SOLUTION: Please help- I have started the problem, just not sure I am going down the right road.... Raise the quantity in parentheses to the indicated exponent, and simplify the resulting

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Question 601287: Please help- I have started the problem, just not sure I am going down the right road....
Raise the quantity in parentheses to the indicated exponent, and simplify the resulting expression. Express answers with positive exponents.
(24y^-1/72x^-3y^4)^2
Here is what I have so far!
72/24=3
(y^-1/3x^-3y^4)^2
(y^-1/3x^-3y^4)^2* y^-4 (which leaves y^-4*y^-1= y^4)
(y^4/3x^-3)^2
PLEASE HELP!!!!!!!!
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
You should have (y^-5/3x^-3)^2 because (y^-1)/(y^4) = y^(-1-4) = y^(-5)

Now square both the numerator and denominator separately

Square the numerator: (y^(-5))^2 = y^(-5*2) = y^(-10)

So the new numerator is now y^(-10). To write the exponent as a positive number, you flip the base to get 1/(y^10)

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Square the denominator: (3x^-3)^2 = 3^2*x^(-3*2) = 9x^-6

So the new denominator is now 9x^-6

Finally, flip the base that has the negative exponent to get 9/(x^6)

So combine the new numerator and denominator to get

(1/(y^10))/(9/(x^6)) = (1/(y^10))*(x^6)/9 = (x^6)/(9y^10)

Giving you the final answer of (x^6)/(9y^10)

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Answer:

So $\left(\frac{24y^{-1}}{72x^{-3}y^4}\right)^2$ completely simplifies to $\frac{x^6}{9y^{10}}$

In other words $\left(\frac{24y^{-1}}{72x^{-3}y^4}\right)^2 = \frac{x^6}{9y^{10}}$ for all allowed values of x and y.