1. Let $x>0$, planar vectors $\overrightarrow{A B}=(0,1), \overrightarrow{B C}=$ $(1,2), \overrightarrow{C D}=\left(\log _{3} x, \log _{9} x\right)$. If $\overrightarrow{A C} \cdot \overrightarrow{B D}=2$, then the value of $x$ is $\qquad$ .
$-1 . \frac{1}{9}$.
Let $\log _{9} x=t$. Then,
$$
\begin{array}{l}
\log _{3} x=2 t, \overrightarrow{C D}=(2 t, t), \overrightarrow{A C}=(1,3), \\
\overrightarrow{B D}=(1+2 t, 2+t) . \\
\text { So } 1 \cdot(1+2 t)+3 \cdot(2+t)=2 \\
\Rightarrow t=-1 \Rightarrow x=\frac{1}{9} .
\end{array}
$$