Find the solution of the equation, \( \ln (2 w+1)=1+\ln (w-1) \), to 4 decimal places.

To solve the equation \( \ln (2w + 1) = 1 + \ln (w - 1) \), we can start by rewriting the right-hand side using properties of logarithms: \[ \ln (2w + 1) = \ln (e) + \ln (w - 1) \] This simplifies to: \[ \ln (2w + 1) = \ln (e(w - 1)) \] Since the logarithmic function is one-to-one, we can drop the logarithm: \[ 2w + 1 = e(w - 1) \] Expanding the right side gives us: \[ 2w + 1 = ew - e \] Now, reorganizing terms: \[ 2w - ew = -e - 1 \] Factoring out \( w \) from the left side: \[ w(2 - e) = -e - 1 \] Solving for \( w \): \[ w = \frac{-e - 1}{2 - e} \] Now we can compute the numerical value. Using the approximation \( e \approx 2.7183 \): \[ w \approx \frac{-2.7183 - 1}{2 - 2.7183} = \frac{-3.7183}{-0.7183} \approx 5.1661 \] Thus, rounding to four decimal places, the solution is: \[ \boxed{5.1661} \]