Determine $f$ such that for $a>1$ we have $f(a^x)\sim a^{x^2} for \ x\ \rightarrow\infty$.

Question

I'm looking for a function $f$ (if it exists) such that for $a>1$ we have $$f(a^x)\sim a^{x^2} for \ x\ \rightarrow\infty.$$ Clearly, polynomials are weak and the exponential function is so strong. Is there any function that we can construct between theme?. Thank you.

Answer 1

We can get $f(a^x) = a^{x^2}$. Just compose $a^{x^2}$ with the inverse of $a^x$ (i.e. $\log_a(x)$) on the right. That is, let $$f(x) = a^{(\log_a(x))^2}.$$

Answer 2

Good answer by @user744868. Also consider$$f(x)=x^{\log_a x}$$