I have problem that I am trying to help my daughter solve. It involves number, letters and negative exponents. Can you give me some pointers

Question

Here is the problem 2b^3 x 3a^4b^3 x 4b^4 divided by 4ba^1

The first and last exponents are both negative and so is the 4ba

Answer 1

I am having a little difficulty understanding exactly what you mean, but I will give it my best shot. I think your problem looks something like this:

(2b^-3 x 3a⁴b³ x 4b^-4)/4ba^-1

If that is correct, I will show you how I would solve it. Even if it is not exactly what you mean, you might pick-up a few tips by following along.

I see that you have terms being multiplied and divided. I also see that some of the exponents are positive, and some are negative. In this situation, remember two rules that can help you solve the problem: 1) To change the exponent from negative to positive or positive to negative, just move the term to the other half of the fraction. For example, move it from the denominator to the numerator. 2) When you multiply like bases, you add the exponents.

= (2b^-3 x 3a⁴b³ x 4b^-4)/4ba^-1

First, lets turn that a^-1 into just an a¹ by moving it up top.

= ((2b^-3 x 3a⁴b³ x 4b^-4)a)/4b

You can do this because 1/a^-x = a^x. Do the same with the b in the denominator.

= ((2b^-3 x 3a⁴b³ x 4b^-4)ab^-1)/4

Now notice you still just have multiplication and division. Let's apply the rule that says when you multiply like bases, you add the exponents.

First, let me rewrite the problem more clearly. Recall that multiplication is commutative, which means the order is not important

= (2 x 3 x 4 x a⁴ x a x b^-3 x b³ x b^-4 x b^-1)/4

I've re-ordered the terms so that the like bases are next to each other. This is only to make the problem more clear and can be skipped once you are comfortable with these types of problems. Now, for all the "a" terms, add the exponents. Do the same for the b's.

= (24 x a⁵ x b^-5)/4

= 6a⁵b^-5 (Answer)

OR

= 6a⁵/b⁵ (Answer, different style)

I hope this helps. Good luck.