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GMAT PREP - algebra/exponents

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by aleph777 » Sun Feb 27, 2011 5:37 am
Had trouble with this one. I think you need to expand 3(2^13) to become 2^13 + 2^13 + 2^13, but then I can't make any progress on the left side of the equation.
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by manpsingh87 » Sun Feb 27, 2011 5:43 am
aleph777 wrote:Had trouble with this one. I think you need to expand 3(2^13) to become 2^13 + 2^13 + 2^13, but then I can't make any progress on the left side of the equation.
2^x-2(2^2 - 1)
3 * 2^x-2= 3 * 2^13
equating both sides
x-2= 13
x=15
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by GMATGuruNY » Sun Feb 27, 2011 6:08 am
If 2^x - 2^(x-2) = 3(2^13), what is the value of x?
a) 9
b) 11
c) 13
d) 15
e) 17
We can plug in the answer choices for x.

Answer choice C: x= 13
2^13 - 2^(13-2) = 3(2^13)
2^13 - 2^11 = 3(2^13)
2^11(2^2 - 1) = 3(2^13)
2^11(3) = 3(2^13)

Plugging in x=13 made the exponent on the left 2^11.
To match 2^13 on the right side of the equation, the exponent needs to be increased by 2.
Thus, x = 13+2 = 15.

The correct answer is D.
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by Anurag@Gurome » Sun Feb 27, 2011 6:18 am
Note that we can write, 2^x = 2^(x - 2 + 2) = (2^2)*(2^(x - 2)) = 4*(2^(x - 2))

Hence, we can rearrange the left side as,
  • ... 2^x - 2^(x - 2)
    = 4*(2^(x - 2)) - 2^(x - 2)
    = 2^(x - 2)*(4 - 1) ............ Taking 2^(x - 2) common from both terms
    = 3*2^(x - 2)
Now comparing with the left side, (x - 2) = 13, i.e. x = 15

The correct answer is D.

Another Method: Note that left side has powers of 2 only, but right side has a 3! How to deal this? We can write 3 = (2^2) - 1. Let's rearrange left side using this,
  • 3*(2^13) = (2^2 - 1)*(2^13) = (2^15) - (2^13) = (2^15) - 2^(15 - 2)
Comparing with left side, x = 15
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by aleph777 » Sun Feb 27, 2011 7:50 am
Thanks everyone.

Mitch, your plug-in method is so easy--I often overlook this approach.

And the same goes for what manpsingh87 did... I see this sort of reverse distribution problem so infrequently, I forget it can be done!

Got to build that into my repertoire for game day!
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