1. **Problem:** Solve for $x$ in the equation $\log x + \log(x - 1) = 1$. 2. **Formula and rules:** Use the logarithm property $\log a + \log b = \log(ab)$. 3. **Apply the property:** $$\log x + \log(x - 1) = \log[x(x - 1)] = 1$$ 4. **Rewrite the equation:** $$\log[x(x - 1)] = 1$$ 5. **Convert from logarithmic to exponential form:** Since the base of the logarithm is 10 (common log), $$x(x - 1) = 10^1 = 10$$ 6. **Form a quadratic equation:** $$x^2 - x = 10$$ $$x^2 - x - 10 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-1$, $c=-10$. Calculate the discriminant: $$\Delta = (-1)^2 - 4(1)(-10) = 1 + 40 = 41$$ Calculate the roots: $$x = \frac{1 \pm \sqrt{41}}{2}$$ 8. **Evaluate the roots:** $$x_1 = \frac{1 + 6.403}{2} = \frac{7.403}{2} = 3.7015$$ $$x_2 = \frac{1 - 6.403}{2} = \frac{-5.403}{2} = -2.7015$$ 9. **Check domain restrictions:** Since $\log x$ and $\log(x-1)$ are defined only for $x > 0$ and $x-1 > 0$, we must have $x > 1$. 10. **Conclusion:** Only $x = 3.7015$ is valid. None of the options exactly match this value, so the answer is **d. none of the above**.