1. Stating the problem: Simplify each logarithmic expression from a) to z). 2. a) $y = \log x$ is the basic common logarithm. No simplification needed. 3. b) $y = \log_3 x$ is the logarithm base 3. 4. c) $y = \log_{\frac{1}{2}} (x-1)$. Recall that $\log_{\frac{1}{2}} a = -\log_2 a$, so $$y = -\log_2 (x-1).$$ 5. d) $y = 3 \log_{\frac{3}{4}} 2x$. Using change of base, $$\log_{\frac{3}{4}} 2x = \frac{\log 2x}{\log \frac{3}{4}},$$ and since $\log \frac{3}{4} < 0$, it's often simpler to leave it or approximate. 6. e) $y = \log_5 x^{-1} = -\log_5 x$ by log power rule. 7. f) $y = 5 - \log_8 x$, just keep as is, or rewrite $5 = \log_8 8^5$, so $$y = \log_8 8^5 - \log_8 x = \log_8 \frac{8^5}{x}.$$ 8. g) $y = \log_{0.5} \frac{x}{2} = -\log_2 \frac{x}{2} = -\left(\log_2 x - 1\right) = 1 - \log_2 x.$ 9. h) $y = \log_3 (2x + 5)^3 = 3 \log_3 (2x + 5)$ by power rule. 10. i) $y = \log 2x + \log 5x = \log (2x) + \log (5x) = \log (2x \times 5x) = \log 10x^2.$ 11. j) $y = \log_4 x + \log_4 \frac{1}{x} = \log_4 x + \log_4 x^{-1} = \log_4 x - \log_4 x = 0.$ 12. k) $y = \log x + \log_{100} x$. Convert second term: $$\log_{100} x = \frac{\log x}{\log 100} = \frac{\log x}{2},$$ so $$y = \log x + \frac{\log x}{2} = \frac{3}{2} \log x.$$ 13. l) $y = \log_3 x + \log_9 x + \log_{27} x$. Bases relate as powers of 3: $$\log_9 x = \frac{\log_3 x}{2}, \quad \log_{27} x = \frac{\log_3 x}{3}.$$ So sum is $$\log_3 x + \frac{\log_3 x}{2} + \frac{\log_3 x}{3} = \log_3 x \left(1 + \frac{1}{2} + \frac{1}{3}\right) = \log_3 x \cdot \frac{11}{6}.$$ 14. m) $y = \frac{\log_7 x^2}{\log_x x^2}$. Compute denominator: $$\log_x x^2 = 2,$$ so $$y = \frac{2 \log_7 x}{2} = \log_7 x.$$ 15. n) $y = -\log_{0.6}(x-1) - 1 = -\log_{0.6}(x-1) - \log_{0.6}(0.6)$ because $\log_a a = 1,$ so $$y = -\log_{0.6} (x-1) - \log_{0.6} (0.6) = -\log_{0.6} ((x-1) \cdot 0.6).$$ 16. o) $y = -\log x + \log 10 = \log 10 - \log x = \log \frac{10}{x}.$ 17. p) $y = \log_3 x - \log_3^{-1} x^{-1} \times \log x^{-1}.$ Here, $\log_3^{-1} x^{-1} = \frac{1}{\log_3 x^{-1}} = \frac{1}{-\log_3 x} = -\frac{1}{\log_3 x}$. And $\log x^{-1} = -\log x$. So, $$y = \log_3 x - \left(-\frac{1}{\log_3 x}\right)(-\log x) = \log_3 x - \frac{1}{\log_3 x} \log x.$$ We cannot simplify easily without numerical values. 18. q) $y = \log_3 x^{-1} - \log_3 9^{-1} = -\log_3 x - (-2) = -\log_3 x + 2.$ 19. r) $y = \log_2 x^2 + \log_4 x^4 + \log_8 x^8.$ Convert to base 2: $$\log_4 x^4 = \frac{\log_2 x^4}{2} = \frac{4 \log_2 x}{2} = 2 \log_2 x,$$ $$\log_8 x^8 = \frac{\log_2 x^8}{3} = \frac{8 \log_2 x}{3}.$$ Sum: $$2 \log_2 x + 2 \log_2 x + \frac{8}{3} \log_2 x = \left(2+2+\frac{8}{3}\right) \log_2 x = \frac{22}{3} \log_2 x.$$ 20. s) $y = \log x + \log_x x + \log_x x^2 + \log_x x^3.$ Note: $$\log_x x = 1, \quad \log_x x^2 = 2, \quad \log_x x^3 = 3,$$ so $$y = \log x + 1 + 2 + 3 = \log x + 6.$$ 21. t) $y = -\log_{0.5} \left( \frac{1}{x^{-1}} \right) = -\log_{0.5} x.$ Since $\log_{0.5} x = -\log_2 x$, we get $$y = -(-\log_2 x) = \log_2 x.$$ 22. u) $y = \log_{7.5} (7-2x)^{12} = 12 \log_{7.5} (7 - 2x).$ 23. v) $y = \log_8 |x|.$ 24. w) $y = 2 \left| -\log_{0.4} x \right| + 7 = 2 \left| \log_{0.4} x \right| + 7.$ Note $\log_{0.4} x = \frac{\log x}{\log 0.4} <0$ so absolute value used. 25. x) $y = -4 \left| -\log_{0.4} x^{-1} \right| = -4 \left| \log_{0.4} x^{-1} \right| = -4 \left| -\log_{0.4} x \right| = -4 |\log_{0.4} x|.$ 26. y) $y = | \log x - \log 10 x | - | \log x - \log x^{10} |.$ Simplify inside absolute: $$\log x - \log (10x) = \log \frac{x}{10x} = \log \frac{1}{10} = -1,$$ $$\log x - \log x^{10} = \log x - 10 \log x = -9 \log x.$$ So $$y = |-1| - | -9 \log x| = 1 - 9 |\log x|.$$ 27. z) $y = \left| \frac{\log_3 x^{-5}}{\log_3 243} - 1 \right| = 5.$ Note $$\log_3 x^{-5} = -5 \log_3 x, \quad \log_3 243 = \log_3 3^5 = 5,$$ so expression inside absolute is $$\frac{-5 \log_3 x}{5} - 1 = -\log_3 x - 1,$$ and $$| -\log_3 x - 1 | = 5.$$ We solve $$\log_3 x + 1 = \pm 5.$$ That is $$\log_3 x = 4 \quad \text{or} \quad \log_3 x = -6,$$ so $$x = 3^4 = 81 \quad \text{or} \quad x = 3^{-6} = \frac{1}{729}.$$