Calculus Help

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ELRA0098321

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Problem: (3x^3/2 y^3 / x^2 y^-1/2)^-2

I did this: (3x^1/2 y^7/2 / 1)^-2 = (1 / 3x^1/2 y^7/2)^2 = 1/9xy^7.

My book has the answer of x / 9y^7.:confused: I don't know what I did wrong.

Thanks for your help!
 
ELRA0098321 said:
Problem: (3x^3/2 y^3 / x^2 y^-1/2)^-2

I did this: (3x^1/2 y^7/2 / 1)^-2 = (1 / 3x^1/2 y^7/2)^2 = 1/9xy^7.

My book has the answer of x / 9y^7.:confused: I don't know what I did wrong.

Thanks for your help!
Click to expand...
Your error is x^(3/2)/x^2 = x^(-1/2), but you have x^(+1/2)

After taking the power (-2), your x wound up in the denominator but should be in the numerator.
 
That's very awkwardly written! I presume you mean \(\displaystyle \left(\dfrac{3x^{3/2}y^3}{x^2y^{-1/2}}\right)^{-2}\). That is the same as \(\displaystyle \left(\dfrac{x^2y^{-1/2}}{3x^{3/2}y^2}\right)^2= \)\(\displaystyle \dfrac{x^2y^{-1}}{9x^3y^6}= \dfrac{1}{9xy^7}\). Your text book apparently has a misprint.
 
HallsofIvy said:
That's very awkwardly written! I presume you mean \(\displaystyle \left(\dfrac{3x^{3/2}y^3}{x^2y^{-1/2}}\right)^{-2}\). That is the same as \(\displaystyle \left(\dfrac{x^2y^{-1/2}}{3x^{3/2}y^2}\right)^2= \)\(\displaystyle \dfrac{x^2y^{-1}}{9x^3y^6}= \dfrac{1}{9xy^7}\). Your text book apparently has a misprint.
Click to expand...

Thanks! I have been trying to find out how to do what latex code you use to do that. What is it?
 
The book is correct. The answer should be \(\displaystyle \dfrac{x}{9y^7}\)

\(\displaystyle \left (\dfrac{x^2y^{-1/2}}{3x^{3/2}y^3} \right)^2=\dfrac{x^4y^{-1}}{9x^3y^6}=\dfrac{x}{9y^7}\)
 
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