1. Problem 15 asks to find $\log_2 x$ for various expressions of $x$ where $a,b,c>0$. 2. Recall the logarithm properties: - $\log_b(mn) = \log_b m + \log_b n$ - $\log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n$ - $\log_b(m^k) = k \log_b m$ - $\log_b(m - n)$ cannot be simplified unless factored. 3. a) $x = 3ab$ $$\log_2(3ab) = \log_2 3 + \log_2 a + \log_2 b$$ 4. b) $x = \frac{7a^5}{11 c^2}$ $$\log_2 \left(\frac{7a^5}{11 c^2}\right) = \log_2 7 + \log_2 a^5 - \log_2 11 - \log_2 c^2$$ Using power rule: $$= \log_2 7 + 5 \log_2 a - \log_2 11 - 2 \log_2 c$$ 5. c) $x = 4(a^2 - b^2)$ Factor difference of squares: $$a^2 - b^2 = (a-b)(a+b)$$ So: $$\log_2 [4(a-b)(a+b)] = \log_2 4 + \log_2 (a-b) + \log_2 (a+b)$$ Since $\log_2 4 = 2$, final: $$= 2 + \log_2 (a-b) + \log_2 (a+b)$$ 6. d) $x = \frac{9 a^2 b \sqrt{3}}{5 c^2 3\sqrt{5}}$ Rewrite radicals: $$\sqrt{3} = 3^{1/2}, \quad 3\sqrt{5} = 3 \times 5^{1/3}$$ So: $$x = \frac{9 a^2 b 3^{1/2}}{5 c^2 3 \times 5^{1/3}} = \frac{9 a^2 b 3^{1/2}}{3 \times 5^{1 + 1/3} c^2}$$ Simplify numerator and denominator: $$= \frac{9 a^2 b 3^{1/2}}{3 c^2 5^{4/3}}$$ Cancel 3: $$= \frac{\cancel{9} a^2 b 3^{1/2}}{\cancel{3} c^2 5^{4/3}} = \frac{3 a^2 b 3^{1/2}}{c^2 5^{4/3}}$$ Combine powers of 3: $$3 \times 3^{1/2} = 3^{1 + 1/2} = 3^{3/2}$$ So: $$x = \frac{a^2 b 3^{3/2}}{c^2 5^{4/3}}$$ Take log base 2: $$\log_2 x = 2 \log_2 a + \log_2 b + \frac{3}{2} \log_2 3 - 2 \log_2 c - \frac{4}{3} \log_2 5$$ 7. e) $x = a^3 + b^3$ Sum of cubes factorization: $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$ So: $$\log_2 (a^3 + b^3) = \log_2 (a+b) + \log_2 (a^2 - ab + b^2)$$ Final answers: - a) $\log_2 3 + \log_2 a + \log_2 b$ - b) $\log_2 7 + 5 \log_2 a - \log_2 11 - 2 \log_2 c$ - c) $2 + \log_2 (a-b) + \log_2 (a+b)$ - d) $2 \log_2 a + \log_2 b + \frac{3}{2} \log_2 3 - 2 \log_2 c - \frac{4}{3} \log_2 5$ - e) $\log_2 (a+b) + \log_2 (a^2 - ab + b^2)$