2. Given the equation in terms of $x$
$$
x^{2}-4 x \log _{2} a+3\left(\log _{2} a\right)^{2}=0
$$

the two roots belong to the intervals $(1,3)$ and $(6,+\infty)$, respectively. Then the range of the real number $a$ is $\qquad$ .
2. $a \in(4,8)$.

The original equation is
$$
\begin{array}{l}
\left(x-\log _{2} a\right)\left(x-3 \log _{2} a\right)=0 \\
\Rightarrow x_{1}=\log _{2} a, x_{2}=3 \log _{2} a=3 x_{1} .
\end{array}
$$

Since both roots are positive, it must be that
$$
\begin{array}{l}
x_{1}=\log _{2} a \in(1,3), \\
x_{2}=3 \log _{2} a \in(6,+\infty) \\
\Rightarrow \log _{2} a \in(2,3) \Rightarrow a \in(4,8) .
\end{array}
$$