$x^y < y^x$ for $y\ll x$? Sorry if this is a naive question; I am not very good at mathematics.
It seems obvious that for many $x$ and $y$, $x^y < y^x$ if $y \ll x$, e.g. $2^{10} > 10^2$. If $x$ and $y$ are very close to each other, this does not hold (simple case: $2^3 < 3^2$).
So - is there really such a regularity? Does it have a name? Is there an exact rule for which $x$ and $y$ it holds (let us, for simplicity, assume both are natural numbers)?
 A: This is an interesting problem. Since you say you're "not very good at mathematics", I'll give the result first and then explain how I came up with it.

If $x$ and $y$ are greater than $2.71828\dots$, then $x^y < y^x$ if and only if $x > y$.
If $x$ and $y$ are less than $2.71828\ldots$, then $x^y < y^x$ if and only if $x < y$.
If $x$ is less than $2.71828\ldots$ and $y$ is greater than $2.71828\ldots$ (or vice versa), you should compute $x^y$ and $y^x$ to see which is bigger.

The number $2.71828\ldots$ is Euler's constant $e$. This result explains why strange things happen around $2$: as you noticed, $2^4 = 4^2$ and the two number $2^3$ and $3^2$ are very close. But in general, if $x$ and $y$ are close to each other but "far away" from about $2.71828\ldots$ then you can just compare them directly to see if $x^y < y^x$.
In more detail, the inequality $x^y < y^x$ is equivalent to $y\ln x < x\ln y$, by taking logarithms. And that inequality is equivalent to $(\ln x)/x < (\ln y)/y$, by dividing by $xy$. So we can reformulate your original problem to the problem of determining whether the function $f(t) = (\ln t)/t$ is increasing or decreasing. You can show, using calculus, (I'll supply the details if you would like them), that the function $f(t)$ is increasing when $t < e$ and decreasing when $t > e$.
A: Take $\ln(\cdot)$ on both sides of the inequality $x^y < y^x$ and then separate the variables, you get the inequality $\ln(x) / x < \ln(y) / y$. So it comes down to investigating where the function $f(x)=\ln(x)/x$ is decreasing, which has a very simple and satisfying answer using calculus.
A: The function $f(x) = \frac{\log x}x$ has one local maximum which is also a global maximum at $x = e$ (considering $(0, \infty)$ as the domain). For the purpose of integer calculcation, we need to consider $2$ and $3$ as they are the integers surrounding $e$. From $f(2) = f(4)$, we know $f(3)$ is the maximum when the domain is restricted to positive integers. From this knowledge, we know that there are only two cases where the strict inequality does not hold: $2^3 < 3^2$ and $2^4 = 4^2$.
