Solve for n in the equation \(2^{2n-1} = {1 \over 8^{n-3}}\)

\(2^{2n-1} = {8^{-(n-3)}}\)

\(2^{2n-1} = 8^{-n+3}\)

\(2^{2n-1} = 2^{3(-n+3)}\)

\(2^{2n-1} = 2^{-3n+9}\)

\(2n-1 = -3n + 9\)

\(n = 2\)