1. **State the problem:** Solve the equation $10\log(x-4) + 10\log(x-2) = \log 3^3$ using algebraic methods. 2. **Recall logarithm properties:** - $a\log b = \log b^a$ - $\log a + \log b = \log(ab)$ - $\log 3^3 = \log 27$ 3. **Apply properties:** Rewrite the left side: $$10\log(x-4) + 10\log(x-2) = \log((x-4)^{10}) + \log((x-2)^{10}) = \log\left((x-4)^{10}(x-2)^{10}\right)$$ 4. **Combine logs:** $$\log\left((x-4)^{10}(x-2)^{10}\right) = \log 27$$ 5. **Set arguments equal:** Since $\log A = \log B$ implies $A = B$, we have: $$ (x-4)^{10}(x-2)^{10} = 27 $$ 6. **Simplify:** $$ \left[(x-4)(x-2)\right]^{10} = 27 $$ 7. **Take 10th root:** $$ (x-4)(x-2) = 27^{\frac{1}{10}} $$ 8. **Expand product:** $$ x^2 - 6x + 8 = 27^{\frac{1}{10}} $$ 9. **Rewrite as quadratic:** $$ x^2 - 6x + (8 - 27^{\frac{1}{10}}) = 0 $$ 10. **Use quadratic formula:** $$ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (8 - 27^{\frac{1}{10}})}}{2} = \frac{6 \pm \sqrt{36 - 32 + 4 \cdot 27^{\frac{1}{10}}}}{2} = \frac{6 \pm \sqrt{4 + 4 \cdot 27^{\frac{1}{10}}}}{2} $$ 11. **Simplify under root:** $$ \sqrt{4(1 + 27^{\frac{1}{10}})} = 2\sqrt{1 + 27^{\frac{1}{10}}} $$ 12. **Final solutions:** $$ x = 3 \pm \sqrt{1 + 27^{\frac{1}{10}}} $$ 13. **Check domain:** - $x-4 > 0 \Rightarrow x > 4$ - $x-2 > 0 \Rightarrow x > 2$ Only solutions with $x > 4$ are valid. 14. **Evaluate approximate values:** Since $27^{1/10} \approx 1.38$, $$ x = 3 \pm \sqrt{1 + 1.38} = 3 \pm \sqrt{2.38} $$ $$ \sqrt{2.38} \approx 1.54 $$ So, - $x_1 = 3 + 1.54 = 4.54$ (valid) - $x_2 = 3 - 1.54 = 1.46$ (invalid, domain restriction) **Answer:** $$ \boxed{x = 3 + \sqrt{1 + 27^{\frac{1}{10}}}} $$