Step-by-step Solution:

To solve the domain of the function \[ \log f(x) = \frac{1}{\sqrt{\log_{\frac{1}{2}}(x^2 - 7x + 13)}} \] we need to consider the constraints on the components of the function. \[\]Domain of the log function The expression inside the logarithm, \(\log_{\frac{1}{2}}(x^2 - 7x + 13)\), is only defined for values of \(x\) such that the argument of the logarithm is positive. Therefore, we must have: \[ x^2 - 7x + 13 > 0 \] Let's analyze the quadratic expression \(x^2 - 7x + 13\). The discriminant of \(x^2 - 7x + 13\) is: \[ \Delta = (-7)^2 - 4(1)(13) = 49 - 52 = -3 \] Since the discriminant is negative, the quadratic expression has no real roots and is always positive. Thus, \(x^2 - 7x + 13 > 0\) for all real values of \(x\). \[\]The square root and denominator Next, we need the expression inside the square root, \(\log_{\frac{1}{2}}(x^2 - 7x + 13)\), to be positive. The logarithmic function \(\log_{\frac{1}{2}}(y)\) is positive if \(0 < y < 1\). This means we need: \[ 0 < \log_{\frac{1}{2}}(x^2 - 7x + 13) < \infty \] The logarithmic function \(\log_{\frac{1}{2}}(y)\) is positive for \(0 < y < 1\), and since the base of the logarithm is \( \frac{1}{2} \), we need: \[ x^2 - 7x + 13 < 1 \] This inequality simplifies to: \[ x^2 - 7x + 12 < 0 \] \[\]Solving the quadratic inequality We solve \(x^2 - 7x + 12 = 0\) to find the boundary points: \[ \Delta = (-7)^2 - 4(1)(12) = 49 - 48 = 1 \] The roots are: \[ x = \frac{7 \pm \sqrt{1}}{2} = \frac{7 \pm 1}{2} \] Thus, the roots are \(x = 3\) and \(x = 4\). Since the quadratic is opening upwards, \(x^2 - 7x + 12 < 0\) between the roots \(3 < x < 4\). \[\]Final Domain Thus, the domain of the function is: \[ 3 < x < 4 \]