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Define $M=\{(x,y) \in \mathbb{R^2}\space|\space x^y=y^x , x>0, y> 0,(x,y)\neq(e,e)\}$. Show that $M$ is a one-dimensional submanifold.

Here's what I have done so far.

Define $$A=\{(x,y) \in \mathbb{R^2}\space|\ x>0, y> 0(x,y)\neq(e,e)\}$$ and $$F:A\longrightarrow\mathbb{R}$$

by putting $F(x,y)=x^y-y^x$.

Then $M=F^{-1}(0)$, and $\nabla F=(yx^{y-1}-y^x\ln y, \space x^y\ln x-xy^{x-1})$.

If we prove that $\nabla F (x,y)\neq (0,0)$ for all $(x,y) \in A$, we are done. Of course, $\nabla F(e,e)=(0,0)$, but $(e,e) \notin A$. My question is:

How to show that the equation $(yx^{y-1}-y^x\ln y, \space x^y\ln x-xy^{x-1})=(0,0)$ has no other solutions?

Célio Augusto
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Note that $M$ is equal to $$M=\lbrace x\in A~\vert~y\ln(x)=x\ln(y)\rbrace,$$ so you should probably take $F(x,y)=y\ln(x)-x\ln(y)$, because the gradient is simpler: $$\nabla F(x,y)=(y/x-\ln(y),\ln(x)-x/y).$$ The equation $\nabla F(x,y)=0$ is equivalent to the system $$\left\lbrace\begin{array}{l} \ln(y)=y/x\\ \ln(x)=x/y \end{array}\right.$$ Using $y=x/\ln (x)$ and the first equation you get $$ \ln(x)-\ln(\ln(x))=1/\ln(x).$$ Taking $u=\ln(x)$ you can rewrite this as $u-\ln(u)-1/u=0$, and by differentiating you see that $$v\mapsto v-\ln(v)-1/v$$ is strictly increasing so $u=1$ which means $x=e$, and then $y=x/\ln(x)=e$. Hence $(e,e)$ is the only solution and you can conclude.

Adam Chalumeau
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