Given that \( a=\lg 2 \) and \( b=\lg 3 \), express each of the following in terms of \( a \) and \( b \). \( \begin{array}{llll}\text { a } \lg 18 & \text { b } \lg 96 & \text { c } \lg \frac{9}{16} & \text { d } \lg 6-\lg 8 \\ \text { e } \lg \sqrt{6} & \text { f } \frac{3}{2} \lg 16+\frac{1}{2} \lg 81 & \text { g } 4 \lg 3-3 \lg 6 & \text { h } \lg 60+\lg 20-2\end{array} \)

To express each logarithmic expression in terms of \( a = \lg 2 \) and \( b = \lg 3 \), we'll make use of logarithm properties such as the product rule, quotient rule, and power rule. 1. **\( \lg 18 \)**: \[ \lg 18 = \lg (2 \cdot 9) = \lg (2 \cdot 3^2) = \lg 2 + 2\lg 3 = a + 2b \] 2. **\( \lg 96 \)**: \[ \lg 96 = \lg (32 \cdot 3) = \lg (2^5 \cdot 3) = 5 \lg 2 + \lg 3 = 5a + b \] 3. **\( \lg \frac{9}{16} \)**: \[ \lg \frac{9}{16} = \lg 9 - \lg 16 = \lg (3^2) - \lg (2^4) = 2\lg 3 - 4\lg 2 = 2b - 4a \] 4. **\( \lg 6 - \lg 8 \)**: \[ \lg 6 - \lg 8 = \lg \frac{6}{8} = \lg \frac{3 \cdot 2}{2^3} = \lg \frac{3}{4} = \lg 3 - \lg (2^2) = b - 2a \] 5. **\( \lg \sqrt{6} \)**: \[ \lg \sqrt{6} = \frac{1}{2} \lg 6 = \frac{1}{2} (\lg 2 + \lg 3) = \frac{1}{2}(a + b) \] 6. **\( \frac{3}{2} \lg 16 + \frac{1}{2} \lg 81 \)**: \[ \frac{3}{2} \lg 16 + \frac{1}{2} \lg 81 = \frac{3}{2}(4 \lg 2) + \frac{1}{2}(4 \lg 3) = 6a + 2b \] 7. **\( 4 \lg 3 - 3 \lg 6 \)**: \[ 4 \lg 3 - 3 \lg 6 = 4b - 3(\lg 2 + \lg 3) = 4b - 3a - 3b = -3a + b \] 8. **\( \lg 60 + \lg 20 - 2 \)**: \[ \lg 60 + \lg 20 - 2 = \lg (60 \cdot 20) - 2 = \lg 1200 - 2 = \lg (2^4 \cdot 3 \cdot 5) - 2 = 4 \lg 2 + \lg 3 + \lg 5 - 2 \] Since \( \lg 5 = \lg \frac{10}{2} = \lg 10 - \lg 2 = 1 - a \), we can express it as: \[ 4a + b + (1 - a) - 2 = 3a + b - 1 \] In summary: - \( \lg 18 = a + 2b \) - \( \lg 96 = 5a + b \) - \( \lg \frac{9}{16} = 2b - 4a \) - \( \lg 6 - \lg 8 = b - 2a \) - \( \lg \sqrt{6} = \frac{1}{2}(a + b) \) - \( \frac{3}{2} \lg 16 + \frac{1}{2} \lg 81 = 6a + 2b \) - \( 4 \lg 3 - 3 \lg 6 = -3a + b \) - \( \lg 60 + \lg 20 - 2 = 3a + b - 1 \)