24) \( 16^{2 p-3} \cdot 4^{-2 p}=2^{4} \)

Let's first rewrite all terms in the equation using base 2. We know that \( 16 = 2^4 \) and \( 4 = 2^2 \). So, we rewrite the equation: \[ 16^{2p-3} \cdot 4^{-2p} = (2^4)^{2p-3} \cdot (2^2)^{-2p} \] This simplifies to: \[ 2^{4(2p-3)} \cdot 2^{-4p} = 2^{4p - 12} \cdot 2^{-4p} \] Now, combine the exponents: \[ 2^{4p - 12 - 4p} = 2^{-12} \] Thus, we have: \[ 2^{-12} = 2^4 \] Since the bases are the same, we can set the exponents equal to each other: \[ -12 = 4 \] But this is clearly not true, so we must have made a mistake or more likely, there are no values of \( p \) that will satisfy this equation. It leads to a contradiction. In summary, this equation has no solution since the exponents cannot equal each other.