Simplify expression with laws of indices

[dmarley]

Helping my daughter with her math and hit this one and not sure how to advise. All help welcome(x^-2y¹⁰)³ / (x^-4yz⁴)^-5

This one throws me off because I don't know how to deal with the z, as only on the right side of the divide

 

[Wilmer]

(x^-2y¹⁰)³ / (x^-4yz⁴)^-5
This one throws me off because I don't know how to deal with the z, as only on the right side of the divide

z^(4*(-5)) = z^(-20)
Now move to numerator:
z^(-20) = z^20

So you'll end up with: y^35 * z^20 / x^26

 

[topsquark]

Helping my daughter with her math and hit this one and not sure how to advise. All help welcome(x^-2y¹⁰)³ / (x^-4yz⁴)^-5

This one throws me off because I don't know how to deal with the z, as only on the right side of the divide

The basic rule is \(\displaystyle a^{-1} = \dfrac{1}{a}\) and \(\displaystyle \left ( a^{-1} \right ) ^{-1} = a\).

Strategy: Get rid of those pesky negative powers.
\(\displaystyle \dfrac{ \left ( x^{-2}y^{10} \right ) ^3 }{ \left ( x^{-4} y z^4 \right ) ^{-5} }\)

\(\displaystyle = \left ( x^{-2}y^{10} \right ) ^3 \left ( x^{-4} y z^4 \right ) ^5\)

\(\displaystyle = \left ( \dfrac{y^{10}}{x^2} \right ) ^3 \left ( \dfrac{yz^4}{x^4} \right ) ^5\)

Can you finish?

-Dan

 

[dmarley]

topsquark - thanks for the help

following your basic rule really helped out and clarified for us.