Competing Exponents

We basically need to check the points (if at all) where $e^{ax}=x^b$ , or, equivalently, $e^{cx}=x$ where $a=\ln 11,\ b=91, c=\frac{a}{b}$ . Now, consider the function $f(x)=e^{cx}-x$ . Then, $f'(x)=ce^{cx}-1,\ f''(x)=c^2e^{cx}$ . If $c>0$ , the function is strictly convex, and hence $f$ has a unique global minima. Also, at the point of minima, we have $f(x)=1/c(1+\ln c)$ . Since here $c<1/e$ , the minimum value of the function is negative. But since $f(0)=1,\lim_{x\to \infty}f(x)=+\infty$ , due to continuity of the function we can apply intermediate value theorem to obtain that the function has two real roots, i.e. $f(x)=0$ for two values of $x$ .

So, it remains to find the two roots $x_1,x_2$ , so that then we have $f(x)<0,\ \forall x\in (x_1,x_2)$ , and hence $x^{b}>a^x, \forall x\in (x_1,x_2)$ . There does not seem to be any known method to explicitly find the solutions, except the use of the Lambert's W function . Using this function, the solution to $e^{cx}=x$ turns out to be $x=-\frac{W(-c)}{c}$ , which produces the lower limit $\approx 1.02744$ , and the upper limit $\approx201.3206$ . Thus, the desired solution is the sum of all the prime numbers between $2$ and $201$ (inclusive), which evaluates to $\boxed{4227}$ .

First we just need to find what real value of $x$ would make $x^{91}=11^x$ and we take every prime integers below that thanks to the list of prime integers provided (great help btw).