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

Official Solution:


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

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

Substitute \(x^y = x\) in the equation given in the stem \(x^{y}*x = 2x^y - 1\):

\(x*x = 2x - 1\);

\(x^2 - 2x + 1= 0\)

\((x - 1)^2 = 0\)

\(x = 1\). Sufficient.

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

\(x=1\) or \(x=-1\)

Check whether these values satisfy the equation given in the stem \(x^{(y + 1)} = 2x^y - 1\):

If \(x = 1\), then \(x^{(y + 1)} = 1^{y+1}=1\) and \(2x^y - 1=2*1^y - 1 = 2-1=1\). LHS = RHS. So, \(x\) can be 1.

If \(x = -1\), then we'd have: \((-1)^{y+1}=2*(-1)^y - 1\):

\((-1)^y*(-1)=2*(-1)^y - 1\);

\((-1)^y= \frac{1}{3}\). This is not possible because -1 in any power is ether 1 or -1. So, \(x\) cannot be -1.

Thus, \(x = 1\). Sufficient.


Answer: D
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Ans is D, both are sufficient to give x=1
see the attached pic.

(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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Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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

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

(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)

My solution is presented in the image attached.

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