1. **Problem (a): Solve the logarithmic equation** $\log_2(x^2 - 2x + 5) = 2$. 2. **Recall the definition of logarithm:** If $\log_b(A) = C$, then $A = b^C$. 3. **Apply the definition:** $$x^2 - 2x + 5 = 2^2 = 4$$ 4. **Form the quadratic equation:** $$x^2 - 2x + 5 = 4 \implies x^2 - 2x + 1 = 0$$ 5. **Simplify:** $$x^2 - 2x + 1 = (x - 1)^2 = 0$$ 6. **Solve for $x$:** $$x - 1 = 0 \implies x = 1$$ 7. **Check domain:** The argument of the logarithm must be positive: $$x^2 - 2x + 5 = 1 - 2 + 5 = 4 > 0$$ so $x=1$ is valid. --- 8. **Problem (b): Simplify** $\log_{10}\left(\frac{30}{16}\right) - 2\log_{10}\left(\frac{5}{9}\right) + \log_{10}\left(\frac{400}{243}\right)$ 9. **Use logarithm properties:** - $a\log_b(c) = \log_b(c^a)$ - $\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)$ - $\log_b(m) + \log_b(n) = \log_b(mn)$ 10. **Rewrite:** $$\log_{10}\left(\frac{30}{16}\right) - \log_{10}\left(\left(\frac{5}{9}\right)^2\right) + \log_{10}\left(\frac{400}{243}\right)$$ 11. **Calculate powers:** $$\left(\frac{5}{9}\right)^2 = \frac{25}{81}$$ 12. **Combine logs:** $$= \log_{10}\left(\frac{30}{16}\right) - \log_{10}\left(\frac{25}{81}\right) + \log_{10}\left(\frac{400}{243}\right)$$ $$= \log_{10}\left(\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25}\right)$$ 13. **Simplify inside the log:** $$= \log_{10}\left(\frac{30 \times 400 \times 81}{16 \times 243 \times 25}\right)$$ 14. **Calculate numerator and denominator:** - Numerator: $30 \times 400 = 12000$, $12000 \times 81 = 972000$ - Denominator: $16 \times 243 = 3888$, $3888 \times 25 = 97200$ 15. **Divide:** $$\frac{972000}{97200} = 10$$ 16. **Final simplification:** $$\log_{10}(10) = 1$$ --- 17. **Problem (c): Simplify** $\frac{3^{n-1} \times 27^{n+1}}{81^n}$ 18. **Express all bases as powers of 3:** - $27 = 3^3$ - $81 = 3^4$ 19. **Rewrite:** $$\frac{3^{n-1} \times (3^3)^{n+1}}{(3^4)^n} = \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}}$$ 20. **Use exponent rules:** $$= 3^{n-1 + 3n + 3 - 4n} = 3^{(n-1 + 3n + 3 - 4n)} = 3^{2}$$ 21. **Simplify exponent:** $$3^2 = 9$$ --- 22. **Problem 3 (a): Is the operation $*$ defined by $a * b = \frac{2a - 1}{2} + b$ closed on $\mathbb{R}$?** 23. **Check closure:** For any real $a,b$, $2a - 1$ is real, dividing by 2 is real, adding $b$ is real. 24. **Conclusion:** The result is always a real number, so $*$ is closed on $\mathbb{R}$. --- 25. **Problem 3 (b): Is $*$ associative on $\mathbb{R}$?** 26. **Recall associativity:** $a * (b * c) = (a * b) * c$ for all $a,b,c \in \mathbb{R}$. 27. **Calculate $b * c$:** $$b * c = \frac{2b - 1}{2} + c$$ 28. **Calculate $a * (b * c)$:** $$a * (b * c) = \frac{2a - 1}{2} + \left(\frac{2b - 1}{2} + c\right) = \frac{2a - 1}{2} + \frac{2b - 1}{2} + c$$ 29. **Calculate $a * b$:** $$a * b = \frac{2a - 1}{2} + b$$ 30. **Calculate $(a * b) * c$:** $$\left(\frac{2a - 1}{2} + b\right) * c = \frac{2\left(\frac{2a - 1}{2} + b\right) - 1}{2} + c$$ $$= \frac{(2a - 1) + 2b - 1}{2} + c = \frac{2a - 1 + 2b - 1}{2} + c = \frac{2a + 2b - 2}{2} + c = (a + b - 1) + c$$ 31. **Compare:** - $a * (b * c) = \frac{2a - 1}{2} + \frac{2b - 1}{2} + c = (a - \frac{1}{2}) + (b - \frac{1}{2}) + c = a + b + c - 1$ - $(a * b) * c = a + b + c - 1$ 32. **Conclusion:** Since both expressions equal $a + b + c - 1$, the operation $*$ is associative on $\mathbb{R}$. --- **Final answers:** - (2a) $x = 1$ - (2b) Simplified expression = 1 - (2c) Simplified expression = 9 - (3a) Operation $*$ is closed on $\mathbb{R}$ - (3b) Operation $*$ is associative on $\mathbb{R}$