Question

Which function dominates x^p or p^x

Answer 1

I'm assuming
`p\in\mathbb R` is fixed. First note that
`x^p=p^x` when
`x=p`.

We can rewrite each expression as


`x^p=e^(p\ln x)`

`p^x=e^(x\ln p)`

Now to compare we only need to consider the exponents. It's easy to show that
`x>\ln x` for all
`x>0`, so it follows that
`x^p<p^x` so long as
`x>p`.