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