If \( 2^{2 x-1}=\frac{1}{8^{x-3}} \) find \( x \)

To solve the equation \( 2^{2x-1} = \frac{1}{8^{x-3}} \), we start by rewriting \( 8^{x-3} \) in terms of base 2, since \( 8 = 2^3 \). Thus, we have: \[ 8^{x-3} = (2^3)^{x-3} = 2^{3(x-3)} = 2^{3x - 9}. \] Now, the right side of the equation becomes: \[ \frac{1}{8^{x-3}} = \frac{1}{2^{3x - 9}} = 2^{-(3x - 9)} = 2^{9 - 3x}. \] So, we can rewrite the original equation as: \[ 2^{2x - 1} = 2^{9 - 3x}. \] Since the bases are the same, we can set the exponents equal to each other: \[ 2x - 1 = 9 - 3x. \] Now, solve for \( x \): \[ 2x + 3x = 9 + 1, \] \[ 5x = 10, \] \[ x = 2. \] Thus, the solution is \( x = 2 \).