$\log_{10}2 = a$、$\log_{10}3 = b$のとき、以下の対数を$a$と$b$で表す問題です。 (1) $\log_{10}\sqrt{216}$ (2) $\log_2 0.15$ (3) $\log_2 \sqrt[3]{108}$ (4) $\log_5 0.03125$

1. 問題の内容

$\log_{10}2 = a$、$\log_{10}3 = b$のとき、以下の対数を$a$と$b$で表す問題です。

(1) $\log_{10}\sqrt{216}$

(2) $\log_2 0.15$

(3) $\log_2 \sqrt[3]{108}$

(4) $\log_5 0.03125$

2. 解き方の手順

(1) $\log_{10}\sqrt{216}$

$\sqrt{216} = (216)^{\frac{1}{2}} = (2^3 \cdot 3^3)^{\frac{1}{2}} = 2^{\frac{3}{2}} \cdot 3^{\frac{3}{2}}$

$\log_{10}\sqrt{216} = \log_{10}(2^{\frac{3}{2}} \cdot 3^{\frac{3}{2}}) = \frac{3}{2}\log_{10}2 + \frac{3}{2}\log_{10}3 = \frac{3}{2}a + \frac{3}{2}b = \frac{3}{2}(a+b)$

(2) $\log_2 0.15$

$0.15 = \frac{15}{100} = \frac{3}{20} = \frac{3}{2^2 \cdot 5}$

$\log_2 0.15 = \log_2 \frac{3}{2^2 \cdot 5} = \log_2 3 - \log_2 (2^2 \cdot 5) = \log_2 3 - (2 + \log_2 5)$

ここで、$\log_2 3 = \frac{\log_{10} 3}{\log_{10} 2} = \frac{b}{a}$

$\log_2 5 = \frac{\log_{10} 5}{\log_{10} 2} = \frac{\log_{10} (10/2)}{\log_{10} 2} = \frac{\log_{10} 10 - \log_{10} 2}{\log_{10} 2} = \frac{1 - a}{a}$

$\log_2 0.15 = \frac{b}{a} - (2 + \frac{1 - a}{a}) = \frac{b}{a} - 2 - \frac{1}{a} + 1 = \frac{b}{a} - \frac{1}{a} - 1 = \frac{b-1}{a} - 1 = \frac{b-1-a}{a}$

(3) $\log_2 \sqrt[3]{108}$

$\sqrt[3]{108} = (108)^{\frac{1}{3}} = (2^2 \cdot 3^3)^{\frac{1}{3}} = 2^{\frac{2}{3}} \cdot 3$

$\log_2 \sqrt[3]{108} = \log_2 (2^{\frac{2}{3}} \cdot 3) = \frac{2}{3} + \log_2 3 = \frac{2}{3} + \frac{\log_{10} 3}{\log_{10} 2} = \frac{2}{3} + \frac{b}{a} = \frac{2a + 3b}{3a}$

(4) $\log_5 0.03125$

$0.03125 = \frac{3125}{100000} = \frac{5^5}{10^5} = \frac{5^5}{(2 \cdot 5)^5} = \frac{5^5}{2^5 \cdot 5^5} = \frac{1}{2^5} = 2^{-5}$

$\log_5 0.03125 = \log_5 2^{-5} = -5 \log_5 2 = -5 \frac{\log_{10} 2}{\log_{10} 5} = -5 \frac{a}{\log_{10} (10/2)} = -5 \frac{a}{1 - a}$

3. 最終的な答え

(1) $\log_{10}\sqrt{216} = \frac{3}{2}(a+b)$

(2) $\log_2 0.15 = \frac{b-a-1}{a}$

(3) $\log_2 \sqrt[3]{108} = \frac{2a+3b}{3a}$

(4) $\log_5 0.03125 = -\frac{5a}{1-a}$