Say you have $\Phi(x)=2^{sA_n}$, for some fixed parameter $s\in \Bbb R$, $n \in \Bbb Z^+ $ with $A_n=1/\log(p_n(x)) $, where $p_n(x) $ is a polynomial with degree $n$.
How many forms $p_n(x)$ yield concave $\Phi$, for all $s$, for, $(-\infty<x<\infty)$ and $\Phi<1.$ So for example $A_1=\{\},$ $A_2=\{p_2(x)=x^2-x+1\}$, $...$ etc.
Let $K=\{A_1,A_2,...\}$. Is $K$ finite or infinite?
Edit: I think the $\log()$ brings unneeded complexity to the problem. You could just have $A_n=1/p_n(x)$ or even $A_n=p_n(x).$
A function $\Phi(x)=2^{sg(x)}$ is concave for $\Phi(x)<1$ if and only if $(\ln 2)s^2 (g')^2+sg'' \geq 0$ whenever $sg(x)<0$. Dividing by $sg$ implies that $(\ln 2)(s/g)(g')^2+g''/g \leq 0$. Given any $x$ with $g(x) \neq 0$, one could choose $s$ such that $sg(x) \rightarrow 0^-$ and so it follows that $g''/g \leq 0$ for all $x$ with $g(x) \neq 0$. Equivalently $g''(x)g(x) \leq 0$ for all $x$.
If $g(x)=p(x)$ is a polynomial, the condition $g''(x)g(x) \leq 0$ for all $x$ cannot hold except if $p(x)$ is linear (this can be seen by taking $x\rightarrow \infty$). In this case, for $g(x)=ax+b$, the function $\Phi(x)=2^{sg(x)}$ is concave for all $x$.
If $g(x)=1/p(x)$ for a polynomial $p(x)$ which is nonzero for all $x$, then $g''(x)g(x)\leq 0$ is equivalent to $$ {2(p')^2} \leq pp'',$$ for all $x$. Comparing the leading terms shows that this is not possible for any nonconstant polynomial.
If $g(x)=1/\log(p(x))$, where $p(x)>0$ for all $x$, then $g''(x)g(x)\leq 0$ is equivalent to $$ (p')^2 \log p+2p'\leq p''p(\log p).$$ Again comparing the largest terms on both sides and their leading coefficients, we see that this inequality cannot hold for all $x$