5. \( 2^{-2 n}\left(2^{n+1}\right)=2^{-2 n} \)

To solve the equation \( 2^{-2 n}(2^{n+1})=2^{-2 n} \), start by simplifying the left side. You have \( 2^{-2 n} \cdot 2^{n + 1} \). When multiplying exponents with the same base, you can add the exponents: \[ 2^{-2n + (n + 1)} = 2^{-2n + n + 1} = 2^{-n + 1}. \] So the equation now becomes: \[ 2^{-n + 1} = 2^{-2n}. \] Since the bases are the same, you can set the exponents equal to each other: \[ -n + 1 = -2n. \] Now, isolate \(n\): \[ 1 = -2n + n, \] \[ 1 = -n, \] \[ n = -1. \] Thus, the solution to the equation is \( n = -1 \).