Explanation

\(2^{(2y+1)} - 5(2^y) + 2\) = 0

Let p = 2\(^y\)
\(2^{2y} (2^1) - 5(2^y)\) + 2 = 0
2p\(^2\) - 5p + 2 = 0
2p\(^2\) - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or \(\frac{1}{2}\)

p = 2\(^y\)
when p = 2
2\(^y\) = 2
y = 1

when p = \(\frac{1}{2}\)

2\(^y\) = \(\frac{1}{2}\)

2\(^y\) = 2\(^{-1}\)

y = -1

y = -1 or 1

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