\( 8^{x / 2}=2^{3 / 8} \times 4^{3 / 4} \) find \( x \)

To solve the equation \( 8^{x / 2} = 2^{3 / 8} \times 4^{3 / 4} \), we start by rewriting the bases of each term in terms of powers of 2. First, recall that: - \( 8 = 2^3 \) and thus \( 8^{x / 2} = (2^3)^{x / 2} = 2^{3x / 2} \). - \( 4 = 2^2 \), so \( 4^{3 / 4} = (2^2)^{3 / 4} = 2^{3/2} \). Now substituting these into the original equation gives us: \[ 2^{3x / 2} = 2^{3 / 8} \times 2^{3/2}. \] We can combine the right side using the laws of exponents: \[ 2^{3 / 8} \times 2^{3/2} = 2^{3 / 8 + 3 / 2}. \] To add \( 3 / 8 \) and \( 3 / 2 \), convert \( 3 / 2 \) to eighths: \[ 3 / 2 = \frac{3 \times 4}{2 \times 4} = \frac{12}{8}. \] Now we can add: \[ \frac{3}{8} + \frac{12}{8} = \frac{15}{8}. \] This gives us: \[ 2^{3x / 2} = 2^{15 / 8}. \] Since the bases are the same, we can set the exponents equal to each other: \[ \frac{3x}{2} = \frac{15}{8}. \] To solve for \( x \), first multiply both sides by 2: \[ 3x = \frac{15}{8} \times 2 = \frac{30}{8} = \frac{15}{4}. \] Now divide both sides by 3: \[ x = \frac{15/4}{3} = \frac{15}{4} \times \frac{1}{3} = \frac{15}{12} = \frac{5}{4}. \] Thus, the solution for \( x \) is: \[ \boxed{\frac{5}{4}}. \]