3. Given that the equation $x^{2}-4 x \log _{2} a+3\left(\log _{2} a\right)^{2}=0$ has two roots, one in the interval $(1,3)$ and the other in the interval $(6,+\infty)$, then the range of the real number $a$ is $\qquad$
3. $(4,8)$.

The original equation is
$$
\left(x-\log _{2} a\right)\left(x-3 \log _{2} a\right)=0,
$$

solving it yields $x_{1}=\log _{2} a, x_{2}=3 \log _{2} a$. Since both roots are positive, it must be that $x_{1}=\log _{2} a \in(1,3), x_{2}=3 \log _{2} a \in$ $(6,+\infty)$, thus $\log _{2} a \in(2,3)$, which means $a \in(4,8)$.