2 . Dftermine the valuer of \( a \) and \( y \) if (4) \[ \frac{3^{y+1}}{32}=\sqrt{96^{x}} \]

To solve for \( a \) and \( y \), we start by simplifying the equation. First, express \( 32 \) and \( 96 \) in terms of powers of \( 2 \) and \( 3 \): - \( 32 = 2^5 \) - \( 96 = 2^5 \cdot 3^1 \) Now, rewrite the equation: \[ \frac{3^{y+1}}{2^5} = \sqrt{(2^5 \cdot 3^1)^x} \] Now, simplify the right side: \[ \sqrt{(2^5 \cdot 3^1)^x} = \sqrt{2^{5x} \cdot 3^x} = 2^{\frac{5x}{2}} \cdot 3^{\frac{x}{2}} \] Equating both sides gives us: \[ \frac{3^{y+1}}{2^5} = 2^{\frac{5x}{2}} \cdot 3^{\frac{x}{2}} \] This leads to two equations after cross-multiplying: 1. For the powers of \( 2 \): \[ -5 = \frac{5x}{2} \] 2. For the powers of \( 3 \): \[ y + 1 = \frac{x}{2} \] Solving the first equation for \( x \): Multiply both sides by \( -2 \): \[ 10 = 5x \quad \Rightarrow \quad x = 2 \] Now plug \( x = 2 \) into the second equation: \[ y + 1 = \frac{2}{2} \Rightarrow y + 1 = 1 \quad \Rightarrow \quad y = 0 \] Finally, we have \( x = 2 \) and \( y = 0 \).