1. **State the problem:** Solve the equation $$\log_3 \sqrt{x} = \frac{1}{2 \log_3 x} + \log_3 (4x^2)$$ where $$x > 0$$. 2. **Recall logarithm properties:** - $$\log_b (a^c) = c \log_b a$$ - $$\sqrt{x} = x^{1/2}$$ - The domain requires $$x > 0$$. 3. **Rewrite terms using properties:** - Left side: $$\log_3 \sqrt{x} = \log_3 x^{1/2} = \frac{1}{2} \log_3 x$$ - Right side: $$\log_3 (4x^2) = \log_3 4 + \log_3 x^2 = \log_3 4 + 2 \log_3 x$$ 4. **Substitute into the equation:** $$\frac{1}{2} \log_3 x = \frac{1}{2 \log_3 x} + \log_3 4 + 2 \log_3 x$$ 5. **Let $$y = \log_3 x$$ (note $$y \neq 0$$ because $$\log_3 x$$ in denominator):** $$\frac{1}{2} y = \frac{1}{2y} + \log_3 4 + 2y$$ 6. **Multiply both sides by $$2y$$ to clear denominators:** $$2y \cdot \frac{1}{2} y = 2y \cdot \frac{1}{2y} + 2y \cdot \log_3 4 + 2y \cdot 2y$$ Simplify: $$y^2 = 1 + 2y \log_3 4 + 4y^2$$ 7. **Bring all terms to one side:** $$y^2 - 4y^2 - 2y \log_3 4 - 1 = 0$$ Simplify: $$-3y^2 - 2y \log_3 4 - 1 = 0$$ Multiply entire equation by $$-1$$: $$3y^2 + 2y \log_3 4 + 1 = 0$$ 8. **Solve quadratic equation:** $$3y^2 + 2y \log_3 4 + 1 = 0$$ Use quadratic formula: $$y = \frac{-2 \log_3 4 \pm \sqrt{(2 \log_3 4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$$ Calculate discriminant: $$\Delta = (2 \log_3 4)^2 - 12 = 4 (\log_3 4)^2 - 12$$ Since $$\log_3 4 \approx 1.2619$$, $$\Delta \approx 4 (1.2619)^2 - 12 = 4 (1.591) - 12 = 6.364 - 12 = -5.636 < 0$$ 9. **Discriminant is negative, so no real solutions for $$y$$.** 10. **Check domain and denominator:** Since $$y = \log_3 x$$ and denominator $$2 \log_3 x$$ cannot be zero, $$y \neq 0$$. 11. **Conclusion:** No real $$y$$ satisfies the equation, so no real $$x > 0$$ solves the original equation. **Final answer:** No real solution for $$x > 0$$.