1. **State the problem:** Solve the equation $$\log_3 \sqrt{x} = \frac{1}{2 \log_2 3} + \log_3 (4x^3)$$ where $$x > 0$$. 2. **Recall logarithm properties:** - $$\log_a b^c = c \log_a b$$ - $$\log_a b + \log_a c = \log_a (bc)$$ - Change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive $$c \neq 1$$. 3. **Rewrite terms:** - $$\log_3 \sqrt{x} = \log_3 x^{1/2} = \frac{1}{2} \log_3 x$$ - $$\log_3 (4x^3) = \log_3 4 + \log_3 x^3 = \log_3 4 + 3 \log_3 x$$ 4. **Substitute into the equation:** $$\frac{1}{2} \log_3 x = \frac{1}{2 \log_2 3} + \log_3 4 + 3 \log_3 x$$ 5. **Group $$\log_3 x$$ terms:** Move $$3 \log_3 x$$ to the left: $$\frac{1}{2} \log_3 x - 3 \log_3 x = \frac{1}{2 \log_2 3} + \log_3 4$$ Simplify left side: $$\left(\frac{1}{2} - 3\right) \log_3 x = \frac{1}{2 \log_2 3} + \log_3 4$$ $$-\frac{5}{2} \log_3 x = \frac{1}{2 \log_2 3} + \log_3 4$$ 6. **Isolate $$\log_3 x$$:** $$\log_3 x = -\frac{2}{5} \left( \frac{1}{2 \log_2 3} + \log_3 4 \right)$$ 7. **Simplify the term $$\frac{1}{2 \log_2 3}$$:** Using change of base, $$\log_2 3 = \frac{\log_3 3}{\log_3 2} = \frac{1}{\log_3 2}$$ So, $$\frac{1}{2 \log_2 3} = \frac{1}{2 \cdot \frac{1}{\log_3 2}} = \frac{\log_3 2}{2}$$ 8. **Substitute back:** $$\log_3 x = -\frac{2}{5} \left( \frac{\log_3 2}{2} + \log_3 4 \right) = -\frac{2}{5} \left( \frac{\log_3 2}{2} + \log_3 2^2 \right)$$ Since $$\log_3 4 = \log_3 2^2 = 2 \log_3 2$$, $$\log_3 x = -\frac{2}{5} \left( \frac{\log_3 2}{2} + 2 \log_3 2 \right) = -\frac{2}{5} \left( \frac{\log_3 2}{2} + \frac{4 \log_3 2}{2} \right) = -\frac{2}{5} \cdot \frac{5 \log_3 2}{2}$$ 9. **Simplify:** $$\log_3 x = -\frac{2}{5} \cdot \frac{5 \log_3 2}{2} = -\log_3 2$$ 10. **Solve for $$x$$:** $$\log_3 x = -\log_3 2 \implies \log_3 x = \log_3 \frac{1}{2}$$ Therefore, $$x = \frac{1}{2}$$. **Final answer:** $$x = \frac{1}{2}$$.