Question:

**Step 1:** Convert the given logarithmic expressions to base 3 using the change of base formula $\log_c M = \frac{\log_b M}{\log_b c}$. $$ \log_{27} a = \frac{\log_3 a}{\log_3 27} = \frac{\log_3 a}{3} $$ $$ \log_9 b = \frac{\log_3 b}{\log_3 9} = \frac{\log_3 b}{2} $$ $$ \log_{27} b = \frac{\log_3 b}{\log_3 27} = \frac{\log_3 b}{3} $$ $$ \log_9 a = \frac{\log_3 a}{\log_3 9} = \frac{\log_3 a}{2} $$ **Step 2:** Substitute these expressions into the given equations. $$ \frac{\log_3 a}{3} + \frac{\log_3 b}{2} = \frac{7}{2} \quad (1) $$ $$ \frac{\log_3 b}{3} + \frac{\log_3 a}{2} = \frac{2}{3} \quad (2) $$ **Step 3:** Let $x = \log_3 a$ and $y = \log_3 b$. Multiply equation (1) by 6 and equation (2) by 6 to clear the denominators. From (1): $$ 6 \left( \frac{x}{3} + \frac{y}{2} \right) = 6 \left( \frac{7}{2} \right) $$ $$ 2x + 3y = 21 \quad (1') $$ From (2): $$ 6 \left( \frac{y}{3} + \frac{x}{2} \right) = 6 \left( \frac{2}{3} \right) $$ $$ 2y + 3x = 4 \quad (2') $$ **Step 4:** Solve the system of linear equations: $$ 2x + 3y = 21 \quad (1') $$ $$ 3x + 2y = 4 \quad (2') $$ Multiply equation (1') by 3 and equation (2') by 2: $$ 3(2x + 3y) = 3(21) \implies 6x + 9y = 63 $$ $$ 2(3x + 2y) = 2(4) \implies 6x + 4y = 8 $$ Subtract the second modified equation from the first modified equation: $$ (6x + 9y) - (6x + 4y) = 63 - 8 $$ $$ 5y = 55 $$ $$ y = 11 $$ Substitute $y=11$ into equation (1'): $$ 2x + 3(11) = 21 $$ $$ 2x + 33 = 21 $$ $$ 2x = 21 - 33 $$ $$ 2x = -12 $$ $$ x = -6 $$ **Step 5:** Determine the value of $ab$. We have $x = \log_3 a = -6$ and $y = \log_3 b = 11$. From the definition of logarithm, we can write: $$ a = 3^{-6} $$ $$ b = 3^{11} $$ Now, calculate $ab$: $$ ab = 3^{-6} \cdot 3^{11} = 3^{-6+11} = 3^5 $$ $$ ab = 243 $$