1. Stated problem: Solve the equation $$3 \log_{10}{(x-3)} + \log_{10}{x} = \log_{10}{12}$$. 2. Use logarithm property: $$a\log_b{c} = \log_b{c^a}$$, so rewrite the equation as $$\log_{10}{(x-3)^3} + \log_{10}{x} = \log_{10}{12}$$. 3. Use logarithm addition property: $$\log_b{A} + \log_b{B} = \log_b{(AB)}$$, so combine left side: $$\log_{10}{\left((x-3)^3 \cdot x\right)} = \log_{10}{12}$$. 4. Since logarithms are equal, arguments must be equal: $$ (x-3)^3 \cdot x = 12 $$. 5. This simplifies to the equation $$ x(x-3)^3 = 12 $$. 6. Domain considerations: $x-3 > 0$ and $x > 0$ imply $x > 3$. 7. To solve $$ x(x-3)^3 = 12 $$ for $x > 3$, expand or use substitution. 8. Let $t = x - 3$, then $x = t + 3$ and equation becomes $$ (t + 3) t^3 = 12 $$ which simplifies to $$ t^4 + 3t^3 = 12 $$. 9. Try real positive values that satisfy this equation, for example, $ t = 1$ gives $1 + 3 = 4$ (too low), $t = 2$ gives $16 + 24 = 40$ (too high), thus root is between 1 and 2. 10. Using approximate numerical methods (bisection or Newton), the root $t \approx 1.5$. 11. Then $x = t + 3 \approx 1.5 + 3 = 4.5$. 12. Verify domain: $4.5 > 3$ is valid. 13. Final answer: $$ x \approx 4.5 $$.