4.2 .3 \[ \frac{16^{2 y+2} 4^{1-y}}{64 y^{y+2}} \] \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) 4.3 Solve \( x \) in the following equations: 4.3.1 \[ 2^{2 x}=8 \] \( \qquad \) \( \qquad \)

Let’s break down the expression \( \frac{16^{2y+2} 4^{1-y}}{64y^{y+2}} \) before diving into the equations. We can simplify the bases: \( 16 \) can be rewritten as \( 4^2 \), and \( 64 \) is \( 4^3 \). Utilizing the properties of exponents will make our expression cleaner, enabling us to solve for \( y \) more easily. For \( 2^{2x} = 8 \), remember that \( 8 \) can be expressed as \( 2^3 \). So, by equating the exponents, we find that \( 2x = 3 \), leading us to the solution: \( x = \frac{3}{2} \). Voila! The world of exponential equations is full of fun surprises!