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Is $a^b$ larger than $b^a$ if $a<b$ and $a,b > 1$?

I tried this out for a few numbers and this seems to be the case. If this is true, could you show me a proof? I would be very interested. If this is not true, can you give me a counter example?

I tested this for some powers, and an example is: $4^7 > 7^4$ as $16384 > 2401$

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Take logs, to get $b\ln a>a\ln b$, or $\frac{\ln a}a>\frac{\ln b}b$.
The function $\frac{\ln x}x$ increases to a maximum at $x=e=2.71828...$, then decreases.
So your question is true if $e<a<b$
If you don't like logs, take the $ab^{th}$ root of both sides, and get $\sqrt[a]a>\sqrt[b]b$ when $e<a<b$.

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  • $\begingroup$ Thanks for your explanation; as I was working in the domain of positive natural numbers, this comes down to making sure that $a > 2$ $\endgroup$ – Héctor van den Boorn Jul 15 '15 at 16:15
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Counter example: $2^3$ and $3^2$ and see what happens

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Not necessarily. Consider $a=2$ and $b=4$ as a case where equality happens as each is 16.

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You can always make it work, if you fix your $b$ than you can always find an $a_0$ such that your inequality holds $\forall a\geq a_0$, please check it here.

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