smc

By definition, the recursion becomes $a_n={\left(n^{\frac{1}{{\log }_7(n-1)}}\right)}^{{\log }_7(a_{n-1})}=n^{\frac{{\log }_7(a_{n-1})}{{\log }_7(n-1)}}$. By the change of base formula, this reduces to $a_n=n^{{\log }_{n-1}(a_{n-1})}$. Thus, we have ${\log }_n(a_n)={\log }_{n-1}(a_{n-1})$. Thus, for each positive integer $m\ge 3$, the value of ${\log }_m(a_m)$ must be some constant value $k$. We now compute $k$ from $a_3$. It is given that $a_3=3♡2=3^{\frac{1}{{\log }_7(2)}}$, so $k={\log }_3(a_3)={\log }_3\left(3^{\frac{1}{{\log }_7(2)}}\right)=\frac{1}{{\log }_7(2)}={\log }_2(7)$. Now, we must have ${\log }_{2019}(a_{2019})=k={\log }_2(7)$. At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with. $\frac{\log a_{2019}}{\log 2019}=\frac{\log 7}{\log 2}⟹\frac{\log a_{2019}}{\log 7}=\frac{\log 2019}{\log 2}⟹{\log }_7(a_{2019})={\log }_2(2019)$ We conclude that ${\log }_7(a_{2019})={\log }_2(2019)\approx \boxed{11}$, or choice $\boxed{11}$.