Does $A>B$ imply $A^B

Question

Does $A>B$ imply $A^B<B^A$? A naive doubt, but I cannot find a proof. Does the property always hold true?

$A>B$ does not necessarily imply that $A^B<B^A$. How do we know if $A^B<B^A$ holds true or not? In other words, can we find $x$ such that for all $y$ with $x \leq y$, the inequality $x^y > y^x$ holds true?

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score 2 · Answer 1 · answered Oct 09 '13 at 18:29

2

HINT:

Assuming $A>B>0$ $$A^B< B^A\iff A^{\frac1A}<B^{\frac1B}$$

Let $f(x)=x^{\frac1x}$

Can you check when $f(x)$ is increasing

answered Oct 09 '13 at 18:29

lab bhattacharjee

279,016

@MayankJha, But this method does not come handy if one of $A<e<B$ To generalize Asaf Karagila's answer $$2^n \begin{cases}>n^2 &\mbox{if } n>4 \ =n^2 & \mbox{if } n=4\ <n^2 & \mbox{if } n<4 \end{cases}$$ as proved here(http://math.stackexchange.com/questions/497092/proof-by-induction-2n-n2-for-all-integer-n-greater-than-4) – lab bhattacharjee Oct 11 '13 at 12:11

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

Note: This answer was given to the original version of the question.

Not at all. Two examples:

$$2^4=4^2\\2^3<3^2$$

edited Apr 13 '17 at 12:20

Community

1

answered Oct 09 '13 at 18:22

Asaf Karagila

405,794

yeah. but in which cases does this hold true ? – Mayank Jha Oct 09 '13 at 18:26
Fix x. Then compare $x^a$ with $a^x$... – The Chaz 2.0 Oct 09 '13 at 18:27
The Chaz, is there some quicker method wherein we can find some a such that for all y>=a the inequality pow(x,y)>pow(y,x) ? – Mayank Jha Oct 09 '13 at 18:31

score 0 · Answer 3 · answered Oct 09 '13 at 18:31

It is true as long as $A,B$ are large enough. If you take logs, you are asking whether $A \gt B \implies B \log A \lt A \log B$ or $\frac B{\log B} \lt \frac A{\log A}$ Since $\log$ grows so slowly, this will be true whenever $A,B$ are reasonable size. Taking the derivative of $\frac x{\log x}=\frac {\log x-1}{\log^2 x}$, which is positive for $x \gt e$, shows this will be true any time $A,B \gt e$

Does $A>B$ imply $A^B

3 Answers3