1. **State the problem:** We want to show that $$\log_{16} (xy) = \frac{1}{2} \log_{4} x + \frac{1}{2} \log_{4} y$$ and then solve the simultaneous equations: (i) $$\log_{16} (xy) = \sqrt{3}$$ (ii) $$\frac{\log_{4} x}{\log_{4} y} = -8$$ 2. **Show the logarithm identity:** - Recall the change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive $c \neq 1$. - Using base 4 for convenience, write $$\log_{16} (xy) = \frac{\log_4 (xy)}{\log_4 16}$$. - Since $$16 = 4^2$$, we have $$\log_4 16 = 2$$. - Next, use the log product rule: $$\log_4 (xy) = \log_4 x + \log_4 y$$. - Substitute back: $$\log_{16} (xy) = \frac{\log_4 x + \log_4 y}{2} = \frac{1}{2} \log_4 x + \frac{1}{2} \log_4 y$$, proving the identity. 3. **Solve the simultaneous equations:** - From the identity, equation (i) is: $$\frac{1}{2} \log_4 x + \frac{1}{2} \log_4 y = \sqrt{3}$$ - Multiply both sides by 2: $$\log_4 x + \log_4 y = 2\sqrt{3}$$ - This can be rewritten as: $$\log_4 (xy) = 2\sqrt{3}$$ - Let $$a = \log_4 x$$ and $$b = \log_4 y$$. - Equation (i) becomes: $$a + b = 2\sqrt{3}$$ - Equation (ii) becomes: $$\frac{a}{b} = -8 \Rightarrow a = -8b$$ - Substitute $a$ in the first equation: $$-8b + b = 2\sqrt{3} \Rightarrow -7b = 2\sqrt{3} \Rightarrow b = -\frac{2\sqrt{3}}{7}$$ - Find $a$: $$a = -8b = -8 \times \left(-\frac{2\sqrt{3}}{7}\right) = \frac{16\sqrt{3}}{7}$$ - Recall definitions: $$x = 4^a = 4^{\frac{16\sqrt{3}}{7}}$$ $$y = 4^b = 4^{-\frac{2\sqrt{3}}{7}}$$ 4. **Final answers:** $$\boxed{x = 4^{\frac{16\sqrt{3}}{7}}, \quad y = 4^{-\frac{2\sqrt{3}}{7}}}$$