(3x^-3*y^-2)^-2

Question

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

Answer 1

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/x³·1/y²)(3/x³·1/y²)

This can be simplified to:

Step 3: (1/3/x³y²)(1/3/x³y²)

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/x³y² as the denominator.  Well, we know that to divide fractions, that we must multiply by the reciprocal.  This leads us to:

Step 4: (1)(x³y²/3)·(1)(x³y²/3)

We know that to multiply exponents that we add them.  So, that leaves us with the answer:

Step 5: x⁶y⁴/9

 

Answer 2

Using properties of exponents remember that

(xⁿ)^m = x^(m*n) and x^-n = 1/xⁿ

So using your equation we have:

(3x^-3y^-2)^-2 = 3^-2x^(-3*-2)y^(-2*-2) = (1/9)x⁶y⁴