与えられた10個の対数の式をそれぞれ簡単にします。

1. 問題の内容

与えられた10個の対数の式をそれぞれ簡単にします。

2. 解き方の手順

(1) $\log_{\frac{1}{2}}4$

$\log_{\frac{1}{2}}4 = \log_{2^{-1}}2^2 = \frac{2}{-1} \log_2 2 = -2$

(2) $\log_2 \sqrt[3]{2}$

$\log_2 \sqrt[3]{2} = \log_2 2^{\frac{1}{3}} = \frac{1}{3} \log_2 2 = \frac{1}{3}$

(3) $\log_{10} 0.0001$

$\log_{10} 0.0001 = \log_{10} 10^{-4} = -4 \log_{10} 10 = -4$

(4) $\log_{0.5} 16$

$\log_{0.5} 16 = \log_{\frac{1}{2}} 2^4 = \log_{2^{-1}} 2^4 = \frac{4}{-1} \log_2 2 = -4$

(5) $\log_2 3 \cdot \log_3 4$

$\log_2 3 \cdot \log_3 4 = \log_2 3 \cdot \log_3 2^2 = \log_2 3 \cdot 2 \log_3 2 = 2 \log_2 3 \cdot \frac{1}{\log_2 3} = 2$

(6) $\log_{10} 4 + \log_{10} 25$

$\log_{10} 4 + \log_{10} 25 = \log_{10} (4 \times 25) = \log_{10} 100 = \log_{10} 10^2 = 2$

(7) $\log_3 63 - \log_3 21$

$\log_3 63 - \log_3 21 = \log_3 \frac{63}{21} = \log_3 3 = 1$

(8) $\log_2 50 - \log_4 25 + \log_2 \frac{8}{5}$

$\log_2 50 - \log_4 25 + \log_2 \frac{8}{5} = \log_2 50 - \frac{\log_2 25}{\log_2 4} + \log_2 \frac{8}{5} = \log_2 50 - \frac{\log_2 5^2}{2} + \log_2 \frac{8}{5} = \log_2 50 - \frac{2\log_2 5}{2} + \log_2 \frac{8}{5} = \log_2 50 - \log_2 5 + \log_2 \frac{8}{5} = \log_2 \frac{50}{5} + \log_2 \frac{8}{5} = \log_2 10 + \log_2 \frac{8}{5} = \log_2 (10 \times \frac{8}{5}) = \log_2 16 = \log_2 2^4 = 4$

(9) $\frac{1}{2}\log_2 18 - \log_2 \frac{3}{8}$

$\frac{1}{2}\log_2 18 - \log_2 \frac{3}{8} = \log_2 18^{\frac{1}{2}} - \log_2 \frac{3}{8} = \log_2 \sqrt{18} - \log_2 \frac{3}{8} = \log_2 3\sqrt{2} - \log_2 \frac{3}{8} = \log_2 \frac{3\sqrt{2}}{\frac{3}{8}} = \log_2 (3\sqrt{2} \times \frac{8}{3}) = \log_2 8\sqrt{2} = \log_2 2^3 \cdot 2^{\frac{1}{2}} = \log_2 2^{\frac{7}{2}} = \frac{7}{2}$

(10) $\frac{\log_4 9}{\log_2 3}$

$\frac{\log_4 9}{\log_2 3} = \frac{\log_4 3^2}{\log_2 3} = \frac{2\log_4 3}{\log_2 3} = \frac{2\frac{\log_2 3}{\log_2 4}}{\log_2 3} = \frac{2\frac{\log_2 3}{2}}{\log_2 3} = \frac{\log_2 3}{\log_2 3} = 1$

3. 最終的な答え

(1) -2

(2) 1/3

(3) -4

(4) -4

(5) 2

(6) 2

(7) 1

(8) 4

(9) 7/2

(10) 1