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If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?

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If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 28 Aug 2017, 02:25
1
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A
B
C
D
E

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Question Stats:

32% (02:38) correct 68% (02:20) wrong based on 171 sessions

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If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 28 Aug 2017, 02:25
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Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 28 Aug 2017, 02:37
Ans is D, both are sufficient to give x=1
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Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 28 Aug 2017, 02:43
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Bunuel wrote:

Fresh GMAT Club Tests' Challenge Question:



If \(xy ≠ 0\) and \(x^{(y + 1)} = 2x^y - 1\), what is the value of x?

(1) \(x^y = x\)

(2) \(x^2=1\)


(1) (x^y) * x = 2x^y - 1
x^2 -2x +1 = 0
(x-1)^2 = 0
x=1
Sufficient

(2)
x=+1 or -1
if x =1 then x^{(y + 1)} = 2x^y - 1 holds for all cases of y (odd,even)
if x= -1 then x^{(y + 1)} = 2x^y - 1 does not hold
hence x can only be 1
sufficient
D
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Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 24 Dec 2018, 01:14
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Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 24 Dec 2018, 06:29
Bunuel wrote:

Fresh GMAT Club Tests' Challenge Question:



If \(xy ≠ 0\) and \(x^{(y + 1)} = 2x^y - 1\), what is the value of x?

(1) \(x^y = x\)

(2) \(x^2=1\)



\(x^{(y + 1)} = 2x^y - 1....X*x^y=2x^y-1....\)

(1) \(x^y = x\)
Substitute this in equation above...
\(x*x=2x-1......x^2-2x+1=0.....
.(x-1)^2=0\)
Therefore, x=1...
Sufficient

(2) \(x^2=1\)
so x=1 or -1..
Now \(x*x^y=2x^y-1....(2-x)x^y=1\)..
Substitute X as 1..
\(.(2-x)x^y=1......(2-1)1^y=1\).. possible so X can be 1
Substitute X as -1
\((2-x)x^y=1......(2-(-1))*(-1)^y=1.......3*(-1)^y=1\).. not possible
So X is 1
Sufficient

D
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If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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New post 22 Oct 2019, 07:47
Bunuel wrote:
If \(xy ≠ 0\) and \(x^{(y + 1)} = 2x^y - 1\), what is the value of x?

(1) \(x^y = x\)

(2) \(x^2=1\)


\(xy ≠ 0…(x,y)≠0\)
\(x^{(y + 1)} = 2x^y - 1…(x^y)(x)=2(x^y)-1…x^y(x-2)=-1\)

(1) \(x^y = x\) sufic.
\(x^y(x-2)=-1…x(x-2)=-1…x^2-2x+1=0…(x-1)^2=0…x=1\)

(2) \(x^2=1\) sufic.
\(x^2=1…|x|=(1,-1)\)
\(x=-1:x^y(x-2)=-1…[-1]([-1]-2)=-1…-1(-3)≠-1\)
\(x=1:x^y(x-2)=-1…[1]([1]-2)=-1…(-1)=-1\)

Answer (D)
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If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?   [#permalink] 22 Oct 2019, 07:47

If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?

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