An exponential symmetric system

Asked Dec 25 '23 at 19:01

Active Dec 25 '23 at 19:56

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I have this symmetric systeme with $x,y\in\Bbb R^+$:

$$\begin{cases} xy=8\\ x^y=y^x \end{cases}$$

Obviously the solution is the couple $(2,4)$ or $(4,2)$ being the symmetrical system. I have never solved such systems. Is there a method of solution to find real solutions?

edited Dec 25 '23 at 19:56

asked Dec 25 '23 at 19:01

Sebastiano

8,290

1

FWIW there’s also $x=y=\sqrt{8}$ – David Quinn Dec 25 '23 at 19:23
@DavidQuinn Is there a method to solve it? Thank you for your comment – Sebastiano Dec 25 '23 at 19:25
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Yes, there is one. Start with the second equation alone, see here. – Dietrich Burde Dec 25 '23 at 19:35
@DietrichBurde Thank you very much. – Sebastiano Dec 25 '23 at 19:38
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The linked dup is about integer solutions, for reals there are more solutions, this was a bit hasted closure. – zwim Dec 25 '23 at 19:54
@zwim reopen the question :-) – Sebastiano Dec 25 '23 at 19:55
@DietrichBurde Read the comment of zwim please. – Sebastiano Dec 25 '23 at 19:57
For real numbers it is also a duplicate, and the first equation then immediately determines the other real solutions. So it reduces to solving $x^y=y^x$. See for example this duplicate, or this one. – Dietrich Burde Dec 25 '23 at 20:26
What is the reason of two downvotes? – Sebastiano Dec 26 '23 at 21:38

An exponential symmetric system

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