7.203. $20 \log _{4 x} \sqrt{x}+7 \log _{16 x} x^{3}-3 \log _{x / 2} x^{2}=0$.

7.203. $20 \log _{4 x} \sqrt{x}+7 \log _{16 x} x^{3}-3 \log _{x / 2} x^{2}=0$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \frac{1}{4}, \\ x \neq \frac{1}{16}, \\ x \neq 2 .\end{array}\right.$

Transition to base 2:

$$
\begin{aligned}
& \frac{20 \log _{2} \sqrt{x}}{\log _{2} 4 x}+\frac{7 \log _{2} x^{3}}{\log _{2} 16 x}-\frac{3 \log _{2} x^{3}}{\log _{2} \frac{x}{2}}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{10 \log _{2} x}{2+\log _{2} x}+\frac{21 \log _{2} x}{4+\log _{2} x}-\frac{6 \log _{2} x}{\log _{2} x-1}=0 \Leftrightarrow \\
& 5 \log _{2}^{3} x+3 \log _{2}^{2} x-26 \log _{2} x=0 \Leftrightarrow \log _{2} x\left(5 \log _{2}^{2} x+3 \log _{2} x-26\right)=0 \Leftrightarrow \\
& \Leftrightarrow \log _{2} x\left(\log _{2} x+\frac{13}{5}\right)\left(\log _{2} x-2\right)=0
\end{aligned}
$$

from which $\left(\log _{2} x\right)_{1}=0,\left(\log _{2} x\right)_{2}=-\frac{13}{5},\left(\log _{2} x\right)_{3}=2$. Therefore,

$x_{1}=1, x_{2}=2^{-\frac{13}{5}}=\frac{1}{4 \sqrt[5]{8}}, x_{3}=4$.

Answer: $1 ; \frac{1}{4 \sqrt[5]{8}} ; 4$.