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Is there a way to compare $a^b$ and $b^a$? If a,b both are greater the $e$( or less than), then one may consider the function $f(x)=lnx/x$. But how to compare these two numbers in general?

Fermat
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  • You may wish to have a look here though it answers only part of your question- https://math.stackexchange.com/questions/1771348/if-xy-yx-show-that-xy2e/1771441#1771441 – MathematicianByMistake Nov 21 '17 at 17:06

2 Answers2

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Nothing is stopping you from using $f$ on any pair of positive numbers. Actually, it's only if one number is larger than $e$ and one number is smaller you really need to use $f$.

If the two numbers are on the same side of $e$, then using the number closest to $e$ as base necessarily gives the largest result (although you would use $f$ to prove this).

Arthur
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  • Thank you for your answer. You are right. But if the two numbers dont lie on the same side of e, what function is good to be considered? – Fermat Nov 21 '17 at 17:02
  • I just said: the $f$ you have in your question works. Use whichever number makes $f$ largest as the base. – Arthur Nov 21 '17 at 17:26
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There are the two well-known inequalities you mentioned.

In general it is impossible to compare these numbers:

$2>1$ and $2^1>1^2$

$5>2$ and $5^2<2^5$

$2\neq 4$ but $2^4=4^2$

tong_nor
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