1. **State the problem:** Solve the logarithmic equation $$\log_2(x^2 - 9x + 4) = 2$$ for all values of $x$. 2. **Recall the logarithm definition:** If $$\log_b(A) = C$$, then $$A = b^C$$. 3. **Apply the definition:** Here, $$b=2$$, $$A = x^2 - 9x + 4$$, and $$C=2$$, so $$x^2 - 9x + 4 = 2^2 = 4$$. 4. **Rewrite the equation:** $$x^2 - 9x + 4 = 4$$ 5. **Subtract 4 from both sides:** $$x^2 - 9x + 4 - 4 = 4 - 4$$ $$x^2 - 9x + \cancel{4} - \cancel{4} = 0$$ $$x^2 - 9x = 0$$ 6. **Factor the quadratic:** $$x(x - 9) = 0$$ 7. **Set each factor equal to zero:** $$x = 0 \quad \text{or} \quad x - 9 = 0$$ $$x = 0 \quad \text{or} \quad x = 9$$ 8. **Check the domain restrictions:** The argument of the logarithm must be positive: $$x^2 - 9x + 4 > 0$$ 9. **Test each solution:** - For $x=0$: $$0^2 - 9(0) + 4 = 4 > 0$$ valid. - For $x=9$: $$9^2 - 9(9) + 4 = 81 - 81 + 4 = 4 > 0$$ valid. 10. **Final solutions:** $$\boxed{x = 0 \text{ or } x = 9}$$