Solve for x and y, if :

$(\sqrt{32})^x ÷ 2^{y + 1} = 1 \text{ and } 8^y - 16^{4 - \dfrac{x}{2}} = 0$

Given,

$\Rightarrow (\sqrt{32})^x ÷ 2^{y + 1} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{32})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{2^5})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow [(2^5)^{\dfrac{1}{2}}]^x = 2^{y + 1} \\[1em] \Rightarrow 2^{5 \times \dfrac{1}{2} \times x} = 2^{y + 1} \\[1em] \Rightarrow \dfrac{5x}{2} = y + 1 \\[1em] \Rightarrow 5x = 2y + 2 \\[1em] \Rightarrow x = \dfrac{2y + 2}{5} \text{ ……(1)}$

Given,

$\Rightarrow 8^y - 16^{4 - \dfrac{x}{2}} = 0 \\[1em] \Rightarrow 8^y = 16^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow (2^3)^y = (2^4)^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow 2^{3y} = (2)^{4\Big(4 - \dfrac{x}{2}\Big)} \\[1em] \Rightarrow 2^{3y} = 2^{16 - 2x} \\[1em] \Rightarrow 3y = 16 - 2x \\[1em] \Rightarrow 2x = 16 - 3y \text{ ……..(2)}$

Substituting value of x from equation (1) in equation (2), we get :

$\Rightarrow 2\Big(\dfrac{2y + 2}{5}\Big) = 16 - 3y \\[1em] \Rightarrow \dfrac{4y + 4}{5} = 16 - 3y \\[1em] \Rightarrow 4y + 4 = 5(16 - 3y) \\[1em] \Rightarrow 4y + 4 = 80 - 15y \\[1em] \Rightarrow 4y + 15y = 80 - 4 \\[1em] \Rightarrow 19y = 76 \\[1em] \Rightarrow y = \dfrac{76}{19} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{2y + 2}{5} = \dfrac{2 \times 4 + 2}{5} \\[1em] = \dfrac{8 + 2}{5} \\[1em] = \dfrac{10}{5} = 2.$

Hence, x = 2 and y = 4.