1. **State the problem:** (i) Given the function $$y = \log_a (x + 3) + \log_a (2x - 1)$$, explain why $$x > \frac{1}{2}$$. (ii) Solve the equation $$\frac{\log_a 6}{\log_a (y + 3)} = 2$$. Also, write the expression $$\log_a 9 + (\log_a b)(\log_b 9a)$$ in the form $$c + d \log_a 9$$ where $$c$$ and $$d$$ are integers. --- 2. **Explain domain restriction for (i):** The logarithm function $$\log_a (z)$$ is defined only for $$z > 0$$. So, for $$\log_a (x + 3)$$, we require: $$x + 3 > 0 \implies x > -3$$ For $$\log_a (2x - 1)$$, we require: $$2x - 1 > 0 \implies x > \frac{1}{2}$$ Since both must be true simultaneously, the domain is: $$x > \max(-3, \frac{1}{2}) = \frac{1}{2}$$ Hence, $$x$$ must be greater than $$\frac{1}{2}$$. --- 3. **Solve equation (ii):** Given: $$\frac{\log_a 6}{\log_a (y + 3)} = 2$$ Multiply both sides by $$\log_a (y + 3)$$: $$\log_a 6 = 2 \log_a (y + 3)$$ Divide both sides by 2: $$\frac{\log_a 6}{2} = \log_a (y + 3)$$ Rewrite left side: $$\log_a 6^{\frac{1}{2}} = \log_a (y + 3)$$ Since $$\log_a$$ is one-to-one: $$y + 3 = \sqrt{6}$$ Solve for $$y$$: $$y = \sqrt{6} - 3$$ --- 4. **Rewrite expression:** Expression: $$\log_a 9 + (\log_a b)(\log_b 9a)$$ Use change of base formula: $$\log_b 9a = \frac{\log_a 9a}{\log_a b}$$ Substitute: $$\log_a 9 + (\log_a b) \times \frac{\log_a 9a}{\log_a b} = \log_a 9 + \log_a 9a$$ Cancel $$\log_a b$$: $$\log_a 9 + \log_a 9a$$ Use log product rule: $$\log_a 9 + \log_a 9a = \log_a (9 \times 9a) = \log_a (81a)$$ Rewrite $$\log_a (81a)$$: $$\log_a 81 + \log_a a$$ Since $$a$$ is the base of the log, $$\log_a a = 1$$. Also, $$81 = 9^2$$, so: $$\log_a 81 = \log_a 9^2 = 2 \log_a 9$$ Therefore: $$\log_a 9 + (\log_a b)(\log_b 9a) = 2 \log_a 9 + 1$$ So, $$c = 1$$ and $$d = 2$$. --- **Final answers:** (i) $$x > \frac{1}{2}$$ because the argument of $$\log_a (2x - 1)$$ must be positive. (ii) $$y = \sqrt{6} - 3$$ Expression rewritten as: $$1 + 2 \log_a 9$$