What is the answer to solvefor x,y=2x^{-3/2} ?

The answer to solvefor x,y=2x^{-3/2} is x=\sqrt[3]{4/(y^2)} {y 0}

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solvefor x,y=2x^{-3/2}

solve for

x, y = 2 x^{- \frac{3}{2}}

Solution steps

y = 2 x^{- \frac{3}{2}}

Expand

2 x^{- \frac{3}{2}} : \frac{2}{x ^{\frac{3}{2}}}

y = \frac{2}{x ^{\frac{3}{2}}}

Square both sides:

y^{2} = \frac{4}{x ^{3}}

y^{2} = \frac{4}{x ^{3}}

Solve

Verify Solutions:

The solution is

v + 6 = 8

y = - y + 20

11 + 10 x + 8 x + 9 = 2

\frac{x + 4}{x - 3} = 0

x - 3 x^{\frac{1}{2}} + 2 = 0

What is the answer to solvefor x,y=2x^{-3/2} ?
The answer to solvefor x,y=2x^{-3/2} is x=\sqrt[3]{4/(y^2)} {y<0\lor y>0}