Given

$\rm 2^y + 2^{y-2} = 5$

By using the law of indices, $\rm a^{m-n} = \frac{a^m}{a^n}$, we get,

$\rm or, 2^y + \frac{2^y}{2^2} = 5$

$\rm or, 2^y + \frac{2^y}{4} = 5$

Multiplying both sides of the equation by 4, we get,

$\rm or, 2^y \cdot 4 + \frac{2^y}{4} \cdot 4 = 5 \cdot 4$

$\rm or, 4 \cdot 2^y + 2^y = 20$

$\rm or, 2^y ( 4 + 1) = 20$

$\rm or, 2^y \cdot 5 = 20$

Dividing both sides of the equation by 5, we get,

$\rm or, 2^y \cdot \frac{5}{5} = \frac{20}{5}$

$\rm or, 2^y = 4$

$\rm or, 2^y = 2^2$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore y = 2$

Hence, the required value of y = 2.