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I am trying to simplify

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

It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.

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

Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.

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

Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.

$4/2 - 3/2 = 1/2$

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

Now I need to get rid of the $y$ exponent.

I am not sure how that is possible.

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  • $\begingroup$ Hint: How about simplifying the expression inside the parnes first and then squaring the remaining items? $\endgroup$
    – Amzoti
    May 17, 2013 at 21:46

6 Answers 6

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So we have $$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$

Yes you are right you can "flip" the fraction to remove the negative exponent.

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

Here you multiply every exponent by 2 to square the expression inside the brackets. This is where you went wrong here.

$$ \left(\frac{x^4 y^{-1}}{9x^{3}y^6}\right)$$

But $y^{-1}=1/y$

so the above is equal to

$$ \left(\frac{x^4}{9x^{3}y^7}\right)$$

Now cancel the $x^3$

$$ \left(\frac{x}{9y^7}\right)$$

And we're done.

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The main problem is that you didn’t actually perform the squaring:

$$\left(\frac{x^2y^{-1/2}}{3x^{3/2}y^3}\right)^2=\frac{(x^2y^{-1/2})^2}{(3x^{3/2}y^3)^2}=\frac{x^4y^{-1}}{9x^3y^6}\;,$$

since squaring entails multiplying each exponent by $2$. Can you finish it from there?

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$$ \left(\frac{3x ^\frac{3}{2}y^3}{x^2 y^{\frac{-1}{2}}}\right)^{-2}=(3x^{3/2-2}y^{3-(-1/2)})^{-2}=$$ $$=(3x^{-1/2}y^{7/2)})^{-2}=3^{-2}x^{(-1/2)(-2)}y^{(7/2)(-2))}=$$ $$=3^{-2}x^1y^{(-7)}=\frac{x}{9y^7}$$

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It may have been easier to simplify the inside first. In dividing, you subtract exponents.

$$(\frac{3x^\frac32y^3}{x^2y^{-\frac12}})^{-2}=(3x^{\frac32-2}y^{3+\frac12})^{-2}=(3x^{-\frac12}y^\frac72)^{-2}$$

Now to clear the parentheses, multiply each exponent by $-2$ to get

$$3^{-2}x^1y^{-7}=\frac x{9y^7}$$

When you attempted to square everything, you squared only the $3$.

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We can flip the equation:

$ \left(\dfrac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $= $\left(\dfrac{x^2y^{-1/2}}{3x^{3/2}y^3}\right)^2$

Multiplying the exponents:

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

We can move the $y^{-1}$ to the bottom and write everything factored:

= $\dfrac{x·x·x·x}{3·3·x·x·x·y·y·y·y·y·y·y}$

Looking at this, we can cancel x's.

= $\dfrac{x}{3·3·y·y·y·y·y·y·y}$

Which simplifies to:

= $\dfrac{x}{9y^7}$

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here are a few simple rules that help me..

if you multiply two numbers with exponents $x^3$ times $x^3$ you don't change the base (x) and you add the exponents (3 and 3 is 6) so the answer would be $x^6$

if you are doing something like this, $(x^3)^3$, you have to multiple the exponents.. so the answer would be x^9 (3x3 is 9)

$x^{1/2}$ is just $1/x^2$, so to bring $y^{1/2}$ to the bottom, you can just put $y^2$ in the denominators place....

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