1. Given the function $f(x)=\log _{a}(2+a x)$ and the function $g(x)=\log _{\frac{1}{a}}(a+2 x)(a>0, a \neq 1)$, the graphs of these functions are symmetric about the line $y=b$ (where $b$ is a constant), then $a+b=$ $\qquad$
1. 2

Analysis: According to the problem, $f(x)+g(x)=2 b$, which means $\log _{a} \frac{2+a x}{a+2 x}=2 b$ holds for any $x$ in the domain, so $\frac{2+a x}{a+2 x}$ is a constant. Let $\frac{2+a x}{a+2 x}=t(t>0)$, simplifying gives $0=t a-2+(2 t-a) x$ holds for any $x$ in the domain. Therefore, $t a-2=2 t-a=0$, solving yields $a=2$, at this time $b=0$, hence $a+b=2$