyzx
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    Not clear. Does $y=(\ln y/\ln x)x -\ln y/\ln x (x^2/y)+x$ stands for: $$y=\frac{\ln y}{\ln x}x -\frac{\ln y}{\ln x (x^2/y)+x}?$$ Or does it stands for: $$y=\frac{\ln y}{\ln x}x -\frac{\ln y}{\ln x} (x^2/y)+x?$$ Other interpretations are feasible!!!! Please, clarify this point, your equation is really ambiguous. – the_candyman Nov 20 '20 at 00:45
  • It stands for the second one. Have edited the question to show this – Zzyzx Nov 20 '20 at 08:38
  • What makes you believe the graph given by desmos is correct? – Servaes Nov 20 '20 at 08:54

1 Answers1

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If we substitute $y=x$ in the equation $$y=-\frac{x^2 \ln y}{y \ln x}+\frac{\ln y}{\ln x}x+x$$ we get $x=-x+x+x\to 0=0$ undefined.

Thus the equation above is not equivalent to $$x^y=y^x\tag{1}$$

I strongly recommend to simply prove by direct inspection that $x=y$ is a set of solution of $(1)$, and use this answer to prove that $(e,e)$ is a self intersection.

Hope this helps

Raffaele
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  • I believe you misread my question. The equation was: y=-lnx/lny (x^2/y)+(lnx/lny)x+x. I think you missed the "x" in the second term of the equation – Zzyzx Nov 20 '20 at 13:26
  • @Zzyzx To a greater extent your equation is not equivalent to the exponential one – Raffaele Nov 20 '20 at 13:45
  • Thanks, this helped a lot. – Zzyzx Nov 23 '20 at 17:50
  • @Zzyzx I had the same curiosity about $x^y=y^x$, a few years ago :) https://math.stackexchange.com/questions/2342821/self-intersection-of-the-implicit-curve-xy-yx-0 – Raffaele Nov 23 '20 at 17:56