5. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=\frac{3}{2}, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Then the number of consecutive 0s after the decimal point in the decimal representation of $a_{6}$ is . $\qquad$
5. 22 .

The conditional recurrence relation can be transformed into $\frac{a_{n+1}+1}{a_{n+1}-1}=\left(\frac{a_{n}+1}{a_{n}-1}\right)^{2}$.
Let $a=\frac{a_{1}+1}{a_{1}-1}=5$. Then $\frac{a_{n}+1}{a_{n}-1}=5^{2 n-1}$.
Thus, the number of consecutive 0s after the decimal point in the decimal representation of $a_{6}$ is
$$
M=\left[\lg \frac{1}{a_{6}-1}\right]=\left[\lg \frac{5^{32}-1}{2}\right] .
$$

Notice that, $5^{32}=\frac{10^{32}}{2^{32}}$.
And $2^{32}=4 \times\left(2^{10}\right)^{3}>4 \times\left(10^{3}\right)^{3}=4 \times 10^{9}$, $2^{32}=\left(2^{16}\right)^{2}=65536^{2}<70000^{2}<5 \times 10^{9}$.
Therefore, $2 \times 10^{22}<5^{32}<2.5 \times 10^{22}$.
Hence, $M=22$.