1. **State the problem:** Solve the equation $$15 \cdot \log_3(2x - 5) - 2 \cdot \log_3 4 = \log_3 x - 1$$ for $x$. 2. **Recall logarithm properties:** - $a \log_b c = \log_b c^a$ - $\log_b A - \log_b B = \log_b \frac{A}{B}$ - $\log_b A = \log_b B \implies A = B$ 3. **Rewrite terms using properties:** $$15 \log_3(2x - 5) = \log_3 (2x - 5)^{15}$$ $$2 \log_3 4 = \log_3 4^2 = \log_3 16$$ 4. **Substitute back:** $$\log_3 (2x - 5)^{15} - \log_3 16 = \log_3 x - 1$$ 5. **Combine left side logs:** $$\log_3 \frac{(2x - 5)^{15}}{16} = \log_3 x - 1$$ 6. **Rewrite $-1$ as a log:** Since $1 = \log_3 3$, then $$-1 = -\log_3 3 = \log_3 \frac{1}{3}$$ 7. **Rewrite right side:** $$\log_3 x - 1 = \log_3 x + \log_3 \frac{1}{3} = \log_3 \left(x \cdot \frac{1}{3}\right) = \log_3 \frac{x}{3}$$ 8. **Set the arguments equal:** $$\frac{(2x - 5)^{15}}{16} = \frac{x}{3}$$ 9. **Cross multiply:** $$3 (2x - 5)^{15} = 16 x$$ 10. **Check domain restrictions:** - $2x - 5 > 0 \implies x > \frac{5}{2} = 2.5$ - $x > 0$ 11. **Solve for $x$ approximately:** This is a transcendental equation; exact algebraic solution is difficult. We test values $x > 2.5$ to find approximate solution. 12. **Check $x=3$:** $$3 (2(3) - 5)^{15} = 3 (6 - 5)^{15} = 3 (1)^{15} = 3$$ $$16 \cdot 3 = 48$$ Left side $3$ vs right side $48$; left < right. 13. **Check $x=5$:** $$3 (2(5) - 5)^{15} = 3 (10 - 5)^{15} = 3 (5)^{15}$$ $$16 \cdot 5 = 80$$ Left side is huge (since $5^{15}$ is very large), so left > right. 14. **By Intermediate Value Theorem, solution lies between 3 and 5.** 15. **Approximate solution numerically (e.g., by trial or calculator):** The solution is approximately $x \approx 3.1$. **Final answer:** $$x \approx 3.1$$