1. **State the problem:** Solve the equation $$\log_9 \sqrt{x} = \frac{1}{2 \log_3 3} + \log_9 (4x^3), \quad x > 0.$$\n\n2. **Recall logarithm rules:**\n- Change of base: $$\log_a b = \frac{\log_c b}{\log_c a}.$$\n- Product rule: $$\log_a (mn) = \log_a m + \log_a n.$$\n- Power rule: $$\log_a (m^k) = k \log_a m.$$\n- Square root as power: $$\sqrt{x} = x^{\frac{1}{2}}.$$\n\n3. **Rewrite terms using these rules:**\n- $$\log_9 \sqrt{x} = \log_9 x^{\frac{1}{2}} = \frac{1}{2} \log_9 x.$$\n- Since $$\log_3 3 = 1,$$ then $$\frac{1}{2 \log_3 3} = \frac{1}{2}.$$\n- Use product rule on right side: $$\log_9 (4x^3) = \log_9 4 + \log_9 x^3 = \log_9 4 + 3 \log_9 x.$$\n\n4. **Substitute back into the equation:**\n$$\frac{1}{2} \log_9 x = \frac{1}{2} + \log_9 4 + 3 \log_9 x.$$\n\n5. **Group like terms:**\nMove all $$\log_9 x$$ terms to one side:\n$$\frac{1}{2} \log_9 x - 3 \log_9 x = \frac{1}{2} + \log_9 4.$$\nSimplify left side:\n$$\left(\frac{1}{2} - 3\right) \log_9 x = \frac{1}{2} + \log_9 4.$$\n$$-\frac{5}{2} \log_9 x = \frac{1}{2} + \log_9 4.$$\n\n6. **Isolate $$\log_9 x$$:**\n$$\log_9 x = -\frac{2}{5} \left( \frac{1}{2} + \log_9 4 \right).$$\n\n7. **Simplify inside parentheses:**\n$$\frac{1}{2} + \log_9 4 = \log_9 9^{\frac{1}{2}} + \log_9 4 = \log_9 \left( 9^{\frac{1}{2}} \times 4 \right) = \log_9 (3 \times 4) = \log_9 12.$$\n\n8. **Substitute back:**\n$$\log_9 x = -\frac{2}{5} \log_9 12.$$\n\n9. **Rewrite as:**\n$$\log_9 x = \log_9 12^{-\frac{2}{5}}.$$\n\n10. **Since logarithm is one-to-one:**\n$$x = 12^{-\frac{2}{5}}.$$\n\n**Explanation of factoring out square root:**\nWe rewrite $$\sqrt{x}$$ as $$x^{\frac{1}{2}}$$ to use the power rule of logarithms, which allows us to bring the exponent $$\frac{1}{2}$$ in front of the logarithm, simplifying the expression and making it easier to combine and solve.