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November 4th, 2017, 11:05 PM   #1
yli
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Need help interpreting and proving this problem.

I'm unsure about how to approach this question. It is confusing me because b is supposed to be larger than a, but the example has two values that equal each other. Any help is appreciated.
In case the picture is difficult to read I have typed the question out as well:

Note that:
$\displaystyle (\frac{1}{2})^\frac{1}{2}=(\frac{1}{4})^\frac{1}{4 }$
Explain why there are infinitely many pairs of numbers a < b such that
$\displaystyle a^a$ = $\displaystyle b^b$.
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November 5th, 2017, 01:11 PM   #2
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If you look at the curve for 0<x<1, you will see a curve with a minimum at x=1/e, so that there are pairs of x's where the function has the same value.
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November 5th, 2017, 01:38 PM   #3
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as they suggest consider $y=x^x$

This function has a minimum of $e^{-1/e}$ at $x = \dfrac 1 e$

and it rises to $(0,1)$ to the left, and off to infinity to the right.

So you can set a horizontal line $y = y_0,~y_0 \in \left(\dfrac 1 e,~1\right]$

and intersect $y=x^x$ in two places, one, $a$, to the left of $\dfrac 1 e$, and $b$, to the right of it.

and from this $a^a = b^b = y_0$

clearly there are an infinite number of $y_0 \in \left(\dfrac 1 e, ~1\right]$ and thus an infinite number of $(a,b)$ pairs.
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