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.

  • $\begingroup$ 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

Let $t=a/b$ be a positive number other than $1$. Then






From this we then get


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

  • 1
    $\begingroup$ 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

Let $f(x)=x^x$ $$\ln(f(x))=x\ln x$$ $$\frac{f'(x)}{f(x)}=(1+\ln x)$$ $$f'(x)=f(x)\ln(ex)$$ $$\text{For }0<ex<1,f'(x)\lt0$$ $$\text{For }ex>1,f'(x)\gt 0 $$ $$\text{Hence, if }a\in\left(0,\frac1e\right),b>\frac1e$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.