8. If $a>0, b>0$, and $a+2 b=6$, then the maximum value of $\lg a+2 \lg b$ is $\qquad$
8. Since $y=\lg x$ is a concave function, we have $\lg a+2 \lg b=3\left(\frac{1}{3} \lg a+\frac{2}{3} \lg b\right) \leqslant 3 \lg \left(\frac{a}{3}+\frac{2 b}{3}\right)=3 \lg 2$. When $a=b=2$, the maximum value $3\lg 2$ is achieved.