1. Given a positive real number $a$ satisfies $a^{a}=(9 a)^{8 a}$. Then the value of $\log _{a} 3 a$ is $\qquad$ .
$$
\text { - 1. } \frac{9}{16} \text {. }
$$

From the condition, we know
$$
\begin{array}{l}
9 a=a^{\frac{1}{8}} \Rightarrow 3 a=\sqrt{9 a \cdot a}=a^{\frac{9}{16}} \\
\Rightarrow \log _{a} 3 a=\frac{9}{16} .
\end{array}
$$