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Manager  Joined: 06 Apr 2010
Posts: 113
Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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2
4 00:00

Difficulty:   35% (medium)

Question Stats: 70% (01:32) correct 30% (01:33) wrong based on 213 sessions

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Is x > k?

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

(2) $$9^x * 3^k = 81$$
Math Expert V
Joined: 02 Sep 2009
Posts: 53657
Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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4
1
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.

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SVP  Joined: 06 Sep 2013
Posts: 1677
Concentration: Finance
Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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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

Cheers
J I'm back and not stopping until I hit 760+
Senior Manager  Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 414
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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

From  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  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?
Senior Manager  Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 414
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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

From  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  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..
Current Student B
Status: DONE!
Joined: 05 Sep 2016
Posts: 372
Re: Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81  [#permalink]

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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
Non-Human User Joined: 09 Sep 2013
Posts: 10130
Re: Is x > k? (1) 2^x 2^k = 4 (2) 9^x 3^k = 81 A. Statement  [#permalink]

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_________________ Re: Is x > k? (1) 2^x 2^k = 4 (2) 9^x 3^k = 81 A. Statement   [#permalink] 21 Feb 2019, 22:22
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# Is x > k? (1) 2^x * 2^k = 4 (2) 9^x * 3^k = 81

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