1. **State the problem:** Solve the logarithmic equation $$\log_2(3x + 5) - \log_2(x - 5) = 3$$. 2. **Recall the logarithm subtraction rule:** $$\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$$. 3. **Apply the rule:** $$\log_2 \left(\frac{3x + 5}{x - 5}\right) = 3$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\frac{3x + 5}{x - 5} = 2^3$$ 5. **Calculate the right side:** $$2^3 = 8$$ 6. **Set up the equation:** $$\frac{3x + 5}{x - 5} = 8$$ 7. **Multiply both sides by $x - 5$ to clear the denominator:** $$\cancel{\frac{3x + 5}{x - 5}} \times (x - 5) = 8 \times (x - 5)$$ $$3x + 5 = 8(x - 5)$$ 8. **Distribute the right side:** $$3x + 5 = 8x - 40$$ 9. **Bring all terms to one side:** $$3x + 5 - 8x + 40 = 0$$ $$-5x + 45 = 0$$ 10. **Isolate $x$:** $$-5x = -45$$ 11. **Divide both sides by $-5$:** $$\cancel{-5}x = \frac{-45}{\cancel{-5}}$$ $$x = 9$$ 12. **Check the domain restrictions:** - Argument of first log: $3x + 5 = 3(9) + 5 = 27 + 5 = 32 > 0$ - Argument of second log: $x - 5 = 9 - 5 = 4 > 0$ Both arguments are positive, so $x=9$ is valid. **Final answer:** $$x = 9$$