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

To solve the equation \( \ln (2 w+1)=1+\ln (w-1) \), we start by rewriting it: 1. Rearranging gives us: \[ \ln (2w + 1) - \ln (w - 1) = 1 \] 2. Using the properties of logarithms, this can be combined: \[ \ln \left( \frac{2w + 1}{w - 1} \right) = 1 \] 3. Exponentiating both sides yields: \[ \frac{2w + 1}{w - 1} = e \] 4. We multiply both sides by \( w - 1 \) to eliminate the fraction: \[ 2w + 1 = e(w - 1) \] 5. Distributing \( e \) gives: \[ 2w + 1 = ew - e \] 6. Rearranging leads to: \[ 2w - ew + e + 1 = 0 \] 7. Factoring out \( w \): \[ (2 - e)w + (e + 1) = 0 \] 8. Solving for \( w \) gives: \[ w = \frac{-(e + 1)}{(2 - e)} \] Now we substitute the approximate value of \( e \approx 2.7183 \) into the equation: - \( w \approx \frac{-(2.7183 + 1)}{(2 - 2.7183)} \) - \( w \approx \frac{-3.7183}{-0.7183} \approx 5.1756 \) (after calculating) So the final answer, rounded to four decimal places is: \( w \approx 5.1756 \)