$x^y=y^x$

Consider the below expression-

$x^y=y^x$

Clearly, this expression has infinitely many trivial (obvious) solutions of the form $x=y$

But what about solutions not of this form i.e. $x\neq y$ . Intuitively, I was able to see that the only solution in the positive integers was (4,2) and (2,4). But I was not able to come up with a rigorous mathematical proof (any help would be greatly appreciated).

But what about solutions in the domain of real numbers or even complex numbers. Do there exist solutions of some generic form? Are there finite or infinitely many solutions? If there do exist solutions, how do we find them (and prove them)?

Thanks for any help or suggestions.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Hint: What is the behavior of the function $f(x) = x^{1/x}$ for $0<x<1$ ?

Pi Han Goh - 3 years, 1 month ago

The trick that will help finding the generator for the solutions is subsituting y=vx this gives us:

(vx)^x = x^vx <=> (vx)^x = (x^v)^x

now we take both sides to the power of 1/x and devide by x which gives us:

v=x^v-1

we rearrange that to solve for x: x= v^(1/v-1)

thus : y=(v^(1/v-1)) :v

Zahraa Bazo - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

xy=yxx^y=y^xxy=yx

Comments

$x^y=y^x$