What are the values of $x$ and $y$ that satisfy the equation $2^{0.7x} \cdot 3^{-1.25y} = \frac{8\sqrt{6}}{27}$?

### Concept & Strategy The key to solving equations with unknown variables in the exponents of different prime bases is to prime factorize both sides of the equation. By expressing all terms as powers of their distinct prime bases (here, 2 and 3), you can then equate the exponents of corresponding bases on both sides. $$a^m \cdot b^n = a^p \cdot b^q \implies m = p \text{ and } n = q$$ ### Step-by-Step Solution * **Given:** $$2^{0.7x} \cdot 3^{-1.25y} = \frac{8\sqrt{6}}{27}$$ * **Calculation / Deduction:** 1. First, break down the Right Hand Side (RHS) into prime bases of 2 and 3. We know that $8 = 2^3$, $27 = 3^3$, and $\sqrt{6} = \sqrt{2 \cdot 3} = 2^{0.5} \cdot 3^{0.5}$. 2. Substitute these back into the RHS: $$\text{RHS} = \frac{2^3 \cdot (2^{0.5} \cdot 3^{0.5})}{3^3}$$ 3. Combine the powers in the numerator and bring the denominator up by changing the sign of its exponent: $$\text{RHS} = 2^{3 + 0.5} \cdot 3^{0.5 - 3}$$ $$\text{RHS} = 2^{3.5} \cdot 3^{-2.5}$$ 4. Now, equate the Left Hand Side (LHS) to the simplified RHS: $$2^{0.7x} \cdot 3^{-1.25y} = 2^{3.5} \cdot 3^{-2.5}$$ 5. Equate the exponents of base 2: $$0.7x = 3.5 \implies x = \frac{3.5}{0.7} = 5$$ 6. Equate the exponents of base 3: $$-1.25y = -2.5 \implies y = \frac{-2.5}{-1.25} = 2$$ ### Exam Strategy & Shortcut **Option Elimination:** Once you simplify the RHS to $2^{3.5} \cdot 3^{-2.5}$, you immediately know that $0.7x = 3.5$. This means $x = 5$. Looking at the options, only option (d) has $x = 5$. You can confidently mark (d) without even calculating for $y$, saving precious seconds. ### Common Pitfall A common mistake is incorrectly expanding $\sqrt{6}$. Students sometimes forget that the square root applies to both prime factors, leading to an incorrect power for either 2 or 3. Always remember $\sqrt{ab} = a^{0.5} \cdot b^{0.5}$. ### Final Answer **Therefore, the correct answer is $x = 5, y = 2$.**