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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: 52123
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
6
00:00

Difficulty:

95% (hard)

Question Stats:

36% (02:01) correct 64% (01:55) wrong based on 97 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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Joined: 02 Sep 2009
Posts: 52123
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) $$x^2=1$$

Previous version had a typo in it. Had to re-post. Sorry about it.
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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.
Attachments

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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
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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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: 7199
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x? &nbs [#permalink] 24 Dec 2018, 06:29
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# If xy ≠ 0 and x^(y + 1) = 2x^y - 1, what is the value of x?

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