If 2^x - 2^(x-2) = 3*2^13, what is the value of x? : GMAT Problem Solving (PS)
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# If 2^x - 2^(x-2) = 3*2^13, what is the value of x?

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If 2^x - 2^(x-2) = 3*2^13, what is the value of x? [#permalink]

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14 Nov 2010, 13:34
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If $$2^x - 2^{x-2} = 3(2^{13})$$, what value has x?

A. 9
B. 11
C. 13
D. 15
E. 17
[Reveal] Spoiler: OA
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14 Nov 2010, 13:51
Pepe wrote:
Another question I have problem with:

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

$$2^x - 2^{x-2} = 3*2^{13}$$ --> factor out $$2^{x-2}$$ --> $$2^{x-2}*(2^2-1)=3*2^{13}$$ --> $$2^{x-2}*3=3*2^{13}$$ --> $$2^{x-2}=2^{13}$$ --> $$x-2=13$$ --> $$x=15$$.

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14 Nov 2010, 13:56
Thank you Bunuel,

it's so simple if you see the solution
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15 Nov 2010, 14:18
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Hey guys,

This has always been a favorite problem of mine - the first time I saw it, an instructor had blanked on how to solve it and emailed me a photo from his phone asking for help. He had excused himself from a tutoring session and needed me to explain how to solve it (correctly, of course) within a few minutes, so the pressure was on!

Bunuel's explanation is perfect (as always), but when the pressure was on and I wasn't exactly thinking about factoring, I did this instead - I looked to see if there were a pattern in the subtraction at left (2 to an exponent minus 2 to another exponent, two less) that would always produce 3*something on the right. So I did:

x = 3 and x-2 = 1
2^3 - 2^1 = 8 - 2 = 6

And 6 = 3(2^1), so I had a start.

x = 4 and x-2 = 2
2^4 - 2^2 = 16 - 4 = 12

And 12 = 3(2^2), so the pattern held

x = 5 and x-2 = 3
2^5 - 2^3 = 32 - 8 = 24

And 24 = 3(2^3), and the pattern became clear...
The operation at left was always producing 3*2^(x-2) as its answer, so if x-2 = 13, then x = 15.

Strategically, using small numbers to establish patterns works pretty well when huge numbers (like 3(2^13)) are in play, and when exponents are involved (exponents are essentially just repetitive multiplication, so there are bound to be some repetitive patterns involved). If you can factor like Bunuel did, that's a great way to go...but I'd recommend having the "prove patterns w/ small numers" ideology in your arsenal!
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06 Feb 2012, 00:21
2^x-2^(x-2)=3(2^13)
=> 2^(x-2) * [2^2 - 1] = 2^(15-2) * [2^2 - 1]
=> x = 15

Option (4)
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Re: If 2^x - 2^(x-2) = 3*2^13, What is the value of x? [#permalink]

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13 May 2012, 05:27
This is how I solved this problem.

2^x-2^x-2=3(2^13)
Factor 2^x 2^x(1-1)*2^-2=3(2^13)
x-2=13
x=15
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Re: If 2^x - 2^(x-2) = 3*2^13, What is the value of x? [#permalink]

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08 Jun 2012, 20:04
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2^x-2^x-2=3*2^13
2^x(1-2^-2)=2^13*3
2^x(1-1/2^2)=3*2^13
2^x(3/4)=3*2^13
2^x=3*2^13*4/3
2^x=2^13*4
2^x=2^13*2^2
2^x=2^15
Hence, x=15
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Re: If 2^x - 2^(x-2) = 3*2^13, What is the value of x? [#permalink]

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08 Jun 2012, 23:44
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This has the look of not factoring easily, so I applied some logic first.

We can rule out A,B and C (as the left hand side would all be 2^13 less a positive number (smaller than 2^13), so could never equal 3 * 2^13 (larger than 2^13)

Try D:
2^15 - 2^13 = 3*2^13

We can work with this easily:

Divide by 2^13
2^2 - 1 = 3
3 = 3

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Re: If 2^x - 2^(x-2) = 3*2^13, What is the value of x? [#permalink]

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09 Jun 2012, 01:39
Hi,

One can always try the options. Values less than 13 can be eliminated. Try others and whenever LHS = RHS, there you get the answer.

Regards,
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22 Jul 2012, 21:43
Hi,

Usually for problems where you have a number at different powers summed you can start by factoring by the smallest power of that number:

$$2^x-2^(x-2)=2^(x-2)*(2^2-1)=2^(x-2)*3$$

Hence

$$x-2=13$$
$$x=15$$

Not sure it is 700+ level
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Re: If 2^x - 2^(x-2) = 3*2^13, what is the value of x? [#permalink]

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15 Jun 2014, 07:14
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Re: If 2^x - 2^(x-2) = 3*2^13, what is the value of x? [#permalink]

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24 Jun 2014, 01:37
$$2^x - 2^{(x-2)} = 3*2^{13}$$

Working on LHS

$$2^x - 2^{(x-2)}$$

$$= 2^x - \frac{2^x}{4}$$

$$= 2^x (1 - \frac{1}{4})$$

$$= 2^x (\frac{3}{4})$$

Equating to RHS

$$2^x (\frac{3}{4}) = 3 * 2^{13}$$

$$2^x * 3 = 3 * 2^{15}$$

x = 15 = Answer = D
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Re: If 2^x - 2^(x-2) = 3*2^13, what is the value of x?   [#permalink] 24 Jun 2014, 01:37
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