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smallsorrow
- Master | Next Rank: 500 Posts
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- Joined: Mon Sep 08, 2008 12:47 am
If you notice that you can factor out 2^(x-2) on the left side, the rest of the problem is fairly straightforward:smallsorrow wrote:2^x - 2 ^x-2 = 3 (2^13), what is x?
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2^x - 2 ^x-2 = 3 (2^13)
[2^(x-2)]*(2^2 - 1) = 3*(2^13)
[2^(x-2)]*(3) = 3*(2^13)
2^(x-2) = 2^13
x-2 = 13
x = 15
We do this kind of factoring all the time- when we are adding or subtracting two terms with the same base and different exponents, it is very often useful to factor out the term with the smaller exponent. We often do this when the base is x:
x^7 - x^5 = x^5(x^2 - 1)
and often need to do this in number theory questions (because it helps to get a prime factorization):
2^30 - 2^28 = 2^28*(2^2 - 1)
The question in the post above is a bit trickier because there's an x in the exponent, but the idea is the same. This kind of 'simple' factoring is one of the most useful techniques in all of algebra, and is very important to know well for the GMAT.
