Simplify with fractional exponents and negative exponents 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.
 A: 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?
A: 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.
A: $$ \left(\frac{3x ^\frac{3}{2}y^3}{x^2 y^{\frac{-1}{2}}}\right)^{-2}=(3x^{3/2-2}y^{3-(-1/2)})^{-2}=$$
$$=(3x^{-1/2}y^{7/2)})^{-2}=3^{-2}x^{(-1/2)(-2)}y^{(7/2)(-2))}=$$
$$=3^{-2}x^1y^{(-7)}=\frac{x}{9y^7}$$
A: 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 $3$.
A: 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:
= $\dfrac{x}{9y^7}$
A: 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....
