Solve the equation $\left|\begin{array}{lll}2^{2 x+1} & 2 & 3 \\ 2^{4 x+1} & 4 & 9 \\ 2 & 1 & 1\end{array}\right|=0$.
$$
\begin{array}{l}
\left|\begin{array}{lll}
2^{2 x+1} & 2 & 3 \\
2^{4 x+1} & 4 & 6 \\
2 & 1 & 1
\end{array}\right|=2\left|\begin{array}{lll}
2^{2 x} & 2 & 3 \\
2^{4 x} & 4 & 9 \\
1 & 1 & 1
\end{array}\right| \\
=2\left|\begin{array}{ccc}
2^{2 x}-2 & -1 & 3 \\
2^{4 x}-4 & -5 & 9 \\
0 & 0 & 1
\end{array}\right|=2\left|\begin{array}{c}
2^{2 x}-2-1 \\
2^{4 x}-4-5
\end{array}\right| \\
=-2\left[5\left(2^{2 x}-2\right)-\left(2^{4 x}-4\right)\right] \\
=2\left(2^{4 x}-5 \cdot 2^{2 x}+6\right) \text {, } \\
\left(2^{2 x}\right)^{2}-5 \cdot 2^{2 x}+6=0 \text {, } \\
\therefore 2^{2 x}=2 \text {, (1) or } 2^{2 x}=3 \text {. } \\
\end{array}
$$
(1) gives $2 x=1, \therefore x=\frac{1}{2}$,

From (2) we get $2 x \lg 2=\lg 3, \therefore x=\frac{\lg 3}{2 \lg 2}$.
$$
\therefore x_{1}=\frac{1}{2}, \quad x_{8}=\frac{\lg 3}{2 \lg 2} .
$$