For $a\neq b,0<a<1,a\in\mathbb R$, find an $a$ such that there is no $b\in\mathbb R$ for $a^a=b^b$

For $a\neq b,0<a<1,a\in\mathbb{R}$, find an $a$ such that there is no $b\in\mathbb R$ for $a^a=b^b$.

Not sure where this belongs.

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I assume you want $b\ne a$. – anon Apr 12 '12 at 22:11
Definitely not number-theory. Perhaps calculus is better. – Aryabhata Apr 12 '12 at 22:33

Try thinking about $x^x$ as a function. If you can find a maximum or minimum of this function in the range $(0,1)$, then it would satisfy the desired property.

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Use the answers here: Where is the highest point of $f(x)=\sqrt[x]{x}$ in the $x$-axis?

to show that for $x \neq e, x \gt 0$,

$$e^{1/e} \gt x^{1/x}$$

Take the reciprocal.

Of course, you could apply the techniques there(like finding critical points) to the current problem too, but they are already worked out there in great detail.

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In fact, I am tending towards casting a close as dupe vote (and convert this answer to a comment). – Aryabhata Apr 12 '12 at 22:41

Indeed, we should consider a function $x^x$. It looks a bit parabolic on interval $(0,1)$. So we can consider it has min or max. So we need to evaluate first and second order derivatives in order to see the possibility of obtaining minimum or maximum.

$(x^x)'=x^x\cdot(\ln(x)+1)$

$(x^x)''={x}^{x} \left( \ln \left( x \right) +1 \right) ^{2}+{\frac {{x}^{x}}{ x}}$

Now we have to solve $x^x\cdot(\ln(x)+1)=0$ the answer is $x=\frac{1}{e}$. This the minimum of the function considering change of the sign from - to +.

Since the second order derivative is positive on the given interval, you have your $a=\frac{1}{e}$.

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