Please solve!
(3x^{-3}·y^{-2})^{-2}
(three 'x' to the power of negative three, 'y' to the power of negative two) to the power of negative two
Please solve!
(3x^{-3}·y^{-2})^{-2}
(three 'x' to the power of negative three, 'y' to the power of negative two) to the power of negative two
On questions like these, I find it easiest to expand everything to be able to show my work. We also have to remember that, since we are dealing with a negative power, that our expression will be in the form of a fraction. So,
Step 1: 1/(3x^{-3}·y^{-2})^{2} = 1/(3x^{-3}·y^{-2})(3x^{-3}·y^{-2})
once it has been expanded out
Then, we need to remember that something raised to a negative power is the same as a fraction where the negative part is on the bottom of a fraction. This means it can be re-written as
Step 2: 1/(3/x^{3}·1/y^{2})(3/x^{3}·1/y^{2})
This can be simplified to:
Step 3: (1/3/x^{3}y^{2})(1/3/x^{3}y^{2})
Yes, this is a fraction over a fraction, but all we need to do is to treat the 1 as the numerator, and the 3/x^{3}y^{2} as the denominator. Well, we know that to divide fractions, that we must multiply by the reciprocal. This leads us to:
Step 4: (1)(x^{3}y^{2}/3)·(1)(x^{3}y^{2}/3)
We know that to multiply exponents that we add them. So, that leaves us with the answer:
Step 5: x^{6}y^{4}/9
Using properties of exponents remember that
(x^{n})^{m} = x^{(m*n)} and x^{-n} = 1/x^{n}
So using your equation we have:
(3x^{-3}y^{-2})^{-2} = 3^{-2}x^{(-3*-2)}y^{(-2*-2)} = (1/9)x^{6}y^{4}