Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
I don't understand the breakdown in the first step, why do that? – user138246 Dec 17 '11 at 18:11
@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. – Arturo Magidin Dec 18 '11 at 0:15

$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}=\frac{3^{-2}\cdot x^{-3}\cdot y^{-6}}{x^2 \cdot y^{\frac{-1}{2}}}=\frac{1}{9}\cdot x^{(-3-2)}\cdot y^{\left(-6+\frac{1}{2}\right)}$$

share|cite|improve this answer

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" $$ {1\over 9x^5y^{11/2}}. $$

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.