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.
Best Answer
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.