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

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

M36-25

sahilvijay · Answer

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

Luckisnoexcuse · Answer

(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

Bunuel · Answer

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

chetan2u · Answer

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

exc4libur · Answer

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… ( -2)=-1…-1(-3)≠-1 
 x=1:x^y(x-2)=-1… ( -2)=-1…(-1)=-1

Answer (D)

analytica233 · Answer

My solution is presented in the image attached.

Bunuel · Answer

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

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