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Playing around on Desmos with the equation $y=x^x$, I noticed that the function, for numbers $0<x<1$, has two $x$ values for every $y$ value. For instance, $\frac12^{\frac12}=\frac14^{\frac14}$, which we can rewrite as $\sqrt[2]\frac12=\sqrt[4]\frac14$.

What I'm trying to find is: Given some number $a$ such that $0<a<1$, find $b$ such that $\sqrt[a]\frac1a=\sqrt[b]\frac1b$.

Is there a general formula for this? I've tried using Newton's Method, but that just approximates the result; it doesn't give an exact answer.

I'm struggling with tagging this question properly, so I'd appreciate some help with that as well.

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  • $\begingroup$ You could try plotting $f(x,y) = (x^x-y^y)^2$ as a surface above $\mathbb{R}^2$. This would be minimum at solutions to your equation. $\endgroup$ – fGDu94 Dec 3 '19 at 2:33
  • $\begingroup$ @George Where does that function come from? How exactly does that help? $\endgroup$ – DonielF Dec 3 '19 at 2:37
  • $\begingroup$ @Oscar Granted that your answer uses parametrics, the question doesn't. I don't feel that that tag is warranted here. $\endgroup$ – DonielF Dec 3 '19 at 3:06
  • $\begingroup$ Ok, couldn't think of anything better. $\endgroup$ – Oscar Lanzi Dec 3 '19 at 3:09
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Let $t=a/b$ be a positive number other than $1$. Then

$(bt)^{bt}=b^b$

$(bt)^t=b$

$bt=b^{1/t}$

$t=b^{(1-t)/t}$

$\color{blue}{b=t^{t/(1-t)}}$

From this we then get

$\color{blue}{a=bt=t^{1/(1-t)}}$

For instance, if $t=2$ then $b=2^{2/(1-2)}=1/4, a=2^{1/(1-2)}=1/2$.

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    $\begingroup$ Your final equality holds for all positive $t\ne1$, where you get a divide-by-zero error in the exponent, where for every solution $a,b$ given some $t>1$, there's an equivalent solution $b,a$ for every $0<t<1$. What's really interesting, though, is that your equality holds even for negative $t$, if you avoid picking $t$ such that you end up with an imaginary answer, yielding a negative $a$ and positive $b$ (with the same feature that for every solution $a,b$ for $t<-1$, there's an equivalent solution $b,a$ for every $-1<t<0$). Thank you so much for this; you've given me much to work with here. $\endgroup$ – DonielF Dec 3 '19 at 3:03
  • $\begingroup$ Try putting $t=-1$. You get no real solutions, but you do get an elegant result. $\endgroup$ – Oscar Lanzi Dec 3 '19 at 13:39
  • $\begingroup$ $a=i, b=\frac1i$? That’s just making my head hurt trying to figure out how to deal with that. :) $\endgroup$ – DonielF Dec 3 '19 at 14:28
  • $\begingroup$ $i^i=(-i)^{(-i)}$! $\endgroup$ – Oscar Lanzi Dec 3 '19 at 19:01
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Let $f(x)=x^x$ $$\ln(f(x))=x\ln x$$ $$\frac{f'(x)}{f(x)}=(1+\ln x)$$ $$f'(x)=f(x)\ln(ex)$$ $$\text{For }0<ex<1,f'(x)\lt0$$ $$\text{For }ex>1,f'(x)\gt 0 $$ $$\text{Hence, if }a\in\left(0,\frac1e\right),b>\frac1e$$

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  • $\begingroup$ How does this help? This doesn't give you an actual pairing of numbers, just what range they're in. $\endgroup$ – DonielF Dec 3 '19 at 3:21
  • $\begingroup$ This says for every a $\in {(0,\frac{1}{e})}$ there exist exactly one b$\gt 1$ that will satisfy the equation. $\endgroup$ – mathsdiscussion.com Dec 3 '19 at 3:26
  • $\begingroup$ That's fine, but it doesn't say that for a specific $a$ how to find that $b$ which will satisfy the equation. $\endgroup$ – DonielF Dec 3 '19 at 3:30
  • $\begingroup$ For that let b=$a^x$ now you end up with equation $$xa^x-a=0$$ now take different a and x you will get b. $\endgroup$ – mathsdiscussion.com Dec 3 '19 at 3:34
  • $\begingroup$ Great, but that's not what it says in your answer. $\endgroup$ – DonielF Dec 3 '19 at 3:36

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