# Find two numbers $a,b$ such that $a^a=b^b$

Playing around on Desmos with the equation $$y=x^x$$, I noticed that the function, for numbers $$0, has two $$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 $$0, find $$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.

• 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. – fGDu94 Dec 3 '19 at 2:33
• @George Where does that function come from? How exactly does that help? – DonielF Dec 3 '19 at 2:37
• @Oscar Granted that your answer uses parametrics, the question doesn't. I don't feel that that tag is warranted here. – DonielF Dec 3 '19 at 3:06
• Ok, couldn't think of anything better. – Oscar Lanzi Dec 3 '19 at 3:09

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 $$b=2^{2/(1-2)}=1/4, a=2^{1/(1-2)}=1/2$$.

• 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. – DonielF Dec 3 '19 at 3:03
• Try putting $t=-1$. You get no real solutions, but you do get an elegant result. – Oscar Lanzi Dec 3 '19 at 13:39
• $a=i, b=\frac1i$? That’s just making my head hurt trying to figure out how to deal with that. :) – DonielF Dec 3 '19 at 14:28
• $i^i=(-i)^{(-i)}$! – Oscar Lanzi Dec 3 '19 at 19:01

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

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