2
$\begingroup$

I am not sure what is meant by exponent notation and therefore how to answer this question is baffling me.

Rewrite this in exponent notation:

$\sqrt[3]{x^2y(z-X)^5}$

  • 0
    Example: $\sqrt[7]{x^2y^3}=x^{2/7}y^{3/7}$. Also, $\sqrt[5]{x/y^2}=x^{1/5}y^{-2/5}$.2012-07-27

1 Answers 1

2

Oh so it is literally just a case of doing this?

$({x^2y(z-X)^5})^\frac{1}{3}$

  • 1
    @Magpie : instead of $\frac {x-1}{x^{\frac {1}{2}}}$, would you prefer $(x-1)x^{-\frac{1}{2}}$ or instead of that $x^{\frac{1}{2}}-x^{-\frac{1}{2}}$ or (just to give them something to think about :) $(x^{\frac{1}{4}}-x^{-\frac{1}{4}})(x^{\frac{1}{4}}+x^{-\frac{1}{4}})$ ( they are all the same(please check for yourself, hope I didnt make a mistake).2012-07-27