5. Let the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3}$, then $\log _{2}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=$
5. From the given conditions, we have
$$
\begin{array}{c}
9=\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}, \\
27=\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
\end{array}
$$

Since $\left|z_{1}\right|=3$, it follows that $\left|z_{2}\right|=3, z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, and
$$
\left|z_{1} \bar{z}_{2}\right|+\left|\bar{z}_{1} z_{2}\right|=9.
$$

Let
$$
z_{1} \bar{z}_{2}=9(\cos \theta+\mathrm{i} \sin \theta),
$$

then
$$
\bar{z}_{1} z_{2}=9(\cos \theta-\mathrm{i} \sin \theta).
$$

From $-9=z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=18 \cos \theta$, we get
$$
\cos \theta=-\frac{1}{2}.
$$

Thus, $z_{1} \bar{z}_{2}=9 \omega$ or $z_{1} \bar{z}_{2}=9 \omega^{2}$, where
$$
\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}.
$$

When $z_{1} \bar{z}_{2}=9 \omega$, we have

Therefore,
$$
\begin{array}{c}
\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}=-9^{2000}, \\
\log \left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=4000.
\end{array}
$$

When $\bar{z}_{1} z_{2}=9 \omega^{2}$, we get the same result, so the answer is 4000.