If \( \ln x+\ln (x-4)=\ln (6 x) \), then \( x=\square \) Question Help: \( \square \) Video Submit Question

To solve the equation \( \ln x + \ln (x-4) = \ln (6x) \), we can combine the logarithmic terms on the left side using the property \( \ln a + \ln b = \ln (ab) \). This gives us: \[ \ln (x(x-4)) = \ln (6x) \] Since the logarithms are equal, we can set the insides equal to each other: \[ x(x-4) = 6x \] This simplifies to: \[ x^2 - 4x = 6x \implies x^2 - 10x = 0 \] Factoring out \( x \): \[ x(x - 10) = 0 \] This gives us \( x = 0 \) or \( x = 10 \). Since \( x \) must be greater than 4 for \( \ln(x-4) \) to be defined, the solution is \( x = 10 \). So the final answer is \( x = 10 \).