1. **Express logarithms in terms of $\ln 2$ and $\ln 3$** Recall the properties of logarithms: - $\ln(ab) = \ln a + \ln b$ - $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$ - $\ln(a^r) = r \ln a$ (a) $\ln 0.75 = \ln \frac{3}{4} = \ln 3 - \ln 4 = \ln 3 - 2\ln 2$ (b) $\ln \frac{4}{9} = \ln 4 - \ln 9 = 2\ln 2 - 2\ln 3$ (c) $\ln \frac{1}{2} = -\ln 2$ (d) $\ln \sqrt[3]{9} = \ln 9^{1/3} = \frac{1}{3} \ln 9 = \frac{1}{3} (2 \ln 3) = \frac{2}{3} \ln 3$ (e) $\ln \sqrt[3]{2} = \frac{1}{3} \ln 2$ (f) $\ln \sqrt{13.5} = \ln (13.5^{1/2}) = \frac{1}{2} \ln 13.5$ Note: $13.5 = \frac{27}{2} = \frac{3^3}{2}$, so $\ln 13.5 = \ln \frac{3^3}{2} = 3 \ln 3 - \ln 2$ Thus, $\ln \sqrt{13.5} = \frac{1}{2} (3 \ln 3 - \ln 2) = \frac{3}{2} \ln 3 - \frac{1}{2} \ln 2$ --- 2. **Express logarithms in terms of $\ln 5$ and $\ln 7$** (a) $\ln \frac{1}{125} = -\ln 125 = -\ln 5^3 = -3 \ln 5$ (b) $\ln 9.8 = \ln \frac{98}{10} = \ln 98 - \ln 10$ $98 = 7^2 \times 2$, so $\ln 98 = 2 \ln 7 + \ln 2$ $\ln 10 = \ln (2 \times 5) = \ln 2 + \ln 5$ Therefore, $\ln 9.8 = (2 \ln 7 + \ln 2) - (\ln 2 + \ln 5) = 2 \ln 7 - \ln 5$ (c) $\ln (7 \sqrt{7}) = \ln 7 + \ln 7^{1/2} = \ln 7 + \frac{1}{2} \ln 7 = \frac{3}{2} \ln 7$ (d) $\ln 1225 = \ln (35^2) = 2 \ln 35 = 2 (\ln 5 + \ln 7) = 2 \ln 5 + 2 \ln 7$ (e) $\ln 0.056 = \ln \frac{56}{1000} = \ln 56 - \ln 1000$ $56 = 7 \times 8 = 7 \times 2^3$, so $\ln 56 = \ln 7 + 3 \ln 2$ $\ln 1000 = \ln 10^3 = 3 \ln 10 = 3 (\ln 2 + \ln 5) = 3 \ln 2 + 3 \ln 5$ Thus, $\ln 0.056 = (\ln 7 + 3 \ln 2) - (3 \ln 2 + 3 \ln 5) = \ln 7 - 3 \ln 5$ (f) $\frac{\ln 35 + \ln \frac{1}{7}}{\ln 25} = \frac{\ln 35 - \ln 7}{\ln 25} = \frac{\ln 5 + \ln 7 - \ln 7}{2 \ln 5} = \frac{\ln 5}{2 \ln 5} = \frac{1}{2}$ --- 3. **Write expressions as a single logarithm** (a) $\ln \sin \theta - \ln \frac{\sin \theta}{5} = \ln \sin \theta - (\ln \sin \theta - \ln 5) = \ln 5$ (b) $\ln (3x^2 - 9x) + \ln \frac{1}{3x} = \ln (3x^2 - 9x) - \ln (3x) = \ln \frac{3x^2 - 9x}{3x} = \ln (x - 3)$ (c) $\frac{1}{2} \ln (4t^4) - \ln b = \ln (4t^4)^{1/2} - \ln b = \ln (2 t^2) - \ln b = \ln \frac{2 t^2}{b}$ (a) $\ln \sec \theta + \ln \cos \theta = \ln (\sec \theta \cos \theta) = \ln 1 = 0$ (b) $\ln (8x + 4) - 2 \ln c = \ln (8x + 4) - \ln c^2 = \ln \frac{8x + 4}{c^2}$ (c) $3 \ln \sqrt[3]{t^2 - 1} - \ln (t + 1) = \ln (t^2 - 1)^{3/3} - \ln (t + 1) = \ln (t^2 - 1) - \ln (t + 1) = \ln \frac{t^2 - 1}{t + 1}$ Note: $t^2 - 1 = (t + 1)(t - 1)$, so expression simplifies to $\ln (t - 1)$ --- 4. **Simplify expressions involving $e$ and $\ln$** (a) $e^{\ln 7.2} = 7.2$ (b) $e^{-\ln x^2} = e^{\ln x^{-2}} = x^{-2} = \frac{1}{x^2}$ (c) $e^{\ln x - \ln y} = e^{\ln \frac{x}{y}} = \frac{x}{y}$ (a) $e^{\ln (x^2 + y^2)} = x^2 + y^2$ (b) $e^{-\ln 0.3} = e^{\ln \frac{1}{0.3}} = \frac{1}{0.3}$ (c) $e^{\ln \pi x - \ln 2} = e^{\ln \frac{\pi x}{2}} = \frac{\pi x}{2}$ (a) $2 \ln \sqrt{e} = 2 \times \frac{1}{2} \ln e = \ln e = 1$ (b) $\ln (\ln e^e) = \ln e = 1$ (c) $\ln (e^{-x^2 - y^2}) = -x^2 - y^2$ (a) $\ln (e^{\sec \theta}) = \sec \theta$ (b) $\ln (e^{e^r}) = e^r$ (c) $\ln (e^{2 \ln x}) = 2 \ln x$ --- **Final answers are given in each step above.**