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I'm a beginner to maths and have trouble simplyfying the following function:

$$\frac{p^y \cdot (pq)^o}{p^{2y+o} \cdot q^{o-2}}$$

The final answer is

$$p^{-y} \cdot q^2$$

But I'm not sure how to get there.

Any help is appreciated.

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1 Answer 1

up vote 5 down vote accepted

Here's the method in general, without actually working out your example. You should do that yourself to seal the concepts.

The intermediate goal is to get all the powers of p and q separated in both the numerator and denominator. In this case, it's almost there, with the exception of (pq)^o. So expand that first, using the principle (x*y)^a = x^a * y^a.

Then gather the p's and q's using the properties of multiplication and exponentiation, x^a * x^b = x^(a+b). Finally, match the p's in the numerator and denominator, likewise the q's, and using the principle x^a / x^b = x^(a-b) calculate the ultimate powers of p and q. Note that there are two ways you could handle the power of p in the final answer, since it is negative.

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Thanks, the only problem I have now is that I do not know how to expend (pq)^o because this is being multiplied with p^y.. p^y * p^o * q^o is different than p^y * (p^o * q^o). I probably am missing something basic. –  Tom Sep 7 '10 at 13:52
    
See my revision. There is a principle for that too. –  David Lewis Sep 7 '10 at 13:53
1  
Remember that multiplication is associative! –  J. M. Sep 7 '10 at 13:53
    
Great, Thanks!! –  Tom Sep 7 '10 at 14:07

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