Solve the following logarithmic equation. \[ \log (1-x)-\log (1+x)=2 \]

Step 1 ：Solve the following logarithmic equation: \(\log (1-x)-\log (1+x)=2\)

Step 2 ：Using the properties of logarithms, the difference of two logarithms with the same base can be written as the logarithm of the quotient of the arguments. So, the equation can be rewritten as \(\log \frac{1-x}{1+x} = 2\)

Step 3 ：Convert the logarithmic equation to an exponential equation to solve for x. The base of the logarithm is 10, so the equation becomes \(\frac{1-x}{1+x} = 10^2\)

Step 4 ：Solve the equation to find \(x = -\frac{99}{101}\)

Step 5 ：Check if this solution is valid. The argument of a logarithm must be greater than zero. So, check if \(1 - x > 0\) and \(1 + x > 0\) for \(x = -\frac{99}{101}\)

Step 6 ：Both checks returned True, which means that \(x = -\frac{99}{101}\) is a valid solution to the original logarithmic equation

Step 7 ：Final Answer: The solution to the logarithmic equation \(\log (1-x)-\log (1+x)=2\) is \(x = \boxed{-\frac{99}{101}}\)