1. **State the problem:** We are given the function $$y = \log_a (x + 3) + \log_a (2x - 1)$$ where $a, b, c, d$ are positive constants. We need to explain why $$x > \frac{1}{2}$$ must hold. 2. **Explain the domain restriction:** The logarithm function $$\log_a (z)$$ is defined only for $$z > 0$$. Therefore, both arguments inside the logarithms must be positive: $$x + 3 > 0 \implies x > -3$$ and $$2x - 1 > 0 \implies 2x > 1 \implies x > \frac{1}{2}$$ Since $$x > \frac{1}{2}$$ is more restrictive than $$x > -3$$, the domain of $$y$$ is $$x > \frac{1}{2}$$. 3. **Solve the equation:** Given: $$\frac{\log_a 6}{\log_a (y + 3)} = 2$$ Multiply both sides by $$\log_a (y + 3)$$: $$\log_a 6 = 2 \log_a (y + 3)$$ Using the logarithm power rule: $$\log_a 6 = \log_a (y + 3)^2$$ Since $$\log_a M = \log_a N \implies M = N$$ for $$a > 0, a \neq 1$$: $$6 = (y + 3)^2$$ Take the square root: $$y + 3 = \pm \sqrt{6}$$ Since $$y + 3 > 0$$ (logarithm argument), we take the positive root: $$y + 3 = \sqrt{6}$$ Therefore: $$y = -3 + \sqrt{6}$$ 4. **Rewrite the expression:** Given: $$\log_a 9 + (\log_a b)(\log_{\sqrt{b}} 9a)$$ Use change of base for $$\log_{\sqrt{b}} 9a$$: $$\log_{\sqrt{b}} 9a = \frac{\log_a 9a}{\log_a \sqrt{b}}$$ Note: $$\log_a 9a = \log_a 9 + \log_a a = \log_a 9 + 1$$ and $$\log_a \sqrt{b} = \log_a b^{1/2} = \frac{1}{2} \log_a b$$ So: $$\log_{\sqrt{b}} 9a = \frac{\log_a 9 + 1}{\frac{1}{2} \log_a b} = \frac{2(\log_a 9 + 1)}{\log_a b}$$ Multiply by $$\log_a b$$: $$(\log_a b)(\log_{\sqrt{b}} 9a) = \log_a b \cdot \frac{2(\log_a 9 + 1)}{\log_a b} = 2(\log_a 9 + 1)$$ Therefore, the entire expression is: $$\log_a 9 + 2(\log_a 9 + 1) = \log_a 9 + 2 \log_a 9 + 2 = 3 \log_a 9 + 2$$ Rewrite as: $$c + d \log_a 9$$ where $$c = 2$$ and $$d = 3$$. **Final answers:** - Domain restriction: $$x > \frac{1}{2}$$ - Exact solution: $$y = -3 + \sqrt{6}$$ - Expression rewritten: $$2 + 3 \log_a 9$$