Solve: \[ 8^{x+1} = 16^{y+2}, \quad \left(\frac{1}{2}\right)^{3+x} = \left(\frac{1}{4}\right)^{3y} \]
Solution
\[ 8^{x+1} = 16^{y+2} \]
\[ \Rightarrow (2^3)^{x+1} = (2^4)^{y+2} \]
\[ \Rightarrow 2^{3x+3} = 2^{4y+8} \]
\[ \Rightarrow 3x + 3 = 4y + 8 \quad …(1) \]
\[ \left(\frac{1}{2}\right)^{3+x} = \left(\frac{1}{4}\right)^{3y} \]
\[ \Rightarrow (2^{-1})^{3+x} = (2^{-2})^{3y} \]
\[ \Rightarrow 2^{-3-x} = 2^{-6y} \]
\[ \Rightarrow -3 – x = -6y \quad …(2) \]
\[ \text{From (2): } x + 3 = 6y \]
\[ \Rightarrow x = 6y – 3 \]
\[ \text{Substitute in (1):} \]
\[ 3(6y – 3) + 3 = 4y + 8 \]
\[ 18y – 9 + 3 = 4y + 8 \]
\[ 18y – 6 = 4y + 8 \]
\[ 14y = 14 \]
\[ y = 1 \]
\[ x = 6(1) – 3 \]
\[ x = 3 \]
Final Answer:
\[ \boxed{x = 3,\; y = 1} \]