1. **State the problem:** Solve the equation $$\ln(2x - 1) = 2 \ln(x + 1) - \ln x$$ and give the answer correct to 3 decimal places. 2. **Rewrite the right side using logarithm properties:** $$2 \ln(x + 1) - \ln x = \ln((x + 1)^2) - \ln x = \ln\left(\frac{(x + 1)^2}{x}\right)$$ 3. **Set the equation:** $$\ln(2x - 1) = \ln\left(\frac{(x + 1)^2}{x}\right)$$ 4. **Since the natural logs are equal, their arguments must be equal:** $$2x - 1 = \frac{(x + 1)^2}{x}$$ 5. **Multiply both sides by $x$ to clear the denominator:** $$x(2x - 1) = (x + 1)^2$$ 6. **Expand both sides:** $$2x^2 - x = x^2 + 2x + 1$$ 7. **Bring all terms to one side:** $$2x^2 - x - x^2 - 2x - 1 = 0$$ Simplify: $$x^2 - 3x - 1 = 0$$ 8. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$$ 9. **Calculate the two roots:** $$x_1 = \frac{3 + \sqrt{13}}{2} \approx \frac{3 + 3.606}{2} = \frac{6.606}{2} = 3.303$$ $$x_2 = \frac{3 - \sqrt{13}}{2} \approx \frac{3 - 3.606}{2} = \frac{-0.606}{2} = -0.303$$ 10. **Check domain restrictions:** - For $\ln(2x - 1)$, argument must be $>0 \Rightarrow 2x - 1 > 0 \Rightarrow x > 0.5$ - For $\ln x$, argument must be $>0 \Rightarrow x > 0$ Only $x_1 = 3.303$ satisfies $x > 0.5$. **Final answer:** $$\boxed{3.303}$$