Question:

Step 1: Convert the logarithms to a common base. We need to express the given logarithmic equations in terms of a single variable system. Using the change of base formula: $$\log_{27} a = \frac{\log a}{\log 27} = \frac{\log a}{3\log 3}$$ $$\log_9 b = \frac{\log b}{\log 9} = \frac{\log b}{2\log 3}$$ Step 2: Introduce substitution variables to simplify. Let us define: $$p = \frac{\log a}{\log 3} = \log_3 a$$ $$q = \frac{\log b}{\log 3} = \log_3 b$$ This allows us to rewrite the logarithmic expressions as $\log_{27} a = \frac{p}{3}$ and $\log_9 b = \frac{q}{2}$. Step 3: Convert the first given equation. The first equation is $\log_{27} a + \log_9 b = \frac{7}{2}$. Substituting our expressions: $$\frac{p}{3} + \frac{q}{2} = \frac{7}{2}$$ Multiplying through by 6 to clear denominators: $$2p + 3q = 21 \quad \text{...(i)}$$ Step 4: Convert the second given equation. The second equation is $\log_{27} b + \log_9 a = \frac{2}{3}$. Substituting our expressions: $$\frac{q}{3} + \frac{p}{2} = \frac{2}{3}$$ Multiplying through by 6 to clear denominators: $$2q + 3p = 4 \quad \text{...(ii)}$$ Step 5: Solve the system of linear equations. We have: $$2p + 3q = 21 \quad \text{...(i)}$$ $$3p + 2q = 4 \quad \text{...(ii)}$$ Multiply equation (i) by 2: $4p + 6q = 42$ Multiply equation (ii) by 3: $9p + 6q = 12$ Subtract the first from the second: $$9p + 6q - (4p + 6q) = 12 - 42$$ $$5p = -30$$ $$p = -6$$ Step 6: Find the value of q. Substitute $p = -6$ into equation (i): $$2(-6) + 3q = 21$$ $$-12 + 3q = 21$$ $$3q = 33$$ $$q = 11$$ Step 7: Convert back to find a and b. Since $p = \log_3 a = -6$: $$a = 3^{-6}$$ Since $q = \log_3 b = 11$: $$b = 3^{11}$$ Step 8: Calculate the product ab. $$ab = 3^{-6} \cdot 3^{11} = 3^{-6+11} = 3^5$$ $$3^5 = 243$$ Therefore, $ab = 243$, which corresponds to **Option 2**.