Question

Rational Exponents again

(x^1/3 / y^-2/3)^9

Answer 1

\[\large \left(\frac{x^{1/3}}{y^{-2/3}}\right)^9 \]

Answer 2

Remember, \[x^{-1} \iff \frac{1}{x}~ , ~ \frac{1}{x^{-1}} \iff x^1\]

Answer 3

So which of those two identities would you use for \(y^{-2/3}\)?

Answer 4

1/y^2/3?

Answer 5

Not quite. The second identity claims this: \[\frac{1}{x^{-1}} = x^{1}\]This meant you can take the reciprocal of the function in the denominator of the fraction and raise it up to the numerator.

Answer 6

oh wait the other one so i can get rid of the denominator

Answer 7

Yes \(\checkmark\)

Answer 8

So what would your equation look like within the parenthesis if you used the second identity on \(y^{-2/3}\)?

Answer 9

(x^1/3 y^2/3)9

Answer 10

so then its just cross multiply again?

Answer 11

Awesome. You're a quick learner, I like that

Answer 12

Don't call it cross multiplication, call it distribution. You're now distributing the exponent of 9 to both x and y terms.

Answer 13

So what is \[\frac{1}{3} \cdot 9 = ~? \]\[\frac{2}{3} \cdot 9 =~? \]

Answer 14

so x would be x^3 and y would be y^6

Answer 15

\(\checkmark\)

Answer 16

awesome thanks

Answer 17

No problem