1. **State the problem:** Solve the equation $$\log_{10}(x - 3) + \log_{10} x = \log_{10} 12$$ for $x$. 2. **Apply logarithm properties:** Recall that the sum of logarithms with the same base is the logarithm of the product, so: $$\log_{10}(x - 3) + \log_{10} x = \log_{10}[(x - 3) \cdot x] = \log_{10}(x^2 - 3x)$$ 3. **Rewrite the equation:** The original equation becomes: $$\log_{10}(x^2 - 3x) = \log_{10} 12$$ 4. **Drop logarithms (since the base and logs are defined for positive arguments):** $$x^2 - 3x = 12$$ 5. **Rewrite as a quadratic equation:** $$x^2 - 3x - 12 = 0$$ 6. **Factor or use the quadratic formula:** The quadratic factors as: $$(x - 6)(x + 2) = 0$$ 7. **Solve for $x$:** $$x = 6 \quad \text{or} \quad x = -2$$ 8. **Check the domain restrictions:** Since $x$ appears inside logarithms, the arguments must be positive: - $x > 0$ - $x - 3 > 0 \Rightarrow x > 3$ Therefore, $x = -2$ is invalid because it does not satisfy $x > 0$. 9. **Final solution:** $$\boxed{x = 6}$$