11.5. Compare the numbers $X=2019^{\log _{2018} 2017}$ and $Y=2017^{\log _{2019} 2020}$.
## Solution.

We have

$$
\begin{gathered}
\ln X=\ln \left(2019^{\log _{2018} 2017}\right)=\log _{2018} 2017 \cdot \ln 2019= \\
=\frac{\ln 2019 \cdot \ln 2017}{\ln 2018}=\log _{2018} 2019 \cdot \ln 2017=\ln \left(2017^{\log _{2018} 2019}\right)
\end{gathered}
$$

which implies $X=2017^{\log _{2018} 2019}$. We use the inequality of means for the distinct positive numbers 2018 and 2020. We obtain the inequality $\sqrt{2018 \cdot 2020}\ln \sqrt{2018 \cdot 2020}=\frac{1}{2} \ln (2018 \cdot 2020)=\frac{\ln 2018+\ln 2020}{2}>\sqrt{\ln 2018 \cdot \ln 2020}
$$

from which it follows that $(\ln 2019)^{2}>\ln 2018 \cdot \ln 2020, \quad$ i.e., $\frac{\ln 2019}{\ln 2018}>\frac{\ln 2020}{\ln 2019}$, which is equivalent to $\log _{2018} 2019>\log _{2019} 2020$. It follows that $X=2017^{\log _{2018} 2019}>2017^{\log _{2019} 2020}=Y$.