Simplify fraction exponents

Hey everyone, I am typically very good at math; however, I have always struggled with fractions. I am currently reviewing pre-calculus so that I can test out of pre-calc for my accounting degree. I have a problem on my review work that I have been attempting to solve and just cannot get it.... I am really hoping someone could walk me through this... Thanks in advance! :)

(49x^{-2}y^{4})^{-1/2}(xy^{1/2})

I can get to what I'm 99% sure is the next step...

__[x√(y)]__

[√(49) √(x^{-2}) √(y^{4})]

Followed by... (which I'm not as sure is correct)

__ x √(y)__

7 √(x^{-2}) y^{2}

And this is where I get stuck... I'm not sure what to do with the negative exponent in the denominator... If someone could explain the next step for me, I would greatly appreciate it.

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Re: Simplify fraction exponents

Re: Simplify fraction exponents

Hello, Cyryn!

Quote:

$\displaystyle \text{Simplify: }\:\left(49x^{\text{-}2}y^4\right)^{\text{-}\frac{1}{2}}\left(xy^{\frac{1}{2}}\right)$

First of all, avoid radicals . . . until the very end.

We have: .$\displaystyle \left(49x^{\text{-}2}y^4\right)^{\text{-}\frac{1}{2}}\left(xy^{\frac{1}{2}}\right)$

. . . . . . $\displaystyle =\;\big(49\big)^{\text{-}\frac{1}{2}}\left(x^{\text{-}2}\right)^{\text{-}\frac{1}{2}}\left(y^4\right)^{\text{-}\frac{1}{2}}\left(xy^{\frac{1}{2}}\right)$

. . . . . . $\displaystyle =\; \left(7^2\right)^{\text{-}\frac{1}{2}} \big(x^1\big)\big( y^{\text{-}2}\big) \cdot x \cdot y^{\frac{1}{2}}$

. . . . . . $\displaystyle =\; 7^{\text{-}1}\big(x\cdot x\big)\left(y^{\text{-}2}\cdot y^{\frac{1}{2}}\right)$

. . . . . . $\displaystyle =\;7^{\text{-}1}\cdot x^2\cdot y^{\text{-}\frac{3}{2}}$

. . . . . . $\displaystyle =\;\frac{1}{7}\cdot x^2\cdot \frac{1}{y^{\frac{3}{2}}}$

. . . . . . $\displaystyle =\;\frac{x^2}{7\sqrt{y^3}} $

Re: Simplify fraction exponents

Thank you, thank you, thank you!!! The examples in my book were putting things into radicals early which I think is where I'm getting confused, but how you did it makes so much more sense! Thanks sooooo much.