1. Which number is greater,

$$
\log _{3} 2019 \text { or } \quad 4+\sqrt{\log _{3} 18171} \quad ?
$$
1. We have $18171=9 \cdot 2019=3^{2} \cdot 2019$, so it follows that $\log _{3} 18171=\log _{3} 3^{2}+\log _{3} 2019=$ $2+\log _{3}$ 2019. Let's introduce the substitution $t=\log _{3}$ 2019. Then we need to compare the numbers $t$ and $4+\sqrt{2+t}$. From $3^{6}=729<2019<2187=3^{7}$ it follows that $6<t<7$. We will prove the inequality $t<4+\sqrt{2+t}$, i.e., $t-4<\sqrt{2+t}$. After squaring, it remains to prove $t^{2}-8 t+16<2+t$, i.e., $t^{2}-9 t+14<0$. The quadratic function on the left side has zeros at points $\frac{9 \pm \sqrt{81-56}}{2}=\frac{9 \pm 5}{2}$, so it is negative for $t \in(2,7)$; since $6<t<7$, the desired function is indeed negative for $t=\log _{3} 2019$, so among the two observed numbers, the larger one is $4+\sqrt{\log _{3} 18171}$.