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I was searching online for some difficult Calculus problems and this one stumped me.

An example of a pair of numbers would be $\left(\frac{1}{2}, \frac{1}{4}\right)$ because

$$\left( \frac{1}{2} \right)^{\frac{1}{2}} = \left( \frac{1}{4} \right)^{\frac{1}{4}} $$

In the question I was looking at, there was hint given that said to consider the function $f(x) = x^x$ for $x > 0$, and to focus on the interval $(0, 1]$.

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Formally, you will be using the Intermediate Value Theorem. –  André Nicolas Feb 12 at 23:47

3 Answers 3

up vote 7 down vote accepted

Consider the function $f(x)=x^x=e^{x\log x}$. Its derivative is $$ f'(x)=e^{x\log x}\left(\log x+x\frac{1}{x}\right)=x^x(1+\log x) $$ Can you argue the intervals where $f$ is increasing or decreasing? What can you conclude?

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$$a^a=b^b\iff\bigg(\frac1A\bigg)^\frac1A=\bigg(\frac1B\bigg)^\frac1B\iff\frac1{\sqrt[A]A}=\frac1{\sqrt[B]B}\iff\sqrt[A]A=\sqrt[B]B$$ Now, plot $f(x)=\sqrt[x]x$, and tell me what you notice. :-)

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A variation on @Lucian's answer: $$a^a=b^b\quad\Leftrightarrow\quad a\ln a=b\ln b\ .$$ Now plot $f(x)=x\ln x$ and see what you notice.

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