$$\log_a(x_1x_2) = \log_a x_1 + \log_a x_2 \quad (2.38a)$$

$$\log_a\left(\frac{x_1}{x_2}\right) = \log_a x_1 - \log_a x_2 \quad (2.38b)$$

$$\log_a x^n = n \log_a x \quad (2.38c)$$

$$x = a^{\log_a x} \qquad (2.38d)$$

$$y^x = a^{x \log_a y} \quad (2.38e)$$

$$\log_a x = \frac{\log_b x}{\log_b a} \quad (2.38f)$$

• 例 2.57 (a) 计算 $\log_2 32$ .
• $$\left(\frac{\sqrt{(10x)}}{y^2}\right)$$

解 (a) 由于 $32 = 2^5$ , $\log_2 32 = \log_2 2^5 = 5 \log_2 2 = 5$ ，因为 $\log_2 2 = 1$ .

• $$\begin{aligned} (b) \quad & \frac{1}{3} \log_2 8 - \log_2 \frac{2}{7} = \log_2 8^{1/3} - \log_2 \frac{2}{7} \ & = \log_2 2 - [\log_2 2 - \log_2 7] = \log_2 7 \end{aligned}$$

• (c) $\ln\left(\frac{\sqrt{10x}}{y^2}\right)$

• $$\begin{aligned} (a) \quad \text{Since } 32 &= 2^2, \log_2 32 = \log_2 2^3 = 5 \log_2 2 = 5, \text{ since } \log_2 2 = 1 \ (b) \quad \frac{1}{3} \log_2 8 - \log_2 \frac{2}{7} &= \log_2 2^{1/3} - \log_2 \frac{2}{7} \ &= \log_2 2 - [\log_2 2 - \log_2 7] = \log_2 7 \ (c) \quad \ln\left(\frac{\sqrt{(10x)}}{y^2}\right) &= \ln(\sqrt{(10x)}) - \ln(y^2) = \frac{1}{2} \ln(10x) - 2 \ln y \ &= \frac{1}{2} \ln(10) + \frac{1}{2} \ln x - 2 \ln y \end{aligned}$$

• (d) $\log_{10}32 = \log_232 \log_{10}2$ ，因此

$$\frac{\log_{10}32}{\log_{10}2} = \log_232 = \log_2 2^5 = 5\log_2 2 = 5$$

• (e) $\log_9 x = \log_3 x \log_9 3$ ，所以

$$\frac{\log_3 x}{\log_9 x} = \frac{\log_3 x}{\log_3 x \log_9 3} = \frac{1}{\log_9 3}$$

但是 $3 = 9^{1/2}$ 从而 $\log_9 3 = \log_9 9^{1/2} = \frac{1}{2} \log_9 9 = \frac{1}{2}$ ，因此

$$\frac{\log_3 x}{\log_9 x} = 2$$