Let $x=64^{t}$, then the original inequality becomes $\log _{14}\left(8^{t}+4^{t}+2^{t}\right) \geqslant t \Rightarrow 8^{t}+4^{t}+2^{t} \geqslant 14^{t}$ $\Rightarrow\left(\frac{8}{14}\right)^{t}+\left(\frac{4}{14}\right)^{t}+\left(\frac{2}{14}\right)^{t} \geqslant 1$. Since the function $f(t)=\left(\frac{8}{14}\right)^{t}+\left(\frac{4}{14}\right)^{t}+\left(\frac{2}{14}\right)^{t}$ is monotonically decreasing and $f(1)=1$, we have $f(t) \geqslant 1 \Rightarrow t \leqslant 1 \Rightarrow 0<x \leqslant 64$.
Therefore, the solution set of the inequality $\log _{14}(\sqrt{x}+\sqrt[3]{x}+\sqrt[6]{x})^{6} \geqslant \log _{2} x$ is $(0,64]$.
7. The solution set of the inequality $\log _{14}(\sqrt{x}+\sqrt[3]{x}+\sqrt[6]{x})^{6} \geqslant \log _{2} x$ is