10. Solve the following inequality.
$$
\log _{1 / 2} x-\sqrt{2-\log _{4} x}+1 \leq 0
$$
Answer: $\frac{1}{\sqrt{2}} \leq x \leq 16$
Solution: Note that $2-\log _{4} x \geq 0 \Longrightarrow 0<x \leq 16$. Let $t=\sqrt{2-\log _{4} x}$. Then
$$
\log _{4} x=2-t^{2} \Longrightarrow \frac{\log _{1 / 2} x}{\log _{1 / 2} 4}=2-t^{2} \Longrightarrow \log _{1 / 2} x=2 t^{2}-4 .
$$

Substituting back to the given inequality, we have
$$
2 t^{2}-4-t+1 \leq 0 \Longrightarrow 2 t^{2}-t-3 \leq 0 \Longrightarrow-1 \leq t \leq \frac{3}{2} \text {. }
$$

Since $t=\sqrt{2-\log _{4} x}$, this means that $0 \leq \sqrt{2-\log _{4} x} \leq \frac{3}{2}$, which has solution $\frac{1}{\sqrt{2}} \leq x \leq 16$.