7.107. $\left(\log _{2} x-3\right) \log _{2} x+2\left(\log _{2} x+1\right) \log _{2} \sqrt[3]{2}=0$.
## Solution.

Domain of definition: $x>0$.

From the condition

$$
\begin{aligned}
& \log _{2} \sqrt[3]{2}=\log _{2} 2^{1 / 3}=\frac{1}{3}, \log _{2}^{2} x-3 \log _{2} x+\frac{2}{3} \log _{2} x+\frac{2}{3}=0 \\
& 3 \log _{2}^{2} x-7 \log _{2} x+2=0
\end{aligned}
$$

Solving the equation as a quadratic in terms of $\log _{2} x$, we have $\left(\log _{2} x\right)_{1}=\frac{1}{3}$ or $\left(\log _{2} x\right)_{2}=2$, from which $x_{1}=\sqrt[3]{2}, x_{2}=4$.

Answer: $\sqrt[3]{2} ; 4$.