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## exponent problem

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### exponent problem

by jfrick90 » Wed Aug 04, 2010 6:08 pm
I was wondering if there was a way to simplify this problem instead of resorting to calculating the problem out by numerator and denominator. Thank you!

:::> [(8^2)(3^3)(2^4)]/(96^2)

I know this is a silly question but just to help brush up on my understanding of exponents would be a great help!

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by KrazyKarl » Wed Aug 04, 2010 7:45 pm
I'm learning to like these problems! With multiplying/dividing exponents you should try to make the bases the same, so break down 96^2 into smaller numbers:

96^2 = (4 * 24)^2 = (2^2 * 4 * 6)^2

Maybe not the nicest way to do it but I think it's easy to look at 100 as 25*4 and then subtract 4 to get 24*4, and then you have the (2^2)^2 term so you can cancel that with 2^4. Now you have:

(8^2 * 3^3) / (4*6)^2

You need to get the 6 broken down to 2 and 3, so:

(8^2 * 3^3) / (4*2*3)^2

And then the 4*2 is 8 so you have:

8^2 * 3^3 / 8^2 * 3^2

The 8^2s cancel and you have 3^3 / 3^2 = 3.

It's harder to type than to write so maybe you can do it faster, but that's how I thought about it. Get the bases to be the same and you can start cancelling.

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by selango » Wed Aug 04, 2010 7:55 pm
[(8^2)(3^3)(2^4)]/(96^2)

=[(2^6)(3^3)(2^4)]/(96^2)]

=[(2^10)(3^3)/((2^5*3)^2]

=[(2^10)(3^3)/((2^10)*(3^2)]

=3
--Anand--

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