Solve the equation
\[
\ln (2 x+1)+\ln (x-7)-2 \ln x=0
\]

The solution set is
(Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)

Step 1 ：Given the equation \(\ln (2 x+1)+\ln (x-7)-2 \ln x=0\)

Step 2 ：Using the properties of logarithms, we can simplify the equation to \(\ln [(2x+1)(x-7)] - \ln x^2 = 0\)

Step 3 ：This simplifies further to \(\ln \left(\frac{(2x+1)(x-7)}{x^2}\right) = 0\)

Step 4 ：Since the natural logarithm of 1 is 0, we can set the argument of the logarithm equal to 1: \(\frac{(2x+1)(x-7)}{x^2} = 1\)

Step 5 ：Cross-multiplying gives us \((2x+1)(x-7) = x^2\)

Step 6 ：Expanding the left side gives us \(2x^2 - 13x - 7 = x^2\)

Step 7 ：Subtracting \(x^2\) from both sides gives us \(x^2 - 13x - 7 = 0\)

Step 8 ：This is a quadratic equation, which we can solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 9 ：Substituting the values of a, b, and c into the quadratic formula gives us \(x = \frac{13 \pm \sqrt{(-13)^2 - 4*1*(-7)}}{2*1}\)

Step 10 ：Solving this gives us \(x = \frac{13 \pm \sqrt{169 + 28}}{2}\) which simplifies to \(x = \frac{13 \pm \sqrt{197}}{2}\)

Step 11 ：So the solutions are \(x = \frac{13 + \sqrt{197}}{2}\) and \(x = \frac{13 - \sqrt{197}}{2}\)

Step 12 ：However, we need to check these solutions in the original equation, because logarithms are only defined for positive arguments.

Step 13 ：Substituting \(x = \frac{13 + \sqrt{197}}{2}\) into the original equation, we find that \(2x+1 > 0\), \(x-7 > 0\), and \(x > 0\), so this solution is valid.

Step 14 ：Substituting \(x = \frac{13 - \sqrt{197}}{2}\) into the original equation, we find that \(2x+1 > 0\), but \(x-7 < 0\), so this solution is not valid.

Step 15 ：Therefore, the only solution to the equation is \(\boxed{x = \frac{13 + \sqrt{197}}{2}}\)