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$y\cdot x^y=x\cdot y^x$; $x\ne y$; $x,y\in\mathbb{R} $; $x,y>0$ has no solutions

$$y\cdot x^y=x\cdot y^x$$ $$\frac{x^y}{x}=\frac{y^x}{y}$$ $$x^{y-1}=y^{x-1}$$ $$\ln(x^{y-1})=\ln(y^{x-1})$$ $$(y-1)\ln x=(x-1)\ln y$$ $$\frac{\ln x}{x-1}=\frac{\ln y}{y-1}$$ $$x^{(x-1)^{-1}}=y^{(y-1)^{-1}}$$ On $x\in\mathbb{R} $; $x>0$; $Z=x^{(x-1)^{-1}}$, $Z$ does not have the same value for any $2$ $x$-values. This means $x=y$ are the only solutions, which aren't possible solutions. $\therefore$ there are no solutions.

How can I word this to make the proof more sound?

Jacob Claassen
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2 Answers2

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You can't prove something is that is false to be true.

$x=1, y=2$ is a solution.

In $\frac{\ln x}{x-1}$, you could have divided by $0$.

Remark:

Now, suppose you change the conditions to be $x \neq y; x, y > 1.$

You may want to prove that $\frac{\ln x}{x-1}$ is indeed monotonic more explicitly. For example, you may want to differentiate it and check that the sign is negative to show that it is decreasing.

Siong Thye Goh
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If $x=1$ we get $\{(1,t)|t>0,t\neq1\}$.

If $y=1$ we get $\{(t,1)|t>0,t\neq1\}$.

We'll prove that our equation has no more solutions.

Indeed, let $f(x)=\frac{\ln{x}}{x-1}$.

We see that $f'(x)=\frac{\frac{x-1}{x}-\ln x}{(x-1)^2}<0$ for all $x>0$, $x\neq1$,

which says that $$\frac{\ln{x}}{x-1}=\frac{\ln{y}}{y-1}$$ implies $x=y$, which is impossible.

Done!

Michael Rozenberg
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