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Best way to simplify 96^2 into other exponents?

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by andydbaines » Sun Apr 12, 2009 1:17 pm
Probably a pretty basic question:

How can one quickly simplify 2 (or 3) digit numbers with exponents to products of more basic numbers with exponents?

i.e. 96^2 = (8^2)(3^2)(2^4)

Obviously 8, 3, 2 are factors of 96...

Any help would be appreciated! Thanks.
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by moutar » Sun Apr 12, 2009 1:36 pm
I would just work with the product of prime factors.

So 96 = 2 x 48
= 2 x 2 x 24
= 2 x 2 x 2 x 12
= 2 x 2 x 2 x 2 x 6
= 2 x 2 x 2 x 2 x 2 x 3 = 2^5 x 3

Obviously its faster if you see that 32 is a factor and 32 = 2^5

Therefore, 96^2 = (2^5 x 3)^2 = 2^10 x 3^2

Is this what you were asking? Is there any sort of question you might want answering?
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Hey thanks! Here's the question...

by andydbaines » Sun Apr 12, 2009 2:44 pm
(8^2)(3^3)(2^4)
______________ =
96^2

The answer provided is as follows:

(8^2)(3^3)(2^4)
______________

(8^2)(3^2)(2^4)

= 3

I was hoping to figure how they factored the denominator so easily to eliminate the numerator.

Cheers!
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by gmat740 » Tue Apr 14, 2009 7:18 pm
Depending upon the Question you are working, you can apply the method suggested by Moutar: break down the number into its prime factors and then do all the exponential calculations.

However, the Question you provided,
I would advice you not to break 96 into its prime factor

Rather break 96 into factors of 8,3 and 2 because by doing that, you will easily cancel out the terms from Numerator and denominator.

so that saves time for you.

Hope this Helps

Karan
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