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# You may have seen this on a Kaplan CD. X= 2/ 3^7

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Eternal Intern
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You may have seen this on a Kaplan CD. X= 2/ 3^7 [#permalink]

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29 Jul 2003, 13:49
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You may have seen this on a Kaplan CD.

X= 2/ 3^7 + 2/ 3^8 + 2/3^9 + 1/ 3 ^9

3^9 is the common denominator.

3^2 * 2
-------
3^9
= Is it legal to cancel the exponents here?
1 *2
-----
3^7

Thank you.

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29 Jul 2003, 14:06
If you had (3^2) * 2 / (3^9) then the answer will be 2/(3^7). You will actually subtract the powers here.

a^m / a^n = a^(m-n)

In this problem though, the answer should be 1/(3^6)
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