7.246. $\log _{2} \sqrt[3]{x}+\sqrt[3]{\log _{2} x}=\frac{4}{3}$.
## Solution.

Domain of definition: $x>0$.

## From the condition we have

$$
\frac{1}{3} \log _{2} x+\sqrt[3]{\log _{2} x}=\frac{4}{3} \Leftrightarrow \log _{2} x+3 \sqrt[3]{\log _{2} x}-4=0
$$

Let $\sqrt[3]{\log _{2} x}=y$. The equation in terms of $y$ becomes

$$
\begin{aligned}
& y^{3}+3 y-4=0 \Leftrightarrow\left(y^{3}-1\right)+(3 y-3)=0 \Leftrightarrow \\
& \Leftrightarrow(y-1)\left(y^{2}+y+1\right)+3(y-1)=0 \Leftrightarrow(y-1)\left(y^{2}+y+4\right)=0
\end{aligned}
$$

from which $y-1=0$, since $y^{2}+y+4>0$. Therefore, $y=1, \sqrt[3]{\log _{2} x}=1$, $\log _{2} x=1, x=2$.

Answer: 2.