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I can't figure this one out on my own either

$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$$

I am a little confused on all the small rules at play here but I know that a negative exponent will flip a fraction so I square the top and then flip it. But before that I should work in the parentheses first since that is the order of operataions. I am not sure how to cancel out each of the numbers though I am a little confused what a negative fraction in the numerator does to a larger positive exponent in the denominator.

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You have: $$\begin{align*} \frac{(3x^{3/2}y^3)^{-2}}{(x^2y^{-1/2})} &= \frac{1}{(3x^{3/2}y^3)^2}\times \frac{1}{x^2} \times y^{1/2}\\ &= \frac{1}{(3^2)(x^{3/2})^2(y^3)^2}\times \frac{1}{x^2}\times\frac{y^{1/2}}{1}\\ &= \frac{1}{9x^3y^6}\times\frac{1}{x^2}\times \frac{y^{1/2}}{1}\\ &= \frac{1}{9x^3x^2}\times\frac{y^{1/2}}{y^6}\\ &= \frac{1}{9x^5}\times \frac{1}{y^{11/2}}\\ &= \frac{1}{9x^5y^{11/2}}. \end{align*}$$

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  • $\begingroup$ I don't understand the breakdown in the first step, why do that? $\endgroup$ – toby yeats Dec 17 '11 at 18:11
  • $\begingroup$ @Jordan: Just to note that I am dealing with the numerator separately (first fraction); then the denominator has two types of exponents: positive and negative. The middle factor is the one with positive exponent in the denominator, the last factor came from the negative exponent in the denominator. I separated them because you said you were confused, so I wanted to deal with them separately. $\endgroup$ – Arturo Magidin Dec 18 '11 at 0:15
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$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}=\frac{3^{-2}\cdot x^{-3}\cdot y^{-6}}{x^2 \cdot y^{\frac{-1}{2}}}=\frac{1}{9}\cdot x^{(-3-2)}\cdot y^{\left(-6+\frac{1}{2}\right)}$$

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I'd distribute the power upstairs first ( $(ab)^x=a^xb^x$) $$ (3x^{3/2} y^3)^{-2}= {3^{-2} (x^{3/2})^{-2} (y^3)^{-2}} $$ Now, on the right hand side of the above use $(a^x)^y=a^{xy}$ $$ 3^{-2} (x^{3/2})^{-2} (y^3)^{-2}=3^{-2} x^{(3/2)\cdot(-2)}y^{ 3\cdot(-2)} =3^{-2} x^{-3}y^{-6} $$

So you have $$ 3^{-2} x^{-3}y^{-6}\over x^2 y^{-1/2} $$ Now use ${a^x\over a^y}=a^{x-y}$

$$ {3^{-2} \color{darkgreen}{x^{-3}}\color{maroon}{y^{-6}}\over\color{darkgreen}{ x^2}\color{maroon} {y^{-1/2}} }=3^{- 2}\color{darkgreen}{x^{-3-2}}\color{maroon}{y^{-6-(-1/2)}}=3^{-1/2}x^{-5}y^{-11/2}. $$ Then "pretty it up" $$ {1\over 9x^5y^{11/2}}. $$

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  • $\begingroup$ Can you explain how y^(-6-(-1/2)) is equal to y^(11/2)? Thanks. $\endgroup$ – user1534664 Aug 25 '14 at 13:45
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    $\begingroup$ @user1534664 $-6-(-1/2)=-6+1/2=-12/2+1/2=(-12+1)/2=-11/2$. Also $y^{-11/2}={1\over y^{11/2}}$. $\endgroup$ – David Mitra Aug 25 '14 at 14:20

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