$7.329 .\left|\log _{2}(3 x-1)-\log _{2} 3\right|=\left|\log _{2}(5-2 x)-1\right|$. 

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$7.329 .\left|\log _{2}(3 x-1)-\log _{2} 3\right|=\left|\log _{2}(5-2 x)-1\right|$.
Solution. The original equation is equivalent to the following:

$\left[\begin{array}{l}\log _{2}(3 x-1)-\log _{2} 3=\log _{2}(5-2 x)-1, \\ \log _{2}(3 x-1)-\log _{2} 3=1-\log _{2}(5-2 x)\end{array} \Leftrightarrow\left[\begin{array}{l}\log _{2} \frac{3 x-1}{3}=\log _{2} \frac{5-2 x}{2}, \\ \log _{2} \frac{3 x-1}{3}=\log _{2} \frac{2}{5-2 x}\end{array} \Leftrightarrow\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-107.jpg?height=197&width=510&top_left_y=110&top_left_x=114)

Answer: $x_{1}=1, x_{2}=\frac{17}{12}, x_{3}=\frac{11}{6}$.

$7.330(2+\sqrt{3})^{x^{2}-2 x+1}+(2-\sqrt{3})^{x^{2}-2 x-1}=\frac{4}{2-\sqrt{3}}$.

Solution. The original equation is equivalent to the following:

$(2+\sqrt{3})^{x^{2}-2 x}+(2-\sqrt{3})^{x^{2}-2 x}=4$.

Let $y=(2-\sqrt{3})^{x^{2}-2 x}$, then $\frac{1}{y}+y=4 \Leftrightarrow$

$\Leftrightarrow y^{2}-4 y+1=0 \Leftrightarrow y=2 \pm \sqrt{3}$.

1) $(2-\sqrt{3})^{x^{2}-2 x}=(2-\sqrt{3}) \Leftrightarrow x^{2}-2 x=1 \Leftrightarrow x=1 \pm \sqrt{2}$.
2) $(2-\sqrt{3})^{x^{2}-2 x}=2+\sqrt{3} \Leftrightarrow(2-\sqrt{3})^{x^{2}-2 x}=(2-\sqrt{3})^{-1} \Leftrightarrow x^{2}-2 x=-1 \Leftrightarrow x=1$.

Answer: $x_{1}=1-\sqrt{2}, x_{2}=1, x_{3}=1+\sqrt{2}$.