2. $11^{\log _{a} 20}\left(12^{\log _{a} 21}-13^{\log _{a} 22}\right)-20^{\log _{a} 11}\left(21^{\log _{a} 12}-22^{\log _{a} 13}\right)=$
2.0

Analysis: Since $\log _{a} x \log _{a} y=\log _{a} x^{\log _{a} y}=\log _{a} y^{\log _{a} x}$, we have $x^{\log _{a} y}=y^{\log _{a} x}$, hence the original expression is 0;