1. **State the problem:** Simplify the expression $$\frac{2 \log_x \sqrt{y} + \log_x y}{2 \log_{x^2} y}$$ where $x > 1$ and $y > 0$. 2. **Recall logarithm rules:** - $\log_x a^m = m \log_x a$ - Change of base: $\log_{x^2} y = \frac{\log_x y}{\log_x x^2}$ - Since $\log_x x^2 = 2$, we have $\log_{x^2} y = \frac{\log_x y}{2}$ 3. **Simplify numerator:** - $\log_x \sqrt{y} = \log_x y^{1/2} = \frac{1}{2} \log_x y$ - So numerator: $2 \times \frac{1}{2} \log_x y + \log_x y = \log_x y + \log_x y = 2 \log_x y$ 4. **Simplify denominator:** - $2 \log_{x^2} y = 2 \times \frac{\log_x y}{2} = \log_x y$ 5. **Combine numerator and denominator:** $$\frac{2 \log_x y}{\log_x y} = 2$$ **Final answer:** $2$