(b) If \( 2^{7 x-2}=64 \sqrt{2} \), find \( x \)

To solve the equation \( 2^{7x - 2} = 64 \sqrt{2} \), we first express both sides in terms of powers of 2. We know that \( 64 = 2^6 \) and \( \sqrt{2} = 2^{1/2} \), so we can rewrite the right side as: \[ 64 \sqrt{2} = 2^6 \cdot 2^{1/2} = 2^{6 + 1/2} = 2^{6.5} = 2^{13/2} \] Now, we can rewrite the equation: \[ 2^{7x - 2} = 2^{13/2} \] Since the bases are the same, we can set the exponents equal to each other: \[ 7x - 2 = \frac{13}{2} \] Next, we solve for \( x \). First, add 2 to both sides. It's easier to work with fractions, so let's convert 2 into a fraction as well: \[ 2 = \frac{4}{2} \] Thus, \[ 7x - 2 = \frac{13}{2} \implies 7x = \frac{13}{2} + \frac{4}{2} = \frac{17}{2} \] Now, divide both sides by 7: \[ x = \frac{17/2}{7} = \frac{17}{14} \] Therefore, the solution for \( x \) is: \[ \boxed{\frac{17}{14}} \]