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

To solve the equation \( 2^{2x-1} = \frac{1}{8^x} \), we first need to express all terms using the same base. We know that \( 8 \) can be expressed as \( 2^3 \), so we have: \[ 8^x = (2^3)^x = 2^{3x}. \] This means that \( \frac{1}{8^x} = \frac{1}{2^{3x}} = 2^{-3x} \). Now, we can rewrite the original equation: \[ 2^{2x-1} = 2^{-3x}. \] Since the bases are the same, we can set the exponents equal to each other: \[ 2x - 1 = -3x. \] Now, let's solve for \( x \): 1. Add \( 3x \) to both sides: \[ 2x + 3x - 1 = 0 \\ 5x - 1 = 0. \] 2. Add \( 1 \) to both sides: \[ 5x = 1. \] 3. Finally, divide by \( 5 \): \[ x = \frac{1}{5}. \] Thus, the value of \( x \) is \[ \boxed{\frac{1}{5}}. \]