# Thread: Tough algebra problem. Fractions and negative exponents

1. ## Tough algebra problem. Fractions and negative exponents

This problem with fractions and negative exponents is killing me.. I thought algebra was supposed to be easy, lol. Any help greatly appreciated. I know the answer but cant get the steps right for some reason..

Simplify: Assume that all variables represent nonzero integers.
$\displaystyle (2^-2)^a x (2^b)^-a / (2^-2)^-b x (2^b)^-2a$

Here is the answer according to the book:

Ans. = 2^(-2a - 2b + ab)

I'm never able to get the final answer correct for some reason. I know it's hard to read these problems using computer text but I'd appreciate if you can bare with me and tell me where I'm messing up here.

Original Numerator: (2^-2)^a x (2^b)^-a

= 2^-2a x 2^-ba

= 1/(4^a) x 1/(2^ba)

= (1 / 8^2ab) <--simplified numerator

Original Denominator: (2^-2)^-b x (2^b)^-2a

= 2^2b x 2^-2ab

= 2^2b x 1/(2^ab)

= (2^2b / 2^ab) <--simplified denominator

So the new problem is:

(1 / 8^2ab) / (2^2b / 2^ab)

Every time I solve this with cross multiplication I get the wrong answer...so I'm guessing I messed up somewhere along the way.. Please heeelp!

I've tried everything I know but never seem to get the correct answer. I know typing formulas is a pain, but if someone could show me the steps involved I'd really appreciate it!

2. your simplified numerator is wrong.it will actually be
1/2^(2a+ab)

3. Thanks alot. I see what I did wrong there but I'm still having trouble following the next steps to get the answer.

I've been out of school for ten years and am trying to relearn this stuff on my own. The textbook had a lot of very simple practice examples of this type of problem, but nothing as complicated as the one I posted. It had no examples that show step by step what is involved in multiplying and dividing the (x-y) or (x+y) type exponents, so I'm wandering in the dark trying to figure this out.

I know it's a pain in the , but when you find time I'd appreciate if you could show the steps involved to get the final answer I posted up top. I really need to see how a problem like this is done to grasp the concept. I'd really appreciate any help or tips. Thanks again.

4. Originally Posted by Libertarian
(2^-2)^a x (2^b)^-a / (2^-2)^-b x (2^b)^-2a
$\displaystyle \frac{(2^{-2})^a(2^b)^{-a}}{(2^{-2})^{-b}(2^b)^{-2a}}$

$\displaystyle = \frac{2^{-2a}2^{-ab}}{2^{2b}2^{-2ab}}$

$\displaystyle = \frac{2^{-2a - ab}}{2^{2b - 2ab}}$

$\displaystyle = 2^{-2a - ab - (2b - 2ab)}$

$\displaystyle = 2^{-2a - ab - 2b + 2ab}$

$\displaystyle = 2^{-2a - 2b + ab}$

Make sure to ask questions about any step that you don't understand.

-Dan

5. wow, thanks dan!

It is so much easier when you just subtract the exponents. I was trying to long way by creating reciprocals for both numerator and denominator then cross multiplying and it was driving me crazy.

My only question is...If the main number in the denominator was something other than 2, this problem would be impossible the way you did it right? Would it be necessary to go the long way (turning ^-3 into 1/3 etc then cross multiplying)...or is there some way to keep it simple the way you did above?

Thanks again

6. Originally Posted by Libertarian
wow, thanks dan!

It is so much easier when you just subtract the exponents. I was trying to long way by creating reciprocals for both numerator and denominator then cross multiplying and it was driving me crazy.

My only question is...If the main number in the denominator was something other than 2, this problem would be impossible the way you did it right? Would it be necessary to go the long way (turning ^-3 into 1/3 etc then cross multiplying)...or is there some way to keep it simple the way you did above?

Thanks again
When you're dealing with exponents, all you need to know are the rules. If the numbers in the denominator had been different, you could still have applied this rule: $\displaystyle \frac{x^a}{x^b} = x^{a-b}$. That tool, in addition to $\displaystyle x^a \cdot x^b = x^{a+b}$ and $\displaystyle (x^a)^b = x^{ab}$, are really all you need to answer any question of this type. But these sorts of questions are a good test of your knowledge of these rules and your ability to apply them.