For the given function, \( f(x)=\ln \left(x^{2}-4 x\right) \), we note that \( x^{2}-4 x \) must be greater than 0 , as the domain of \( \ln (u) \) is \( u \geq 0 \). Therefore, the domain is found by first solving the equation below for \( x \). \[ x^{2}-4 x=0 \] Solving the above for \( x \) gives us the following values. (Enter your answers as a comma-separated list.) \( x=\square \)

To solve the equation \( x^{2} - 4x = 0 \), we can factor it: \[ x(x - 4) = 0 \] Setting each factor equal to zero gives us: 1. \( x = 0 \) 2. \( x - 4 = 0 \) which simplifies to \( x = 4 \) Thus, the solutions to the equation \( x^{2} - 4x = 0 \) are: \[ x = 0, 4 \] Now, we need to determine the intervals where \( x^{2} - 4x > 0 \). The critical points are \( x = 0 \) and \( x = 4 \). We can test the intervals: 1. For \( x < 0 \) (e.g., \( x = -1 \)): \[ (-1)^{2} - 4(-1) = 1 + 4 = 5 > 0 \] 2. For \( 0 < x < 4 \) (e.g., \( x = 2 \)): \[ 2^{2} - 4(2) = 4 - 8 = -4 < 0 \] 3. For \( x > 4 \) (e.g., \( x = 5 \)): \[ 5^{2} - 4(5) = 25 - 20 = 5 > 0 \] From this analysis, we find that \( x^{2} - 4x > 0 \) in the intervals \( (-\infty, 0) \) and \( (4, \infty) \). Therefore, the values of \( x \) that make \( x^{2} - 4x = 0 \) are: \[ \boxed{0, 4} \]