[Algebra]
Problem: Let $x$ be a positive real number such that $\log _{\sqrt{2}} x=20$. Compute $\log _{2} \sqrt{x}$.

Solution: The identity $\log _{b^{n}} x=\frac{1}{n} \log _{b} x$ yields $\log _{2} x=10$. Then $\log _{2} \sqrt{x}=\log _{2} x^{1 / 2}=\frac{1}{2} \log _{2} x=5$.
Use the definition of $\log$ to obtain $x=(\sqrt{2})^{20}=\left(2^{1 / 2}\right)^{20}=2^{10}$. Thus $\log _{2} \sqrt{x}=\log _{2} 2^{5}=\mathbf{5}$.
Use the change of base formula to obtain $\frac{\log x}{\log \sqrt{2}}=20$, so $\log x=$ $20 \log \sqrt{2}=20 \log 2^{1 / 2}=10 \log 2$. Thus $x=2^{10}$, and $\log _{2} \sqrt{x}=\log _{2} 2^{5}=5$.

Final answer: 5