9. Given that $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2, and $\left\{b_{n}\right\}$ is an arithmetic sequence with the first term 2 and common difference 5, the numbers that appear in both sequences are arranged in ascending order to form the sequence $\left\{x_{n}\right\}$. Then $x_{100}=$ $\qquad$.
$a_{n}=2^{n-1}, b_{n}=5 n-3$, Note that $2^{4} \equiv 1(\bmod 5)$, so $a_{n}=b_{m} \Leftrightarrow n=4 k-2$, thus $x_{n}=a_{4 n-2}=2^{4 n-3}$. Therefore, $x_{100}=a_{398}=2^{397}$.