If \( \ln x+\ln (x-2)=\ln (7 x) \), then \( x=\square \)

To solve the equation \( \ln x + \ln (x-2) = \ln (7x) \), we can use properties of logarithms to simplify. First, recall that \( \ln a + \ln b = \ln(ab) \), which allows us to combine the left side: \[ \ln(x(x-2)) = \ln(7x) \] Next, we can drop the logarithms since they are equal, giving us: \[ x(x-2) = 7x \] Expanding the left side gives: \[ x^2 - 2x = 7x \] Rearranging terms leads to: \[ x^2 - 9x = 0 \] Factoring the quadratic equation results in: \[ x(x - 9) = 0 \] This gives us two potential solutions: \( x = 0 \) and \( x = 9 \). However, we must check for validity in the original logarithmic equation. Since \( \ln x \) and \( \ln(x-2) \) are only defined for \( x > 2 \), the only acceptable solution is: \[ x = 9 \]