Solve the following equation :

8^{x + 1} = 16^{y + 2}, $\Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y}$

Given,

⇒ 8^{x + 1} = 16^{y + 2}

⇒ (2³)^{x + 1} = (2⁴)^{y + 2}

⇒ 2^{3x + 3} = 2^{4y + 8}

⇒ 3x + 3 = 4y + 8

⇒ 3x - 4y = 5 …….(i)

Given,

$\Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big[\Big(\dfrac{1}{2}\Big)^2\Big]^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{2}\Big)^{6y} \\[1em] \Rightarrow 3 + x = 6y \\[1em] \Rightarrow 6y - x = 3 …….(ii)$

Multiplying (ii) by 3 we get,

⇒ 18y - 3x = 9 …….(iii)

Adding (i) and (iii) we get,

⇒ 3x - 4y + (18y - 3x) = 5 + 9
⇒ 14y = 14
⇒ y = 1.

Substituting value of y in (i) we get,

⇒ 3x - 4(1) = 5
⇒ 3x - 4 = 5
⇒ 3x = 9
⇒ x = 3.

Hence, x = 3 and y = 1.