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If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2?

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If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2?  [#permalink]

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New post 01 Oct 2018, 03:40
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

34% (02:41) correct 66% (02:43) wrong based on 64 sessions

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Re: If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2?  [#permalink]

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New post 01 Oct 2018, 03:48
Submitted option C as an answer but feel like B is the answer, if anyone is able to solve with B then could you please post the solution.
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Re: If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2?  [#permalink]

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New post 11 Oct 2018, 10:58
Bunuel could you please provide assist to understand why D is answer.
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Re: If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2?  [#permalink]

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New post 22 Oct 2018, 15:29
I can see how statement 1 is sufficient, but I am getting not sufficient for statement 2. Can you show how statement two is sufficient? Perhaps I made an algebraic error.

I began by equating the exponents, so we know that k*y = l*y*2 - 8 (is that the correct reading or is that supposed to be y^2?)

Statement one states that k = - 6
Plugging in, we get -6y = 2ly-8
y(2l+6) = 8
y can be either 2 or 4, which means that l can be -1 or -2
In both cases when multiplied by k = -6, we have a result > 2
Sufficient

Statement 2 states that
3l−k=3
Solving for k and plugging in, we get (3l - 3)*y = 2ly - 8
3ly - 3y = 2ly - 8
ly -3y = -8
y can be either 2 or 4
when y is 2, l = -1 and therefore k = -6. lk > 2
when y is 4, l = 1 and therefore k = 0 lk is not > 2
Two differing answers. Not sufficient.

Hence, my answer of A.
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Re: If x and y belong to the set {2, 4}, and x^ky = x^(ly2−8), is kl > 2? &nbs [#permalink] 22 Oct 2018, 15:29
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