Solve for \( x \) (i) \( 4^{x+1}=32 \)

To solve the equation \( 4^{x+1} = 32 \), we can express both sides with the same base. We know that \( 4 \) can be rewritten as \( 2^2 \) and \( 32 \) can be rewritten as \( 2^5 \). This gives us: \[ (2^2)^{x+1} = 2^5 \] Using the power of a power property, we simplify the left side: \[ 2^{2(x+1)} = 2^5 \] Now we can set the exponents equal to each other because the bases are the same: \[ 2(x+1) = 5 \] Next, we distribute the \( 2 \): \[ 2x + 2 = 5 \] Now, we can isolate \( x \): \[ 2x = 5 - 2 \] \[ 2x = 3 \] Finally, divide by \( 2 \): \[ x = \frac{3}{2} \] So, the solution is \( x = \frac{3}{2} \).