Question

What are the values of $x$ and $y$ that satisfy both equations? $$2^{0.7x} \cdot 3^{-1.25y} = \frac{8\sqrt{6}}{27}$$ $$4^{0.3x} \cdot 9^{0.2y} = 8 \cdot (81)^{1/5}$$

Accepted Answer

x = 5, y = 2. From equation 2: $2^{0.6x} \cdot 3^{0.4y} = 2^3 \cdot 3^{4/5}$. Equating powers of 2 and 3: $0.6x = 3 \implies x = 5$. $0.4y = 4/5 = 0.8 \implies y = 2$. Verify in equation 1: $2^{0.7(5)} \cdot 3^{-1.25(2)} = 2^{3.5} \cdot 3^{-2.5} = \frac{2^3 \cdot 2^{0.5}}{3^{2.5}} = \frac{8\sqrt{2}}{9\sqrt{3}} = \frac{8\sqrt{6}}{27}$. Correct!

System of Exponential Equations

Explanation

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