# Rules of Exponents - Problem 3

Simplify an expression using the laws of exponents. So what I have here is a fairly ugly statement that we want to clean up and the first thing I'm going to do is simplify what's inside my parenthesis.

Whenever we have more than one term to a power, that power has to get distributed into everything, so I know that this 2 has to come in to both the x and the y and this 3 has to come in to both the x and the z. The 5 stays as it is and then we have x² squared, when you have a power to a power you have to multiply so this turns out to be x to the fourth. Y to the first squared that just turns into y² and then z stays on the outside.

10 stays on the outside as well, xz to the third there's no powers on either x or z, so we end up with just a single x³ and z³ and then the y on the outside.

Now we have to simplify. So the numeric part is the easy part, we have 5 over 10 that just cancels into 1 over 2, so we know we have a 2 in the denominator. Then I have to pair up our xs, our ys, and our zs. We have x to the fourth over x to the third. When we are dividing, I know I have to subtract my power when the numerator is larger than the denominator so that tells me I'm going to have my term left with numerator, 4 minus 3 is 1, so I just have a single x in the numerator.

Going on to the ys, y² over y again division we can subtract there's a little imaginary 1 here, so our numerator wins out, 2 minus 1 is 1 we end up with just a single y in the numerator.

And our last is our z, we have a single z in the numerator and 3z's in the denominator so 1 minus 3 is -2, the negative tells me that it's going to be in the denominator so we have a z² in the bottom.

So using our rules of exponents, we were able to expand this out and make it a little bit ugly, but then combine all our like terms just using our tricks from division.

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