1. **Problem:** Given the logarithmic equations $\log_g xy = \frac{5}{2}$ and $(\log_3 x)(\log_3 y) = -6$, find the values of $x$ and $y$ or relate them. 2. **Step 1: Understand the problem and notation.** - $\log_g xy = \frac{5}{2}$ means the logarithm base $g$ of the product $xy$ equals $\frac{5}{2}$. - $(\log_3 x)(\log_3 y) = -6$ means the product of the logarithms base 3 of $x$ and $y$ equals $-6$. 3. **Step 2: Use logarithm properties.** - Recall that $\log_g xy = \log_g x + \log_g y$. - So, $\log_g x + \log_g y = \frac{5}{2}$. 4. **Step 3: Introduce variables for simplification.** - Let $a = \log_3 x$ and $b = \log_3 y$. - Then, from the second equation, $ab = -6$. 5. **Step 4: Express $\log_g x$ and $\log_g y$ in terms of $a$ and $b$.** - Using change of base formula: $\log_g x = \frac{\log_3 x}{\log_3 g} = \frac{a}{\log_3 g}$. - Similarly, $\log_g y = \frac{b}{\log_3 g}$. 6. **Step 5: Substitute into the first equation.** - $\log_g x + \log_g y = \frac{a}{\log_3 g} + \frac{b}{\log_3 g} = \frac{a + b}{\log_3 g} = \frac{5}{2}$. 7. **Step 6: Solve for $a + b$.** - Multiply both sides by $\log_3 g$: $$a + b = \frac{5}{2} \log_3 g$$ 8. **Step 7: Recall $ab = -6$.** - We have a system: $$a + b = \frac{5}{2} \log_3 g$$ $$ab = -6$$ 9. **Step 8: Interpret $a$ and $b$ as roots of quadratic equation.** - The quadratic with roots $a$ and $b$ is: $$t^2 - (a+b)t + ab = 0$$ - Substitute values: $$t^2 - \frac{5}{2} \log_3 g \cdot t - 6 = 0$$ 10. **Step 9: Final expression.** - The values of $a$ and $b$ satisfy this quadratic. - Since $a = \log_3 x$ and $b = \log_3 y$, $x = 3^a$ and $y = 3^b$. **Answer:** The logarithms $a$ and $b$ satisfy $$t^2 - \frac{5}{2} \log_3 g \cdot t - 6 = 0,$$ and $x = 3^a$, $y = 3^b$. This relates $x$ and $y$ through the base $g$ and the given equations.