Question 817811: (2x^-2)/(3y^3) ^-3
simplify leaving positive exponents only using rules of exponents
Answer by FightinBlueHens(27)   (Show Source):
You can put this solution on YOUR website! (2x^-2)/(3y^3) ^-3
Remember: When you raise a number to a negative power, first you have to make that number into its reciprocal, and then you raise it to that power.
When you raise a power to a power (ex. (a^m)^n) you multpily the the powers that you are raising the number to because you are multiplying that power by itself a certain number of times. For example:
(x^3)^2 = xxx*xxx = x*x*x*x*x*x which equals x^6
Just to be clear, the way it's written:
1. in 2x^−2, you are raising x to the power of -2 and not the whole expression 2x to the -2
2. you are raising y to the power of 3, and not 3y to the power of 3
3. you are raising 3y^3 to the power of -3
First solve for the numerator: (2x^-2) = (2/x^2)
The the denominator: For this the expression inside the parentheses is already in simplest form, so you just need to raise 3y^3 to the power of -3.
This will become (1/3y^3)^3 (Now we don't have to deal with a negative value anymore.)
1^3 = 1, 3^3 = 3*3*3 = 9*3 = 27
(y^3)^3 = y^9
(1/3y^3)^3 = 1/(27y^9).
(2/x^2)/ 1/(27y^9)
When you divide by a fraction, you are multiplying by its reciprocal (you flip the fraction and then multiply by it), so when you divide by 1/(27y^9), you are multiplying by (27y^9)/1 whch equals 27y^9.
Multiply the numerators: 2*27y^9 = 54y^9
Multiply the denominators: x^2*1 = x^2
Answer: 54y^9/x^2
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