1. **State the problem:** Solve the equation $$\log_2(x^2 - 6x) = 3 + \log_2(1 - x)$$ for real numbers $x$. 2. **Recall logarithm properties:** - The domain of $\log_2(a)$ requires $a > 0$. - The property $\log_b A + \log_b B = \log_b(AB)$ and $\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)$. 3. **Rewrite the equation:** $$\log_2(x^2 - 6x) = 3 + \log_2(1 - x)$$ can be rewritten as $$\log_2(x^2 - 6x) - \log_2(1 - x) = 3$$ 4. **Use logarithm subtraction rule:** $$\log_2\left(\frac{x^2 - 6x}{1 - x}\right) = 3$$ 5. **Convert logarithmic to exponential form:** $$\frac{x^2 - 6x}{1 - x} = 2^3 = 8$$ 6. **Solve the rational equation:** $$\frac{x^2 - 6x}{1 - x} = 8$$ Multiply both sides by $1 - x$ (noting $1 - x \neq 0$): $$x^2 - 6x = 8(1 - x)$$ 7. **Expand and simplify:** $$x^2 - 6x = 8 - 8x$$ Bring all terms to one side: $$x^2 - 6x - 8 + 8x = 0$$ $$x^2 + 2x - 8 = 0$$ 8. **Solve quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=2$, $c=-8$: $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}$$ $$x = \frac{-2 \pm 6}{2}$$ 9. **Calculate roots:** - $$x = \frac{-2 + 6}{2} = \frac{4}{2} = 2$$ - $$x = \frac{-2 - 6}{2} = \frac{-8}{2} = -4$$ 10. **Check domain restrictions:** - For $\log_2(x^2 - 6x)$, require $x^2 - 6x > 0$. - For $\log_2(1 - x)$, require $1 - x > 0$ or $x < 1$. Check $x=2$: - $2^2 - 6 \cdot 2 = 4 - 12 = -8 < 0$ (invalid) - $1 - 2 = -1 < 0$ (invalid) Check $x=-4$: - $(-4)^2 - 6(-4) = 16 + 24 = 40 > 0$ (valid) - $1 - (-4) = 5 > 0$ (valid) Only $x = -4$ satisfies the domain. **Final answer:** $$\boxed{-4}$$