1. If $a>0, b>0$, and $a+2b=6$, then the maximum value of $\ln a+2 \ln b$ is $\qquad$
$\ln a+2 \ln b=\ln \left(a b^{2}\right) \leqslant \ln \left(\frac{a+2 b}{3}\right)^{3}=3 \ln 2$, equality holds when $a=b=2$. Therefore, the maximum value of $\ln a+2 \ln b$ is $3 \ln 2$.