What are the values of x and y that satisfy both the equations?
$$2^{0.7x}\cdot3^{-1.25y}=\frac{8\sqrt{6}}{27}$$
$$4^{0.3x}\cdot9^{0.2y}=8\cdot81^{\frac{1}{5}}$$
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!
What are the values of x and y that satisfy both the
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$$2^{0.7X}\ .\ 3^{-1.25Y}=\ \frac{\sqrt[8]{6}}{27}$$
The product on the right hand side is the answer to the question on the left hand side, by keeping the focus on the right hand side
=\ \frac{\sqrt[8]{6}}{27}$$
$$\ \frac{^{2^3\cdot\ \sqrt{2}\cdot\sqrt{3}}}{3^3}$$
$$\ \frac{^{2^{\frac{7}{2}}\cdot\ \sqrt{3}}}{3^3}$$
The power of 2 = $$\frac{7}{2}$$ of the left hand side $$2^{0.7x}
2^{0.7x\ =\ 2^{^{\frac{7}{2}}}}
frac{0.7x}{0.7\ }=\ \frac{7}{2}
frac{7}{2\ }=0.7
x\ =\ 3.5\ \frac{0.7\ }{ }
\ x=\ 5
Option\ E\ is\ the\ only\ answer\ with\ x=\ 5$$
The product on the right hand side is the answer to the question on the left hand side, by keeping the focus on the right hand side
=\ \frac{\sqrt[8]{6}}{27}$$
$$\ \frac{^{2^3\cdot\ \sqrt{2}\cdot\sqrt{3}}}{3^3}$$
$$\ \frac{^{2^{\frac{7}{2}}\cdot\ \sqrt{3}}}{3^3}$$
The power of 2 = $$\frac{7}{2}$$ of the left hand side $$2^{0.7x}
2^{0.7x\ =\ 2^{^{\frac{7}{2}}}}
frac{0.7x}{0.7\ }=\ \frac{7}{2}
frac{7}{2\ }=0.7
x\ =\ 3.5\ \frac{0.7\ }{ }
\ x=\ 5
Option\ E\ is\ the\ only\ answer\ with\ x=\ 5$$