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Start the plan$$2^{0.7x}\cdot3^{-1.25y}=\frac{8\sqrt{6}}{27}$$
$$4^{0.3x}\cdot9^{0.2y}=8\cdot81^{\frac{1}{5}}$$
A. x = 2, y = 5
B. x = 2.5, y = 6
C. x = 3, y = 5
D. x = 3, y = 4
E. x = 5, y = 2
The OA is E.
$$2^{0.7x}\cdot3^{-1.25y}=\frac{8\sqrt{6}}{27}$$
$$\Rightarrow \quad 2^{0.7x}\cdot3^{-1.25y}=2^3\cdot 2^{1/2}\cdot 3^{1/2}\cdot 3^{-3}$$
$$\Rightarrow \quad 2^{0.7x}\cdot3^{-1.25y}=2^{7/2}\cdot 3^{-5/2}$$
$$\Rightarrow \quad 0.7x = 7/2 \quad \Rightarrow \quad x = 5$$
Only E has a option for x = 5. Hence, the correct answer is E.
Has anyone another approach to solve this PS question? Regards!













