Which is larger :: $y!$ or $x^y$, for numbers $x,y$.

Question

This is a generalization of this question :: Which is larger? $20!$ or $2^{40}$?.

No explicit general solution was presented there and I'm just curious :D

Thank-you.

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Edit :: I want a most-general solution lfor arbitrary $x$ and $y$; not some specfic cases which can then be solved by direct computation. Below, Ahaan-Rungta shows that the case $x < y$ is the one to be considered.

Answer 1

"Just compute them and compare" is the only fully failsafe method.

In most cases, however, estimating the logarithms of both of $x^y$ and $y!$ using Stirling's formula will yield a conclusive result without needing to compute the two values in high detail.

Answer 2

For a fixed $x,$ $y!$ will eventually be larger, as you can tell by using Stirling's formula.

Answer 3

Hint:

In general, for positive integer $x,y$ we have:

$y!=y(y-1)(y-2) \cdot\cdot\cdot2\cdot1$

$x^y=xxx\cdot\cdot\cdot xx$

1. If $x\geq y$; then $ x^y\ge y! $
2. If $x<y $, ?! (Now this is the question!)

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Extra:

If $ \left\lceil{\frac{y}{2}}\right\rceil+1< x<y$; then $x^y\ge y!.$ (easy proof)

Now, we have only the case $ x\le\lceil{\frac{y}{2}}\rceil +1 $

Answer 4

$$\frac{20!}{2^{40}}= \frac{20!}{4^{20}} = \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right) \left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)\cdots \left(\frac{20}{4}\right)>\cdots$$

$$ \left(\frac{1}{4}\right)\left(\frac{2}{4}\right)\left(\frac{3}{4}\right) \left(\frac{4}{4}\right)\left(\frac{5}{4}\right)\left(\frac{6}{4}\right)\left(\frac{7}{4}\right)\left(\frac{8}{4}\right)\left(\frac{9}{4}\right)=\frac{9!}{4^9}=\frac{2835}{2048}>1,$$

so $20!>2^{40}$.