1. **State the problem:** Solve the equation $$\log_9(x - 5) + \log_9(x + 3) = 1$$ for $x$. 2. **Recall the logarithm property:** The sum of logarithms with the same base can be combined as the logarithm of the product: $$\log_b A + \log_b B = \log_b (A \times B)$$ 3. **Apply the property:** $$\log_9((x - 5)(x + 3)) = 1$$ 4. **Rewrite the logarithmic equation in exponential form:** Since $\log_9 y = 1$ means $y = 9^1 = 9$, we have $$(x - 5)(x + 3) = 9$$ 5. **Expand the left side:** $$x^2 + 3x - 5x - 15 = 9$$ $$x^2 - 2x - 15 = 9$$ 6. **Bring all terms to one side:** $$x^2 - 2x - 15 - 9 = 0$$ $$x^2 - 2x - 24 = 0$$ 7. **Factor the quadratic:** We look for two numbers that multiply to $-24$ and add to $-2$, which are $-6$ and $4$. $$ (x - 6)(x + 4) = 0 $$ 8. **Solve for $x$:** $$x - 6 = 0 \Rightarrow x = 6$$ $$x + 4 = 0 \Rightarrow x = -4$$ 9. **Check the domain restrictions:** The arguments of the logarithms must be positive: $$x - 5 > 0 \Rightarrow x > 5$$ $$x + 3 > 0 \Rightarrow x > -3$$ Only $x = 6$ satisfies both conditions. **Final answer:** $$x = 6$$