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

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Math Expert
Joined: 02 Sep 2009
Posts: 64222
If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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28 Aug 2017, 02:25
1
1
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Difficulty:

95% (hard)

Question Stats:

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

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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) $$x^2=1$$

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Math Expert
Joined: 02 Sep 2009
Posts: 64222
If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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28 Aug 2017, 02:25
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) $$x^2=1$$

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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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28 Aug 2017, 02:37
Ans is D, both are sufficient to give x=1
see the attached pic.
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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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28 Aug 2017, 02:43
1
1
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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Luckisnoexcuse
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Posts: 64222
Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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24 Dec 2018, 01:14
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) $$x^2=1$$

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 8602
Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?  [#permalink]

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

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