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 = ax. 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)
= 6a5/b5 (Answer, different style)
I hope this helps. Good luck.