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Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81
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udaymathapati
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Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  09 Dec 2010, 08:51
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Is x > k?


(1) \(2^x * 2^k = 4\)

(2) \(9^x * 3^k = 81\)
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Bunuel
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  09 Dec 2010, 09:05
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udaymathapati wrote:
Is x > k?
(1) 2x • 2k = 4
(2) 9x • 3k = 81


In its current form statement (1) xk=1 and statement (2) xk=3 clearly contradict each other and we know that on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

So I think the question should be:

Is x > k?
(1) 2^x * 2^k = 4 --> \(2^{x+k}=2^2\) --> \(x+k=2\) --> not sufficient to determine which one is greater.
(2) 9^x * 3^k = 81 --> \(3^{2x+k}=3^4\) --> \(2x+k=4\) --> not sufficient to determine which one is greater.

(1)+(2) \(x+k=2\) and \(2x+k=4\) --> subtract 1 from 2: \(x=2\) --> \(k=0\) --> so, \(x>k\). Sufficient.

Answer: C.
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jlgdr
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  31 Mar 2014, 07:51
From Statement 1 we get that x+k=2.

From Statement 2 we get that 2x+k=4. This is also insufficient.

From both statements we get that x=2, and k=0.

Therefore C is the correct answer

Please ask if anything remains unclear
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pacifist85
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  27 Jul 2015, 02:02
Hello,

I ended up in the same but in a more complicated way perhaps. I don't know why, as I always use this proposed way by bunuel. But, this is a good oportunity to ask something.

So, from [1] I ended up that x=1, k=1 or that x=2 and k=0 (or the opposite). So, NS.

From [2] I concluded that k should be zero, so x will be greater than k, like this:
9^x*3^k=81
3*(3^x*1^k)=3^4
3*3^x=3^4, so again not sufficient.

From [1 and 2], as 1 in whichever power is 1, and because in [2] we have 1^k, I accepted that k=0 and x=2.

Anyway, this could be wrong, but would like to ask about the part in red. Can you take the common factor 3 out like this?
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pacifist85
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  27 Jul 2015, 02:17
pacifist85 wrote:
Hello,

I ended up in the same but in a more complicated way perhaps. I don't know why, as I always use this proposed way by bunuel. But, this is a good oportunity to ask something.

So, from [1] I ended up that x=1, k=1 or that x=2 and k=0 (or the opposite). So, NS.

From [2] I concluded that k should be zero, so x will be greater than k, like this:
9^x*3^k=81
3*(3^x*1^k)=3^4
3*3^x=3^4, so again not sufficient.

From [1 and 2], as 1 in whichever power is 1, and because in [2] we have 1^k, I accepted that k=0 and x=2.

Anyway, this could be wrong, but would like to ask about the part in red. Can you take the common factor 3 out like this?


Or perhaps you cannot do that due to the order of calculations? So, we would need to resolve the powers first, and then take the common factor out..
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law258
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  19 Sep 2016, 06:28
C is correct. Here's why:

(1) (2^x)(2^k) = 4 --> Rewrite as (2^x)(2^k) = 2^2

x+k = 2

NOT SUFFICIENT - x and k could be any values for all we know at this point

(2) (9^x)(3^k) =81 --> Rewrite as (3^2x)(3^k) = 3^4

2x+k = 4

NOT SUFFICIENT - same as (1)

A,B,D eliminated

Together - (1) + (2) --> Plug equation from (1) (i.e. x = 2-k) into equation from (2) (i.e. 2x+k = 4)

2(2-k)+k = 4
4-2k+k = 4
4-k = 4
k = 0 --> x = 2

SUFFICIENT
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GMATwithAK
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Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81 [#permalink] New post  16 Feb 2021, 08:05
Can you take the common factor 3 out like this?

No. Because it is a multiplication. Had the expression been 9^x+3^k, you can take out 3 as a common factor.



pacifist85 wrote:
Hello,

I ended up in the same but in a more complicated way perhaps. I don't know why, as I always use this proposed way by bunuel. But, this is a good oportunity to ask something.

So, from [1] I ended up that x=1, k=1 or that x=2 and k=0 (or the opposite). So, NS.

From [2] I concluded that k should be zero, so x will be greater than k, like this:
9^x*3^k=81
3*(3^x*1^k)=3^4
3*3^x=3^4, so again not sufficient.

From [1 and 2], as 1 in whichever power is 1, and because in [2] we have 1^k, I accepted that k=0 and x=2.

Anyway, this could be wrong, but would like to ask about the part in red. Can you take the common factor 3 out like this?

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