The number $16$ can be represented as $2^4=4^2$. Is there a (real) number after $16$ that can be represented as $A^B=B^A$? Where A and B are two distinct real numbers.I've checked many numbers and I'm convinced that there are no other numbers with such property, I conjecture that $16$ is the only real number with such property. But maybe I'm wrong. Is it true that $16$ is the only number with such property ? (My question here is much more specific than the other similar question has been posted.)
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Is the number required to be an integer or just a real number $\gt16$? – awllower Jul 13 '16 at 12:02
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1See also http://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y if you just want integers. – Najib Idrissi Jul 13 '16 at 12:03
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I mean any real numbers : ) – sumpelo miyapah Jul 13 '16 at 12:04
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1I think my question is much more specific than the other question. – sumpelo miyapah Jul 13 '16 at 12:07
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"Is it true that 16 is the only number with such property ?" - No, at least $1$ and $4$ are also with such property. – barak manos Jul 13 '16 at 12:07
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@barakmanos $1^4\ne4^1$. here I mean 2 distinct real numbers A and B – sumpelo miyapah Jul 13 '16 at 12:08
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@sumpelomiyapah: $1^1=1=1^1,2^2=4=2^2$. And in fact, every positive number has that property. – barak manos Jul 13 '16 at 12:08
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(DAA..) 2 distinct real numbers A and B – sumpelo miyapah Jul 13 '16 at 12:10
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Then you should by the least rephrase that last part of the question (namely, the part where you state "$16$"). – barak manos Jul 13 '16 at 12:10
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$16=2^4=4^2$. two distinct real numbers A and B – sumpelo miyapah Jul 13 '16 at 12:15
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@sumpelomiyapah: This property is not a property of $16$, but a property of the pair $[2,4]$. – barak manos Jul 13 '16 at 12:16
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@barakmanos, Number $16$ has a property that.... (you guess it) – sumpelo miyapah Jul 13 '16 at 12:17
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1BTW, it is also one of the only two numbers with the property of being equal to the sum of squares of its prime factors ($16=2^2+2^2+2^2+2^2$, and the other one being $27=3^2+3^2+3^2$). – barak manos Jul 13 '16 at 12:19
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@barakmanos, That's also nice, but that's out of context. – sumpelo miyapah Jul 13 '16 at 12:21
2 Answers
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Yes there are such numbers.
By this question, we know that if $x^y=y^x$ and $x\ne y,$ then $x=a^{1/(a-1)},y=a^{a/(a-1)}$ for some $a\not=1.$
Then we see that $$y\ln x=a^{a/(a-1)}\ln a/(a-1)\gt\ln a\frac{a}{a-1}\gt\ln a,$$ when $a\gt1.$ So if we choose $a\gt16,$ then $x^y=y^x$ so produced will satisfy the requirement.
For example, taking $a=100,$ we see $x=100^{1/99}\cong 1.047615, y=100^{100/99}\cong104.761575$ and $x^y=y^x\cong130.726237\gt16.$
Hope this helps.
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Quoting the answer: "However, I'm going to go a step further. I will say that I want my solution $(x,y)$ of $x^y = y^x$ to be of the form $(x,ax)$ so that $y = ax$ for some value $a$. Furthermore, I'll assume that $x \neq y$ so that $a \neq 1$ (of course, $x = y$ is always a solution if $x^x$ is defined). With that, we have... " – awllower Jul 13 '16 at 13:47
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Does this kind of number has been named by someone ? Partner numbers ? : ) – sumpelo miyapah Jul 13 '16 at 13:52
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Yes. The power can be larger than $16$.
Consider the set of points on the plane satisfying the equation
$$x^y=y^x$$
You see that those points form a union of two curves: the line $x=y$ and the more interesting part.
By implicit differentiation we get that on that other curve we have $$ \frac{dy}{dx}=\frac{y(x\ln y-y)}{x(y\ln x-x)}. $$
Using that and logarithmic differentiation we get that on that curve
$$ \frac d{dx}\,y^x=y^x\left[\ln y+\frac{x\ln y-y}{y\ln x-x}\right]. $$
Plugging in $x=4, y=2$ shows that at the point $(x,y)=(4,2)$ we arrive at
$$\frac{d(y^x)}{dx}=\frac{16 (1-\ln2\ln4)}{2-\ln4}\approx1.019.$$
Therefore when $x$ is a little bit larger than $4$ the power $y^x=x^y$ will be a little bit larger than $16$.
I hazard a guess that along that curve $y^x=x^y$ attains a local minimum $e^e\approx 15.15$ at the point $x=y=e$. The formula for the derivative breaks down there (which may explain why Mathematica left a tiny gap there in the above figure).
Jyrki Lahtonen
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