9．在实际应用中，通常用吸光度$A$和透光率$T$来衡量物体的透光性能，它们之间的换算公式为$A = \lg\frac{1}{T}$，下表为不同玻璃材料的透光率：
| 玻璃材料 | 材料 1 | 材料 2 | 材料 3 |
| --- | --- | --- | --- |
| $T$ | 0.7 | 0.8 | 0.9 |
设材料 1、材料 2、材料 3 的吸光度分别为$A_{1}$，$A_{2}$，$A_{3}$，则（ ）
A．$A_{1} > A_{2}$ B．$A_{2} > 3A_{3}$
C．$A_{1} + A_{3} > 2A_{2}$ D．$A_{1}A_{3} > A_{2}^{2}$
解 $A_{1} = \lg\frac{1}{0.7} = - \lg 0.7$
$A_{2} = \lg\frac{1}{0.8} = - \lg 0.8$
$A_{3} = \lg\frac{1}{0.9} = - \lg 0.9$
$\therefore A_{1}A_{3} = \lg 0.7\lg 0.9$，$A_{2}^{2} = \left( \lg 0.8 \right)^{2}$
法一：
$\frac{A_{1}A_{3}}{\lg 0.9\lg 0.8} = \frac{\lg 0.7}{\lg 0.8} = \log_{0.8}0.7$
$\frac{A_{2}^{2}}{\lg 0.9\lg 0.8} = \frac{\lg 0.8}{\lg 0.9} = \log_{0.9}0.8$
$\because \log_{0.8}(0.7) = 1 + \log_{0.8}\left( \frac{7}{8} \right) = 1 + \log_{0.8}\left( \frac{35}{40} \right) < 1 + \log_{0.8}\left( \frac{32}{40} \right) = 1 + \log_{0.8}(0.8) = 2$
$\log_{0.9}(0.8) = 1 + \log_{0.9}\left( \frac{8}{9} \right) = 1 + \log_{0.9}\left( \frac{80}{90} \right) > 1 + \log_{0.9}\left( \frac{81}{90} \right) = 1 + \log_{0.9}(0.9) = 2$
$\therefore A_{1}A_{3} < A_{2}^{2}$
法二：
$\because \log_{0.8}(0.7) - 2 = \log_{0.8}\left( \frac{0.7}{0.8^{2}} \right) = 2\log_{0.8}\left( \sqrt{\frac{0.7}{0.64}} \right) < 0$
$\log_{0.9}(0.8) - 2 = \log_{0.9}\left( \frac{0.8}{0.9^{2}} \right) = 2\log_{0.9}\left( \sqrt{\frac{0.8}{0.81}} \right) > 0$
$\therefore \log_{0.8}(0.7) < \log_{0.9}(0.8)$
$\therefore A_{1}A_{3} < A_{2}^{2}$