与えられた10個の対数を含む式の値をそれぞれ求める問題です。

1. 問題の内容

与えられた10個の対数を含む式の値をそれぞれ求める問題です。

2. 解き方の手順

各問題について、以下の手順で計算します。

(1) $\log_8 2 + \log_8 32$

$\log_8 2 + \log_8 32 = \log_8 (2 \times 32) = \log_8 64 = \log_8 8^2 = 2$

(2) $\log_3 135 - \log_3 5$

$\log_3 135 - \log_3 5 = \log_3 \frac{135}{5} = \log_3 27 = \log_3 3^3 = 3$

(3) $\log_6 3 - \log_6 108$

$\log_6 3 - \log_6 108 = \log_6 \frac{3}{108} = \log_6 \frac{1}{36} = \log_6 6^{-2} = -2$

(4) $\log_2 6 + \log_2 12 - 2\log_2 3$

$\log_2 6 + \log_2 12 - 2\log_2 3 = \log_2 6 + \log_2 12 - \log_2 3^2 = \log_2 6 + \log_2 12 - \log_2 9 = \log_2 \frac{6 \times 12}{9} = \log_2 \frac{72}{9} = \log_2 8 = \log_2 2^3 = 3$

(5) $\log_3 24 - \log_3 8 + \log_3 \sqrt{3}$

$\log_3 24 - \log_3 8 + \log_3 \sqrt{3} = \log_3 \frac{24}{8} + \log_3 \sqrt{3} = \log_3 3 + \log_3 \sqrt{3} = \log_3 3 + \log_3 3^{1/2} = 1 + \frac{1}{2} = \frac{3}{2}$

(6) $2\log_{10} \frac{3}{14} + \log_{10} \frac{28}{45} - \log_{10} \frac{2}{7}$

$2\log_{10} \frac{3}{14} + \log_{10} \frac{28}{45} - \log_{10} \frac{2}{7} = \log_{10} (\frac{3}{14})^2 + \log_{10} \frac{28}{45} - \log_{10} \frac{2}{7} = \log_{10} \frac{(\frac{3}{14})^2 \times \frac{28}{45}}{\frac{2}{7}} = \log_{10} \frac{\frac{9}{196} \times \frac{28}{45}}{\frac{2}{7}} = \log_{10} \frac{9 \times 28 \times 7}{196 \times 45 \times 2} = \log_{10} \frac{9 \times 4 \times 7 \times 7}{4 \times 49 \times 45 \times 2} = \log_{10} \frac{9}{45 \times 2} = \log_{10} \frac{1}{10} = \log_{10} 10^{-1} = -1$

(7) $\log_{\frac{1}{2}} \frac{8}{13} - 2\log_{\frac{1}{2}} \frac{2}{3} + \log_{\frac{1}{2}} \frac{26}{9}$

$\log_{\frac{1}{2}} \frac{8}{13} - 2\log_{\frac{1}{2}} \frac{2}{3} + \log_{\frac{1}{2}} \frac{26}{9} = \log_{\frac{1}{2}} \frac{8}{13} - \log_{\frac{1}{2}} (\frac{2}{3})^2 + \log_{\frac{1}{2}} \frac{26}{9} = \log_{\frac{1}{2}} \frac{\frac{8}{13} \times \frac{26}{9}}{(\frac{2}{3})^2} = \log_{\frac{1}{2}} \frac{\frac{8}{13} \times \frac{26}{9}}{\frac{4}{9}} = \log_{\frac{1}{2}} \frac{8 \times 26 \times 9}{13 \times 9 \times 4} = \log_{\frac{1}{2}} \frac{8 \times 2}{4} = \log_{\frac{1}{2}} 4 = \log_{\frac{1}{2}} (\frac{1}{2})^{-2} = -2$

(8) $\frac{1}{2} \log_3 5 - \log_3 \sqrt{\frac{5}{3}}$

$\frac{1}{2} \log_3 5 - \log_3 \sqrt{\frac{5}{3}} = \log_3 5^{\frac{1}{2}} - \log_3 (\frac{5}{3})^{\frac{1}{2}} = \log_3 \frac{5^{\frac{1}{2}}}{(\frac{5}{3})^{\frac{1}{2}}} = \log_3 \frac{5^{\frac{1}{2}}}{5^{\frac{1}{2}} \times 3^{-\frac{1}{2}}} = \log_3 3^{\frac{1}{2}} = \frac{1}{2}$

(9) $\log_2 \sqrt{\frac{7}{48}} + \log_2 12 - \frac{1}{2} \log_2 42$

$\log_2 \sqrt{\frac{7}{48}} + \log_2 12 - \frac{1}{2} \log_2 42 = \log_2 (\frac{7}{48})^{\frac{1}{2}} + \log_2 12 - \log_2 (42)^{\frac{1}{2}} = \log_2 \frac{(\frac{7}{48})^{\frac{1}{2}} \times 12}{(42)^{\frac{1}{2}}} = \log_2 \sqrt{\frac{\frac{7}{48} \times 12^2}{42}} = \log_2 \sqrt{\frac{7 \times 144}{48 \times 42}} = \log_2 \sqrt{\frac{7 \times 3}{1 \times 42}} = \log_2 \sqrt{\frac{1}{2}} = \log_2 2^{-\frac{1}{2}} = -\frac{1}{2}$

(10) $\log_{16} (\sqrt{5 + \sqrt{24}} - \sqrt{5 - \sqrt{24}})$

まず、$\sqrt{5 + \sqrt{24}} - \sqrt{5 - \sqrt{24}} = \sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}} = \sqrt{(\sqrt{3} + \sqrt{2})^2} - \sqrt{(\sqrt{3} - \sqrt{2})^2} = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) = 2\sqrt{2}$

したがって、$\log_{16} (2\sqrt{2}) = \log_{16} (2 \times 2^{\frac{1}{2}}) = \log_{16} 2^{\frac{3}{2}} = \log_{16} (16^{\frac{1}{4}})^3 = \log_{16} 16^{\frac{3}{4}} = \frac{3}{4}$

3. 最終的な答え

(1) 2

(2) 3

(3) -2

(4) 3

(5) 3/2

(6) -1

(7) -2

(8) 1/2

(9) -1/2

(10) 3/4