【高校数学Ⅱ】5-2-1 対数｜問題集

\(=\log_{2}2^5\)
\(=5\)

\(=\log_{5}5^2\)
\(=2\)

\(=\log_{3}3^1\)
\(=1\)

\(=\log_{3}3^0\)
\(=0\)

\(=\log_{3}3^{-3}\)
\(=-3\)

\(\displaystyle =\log_{\frac{1}{2}}\left(\frac{1}{2}\right)^4\)
\(=4\)

\(=\log_{10}10^{-1}\)
\(=-1\)

\(\displaystyle =\log_{\frac{1}{3}}\left(\frac{1}{3}\right)^{-1}\)
\(=-1\)

\(\displaystyle =\log_{2}2^{\frac{1}{3}}\)
\(\displaystyle =\frac{1}{3}\)

\(=\log_{\sqrt{5}}(\sqrt{5})^{2}\)
\(=2\)

\(=\log_{4}(2×8)\)
\(=\log_{4}16\)
\(=\log_{4}4^2\)
\(=2\)

\(\displaystyle =\log_{3}\frac{2}{18}\)
\(\displaystyle =\log_{3}\frac{1}{9}\)
\(=\log_{3}3^{-2}\)
\(=-2\)

\(=\log_{3}4+\log_{3}18-\log_{3}2^3\)
\(\displaystyle =\log_{3}\frac{4・18}{8}\)
\(=\log_{3}9\)
\(=\log_{3}3^2\)
\(=2\)

\(=\log_{2}12^{\frac{1}{3}}-\log_{2}3^{\frac{1}{3}}\)
\(\displaystyle =\log_{2}\frac{12^{\frac{1}{3}}}{3^{\frac{1}{3}}}\)
\(\displaystyle =\log_{2}\left(\frac{12}{3}\right)^{\frac{1}{3}}\)
\(=\log_{2}4^{\frac{1}{3}}\)
\(=\log_{2}2^{\frac{2}{3}}\)
\(\displaystyle =\frac{2}{3}\)

\(=\log_{12}(3×4)\)
\(=\log_{12}12\)
\(=\log_{12}12^1\)
\(=1\)

\(\displaystyle =\log_{2}\frac{3}{6}\)
\(\displaystyle =\log_{2}\frac{1}{2}\)
\(=\log_{2}2^{-1}\)
\(=-1\)

\(\displaystyle =\log_{4}\frac{8}{2}\)
\(=\log_{4}4\)
\(=\log_{4}4^1\)
\(=1\)

\(=\log_{12}(8×18)\)
\(=\log_{12}144\)
\(=\log_{12}12^2\)
\(=2\)

\(\displaystyle =\frac{\log_{2}8}{\log_{2}4}\)
\(\displaystyle =\frac{\log_{2}2^3}{\log_{2}2^2}\)
\(\displaystyle =\frac{3}{2}\)

\(\displaystyle =\frac{\log_{3}3}{\log_{3}9}\)
\(\displaystyle =\frac{\log_{3}3^1}{\log_{3}3^2}\)
\(\displaystyle =\frac{1}{2}\)

\(\displaystyle =\frac{\log_{2}2}{\log_{2}3}・\log_{2}3^3\)
\(\displaystyle =\frac{1}{\log_{2}3}・3\log_{2}3\)
\(=3\)

\(\displaystyle =\log_{3}8・\frac{\log_{3}3}{\log_{3}4}\)
\(\displaystyle =\log_{3}2^3・\frac{\log_{3}3}{\log_{3}2^2}\)
\(\displaystyle =3\log_{3}2・\frac{1}{2\log_{3}2}\)
\(\displaystyle =\frac{3}{2}\)

\(\displaystyle =\frac{\log_{3}9}{\log_{3}27}\)
\(\displaystyle =\frac{\log_{3}3^2}{\log_{3}3^3}\)
\(\displaystyle =\frac{2}{3}\)

\(\displaystyle =\log_{2}5・\frac{\log_{2}4}{\log_{2}5}\)
\(=\log_{2}2^2\)
\(=2\)

\(=\log_{2}2^2+\log_{2}3-2\)
\(=2+\log_{2}3-2\)
\(=\log_{2}3\)

\(\displaystyle =\frac{\log_{2}\frac{1}{4}}{\log_{2}\sqrt{2}}\)
\(\displaystyle =\frac{\log_{2}2^{-2}}{\log_{2}2^\frac{1}{2}}\)
\(\displaystyle =\frac{-2}{\frac{1}{2}}\)
\(=-4\)

\(\displaystyle =\frac{\log_{2}9}{\log_{2}4}・\frac{\log_{2}8}{\log_{2}3}\)
\(\displaystyle =\frac{\log_{2}3^2}{\log_{2}2^2}・\frac{\log_{2}2^3}{\log_{2}3}\)
\(\displaystyle =\frac{2\log_{2}3}{2}・\frac{3}{\log_{2}3}\)
\(=3\)

\(\log_{10}35\)
\(\displaystyle =\frac{\log_{2}35}{\log_{2}10}\)
\(\displaystyle =\frac{\log_{2}5+\log_{2}7}{\log_{2}2+\log_{2}5}\)
\(\displaystyle =\frac{a+\log_{2}7}{1+a}\)
\(b=\log_{5}7\)より、
\(\displaystyle b=\frac{\log_{2}7}{\log_{2}5}\)
\(\displaystyle \log_{2}7=ab\)
よって、
\(\displaystyle \log_{10}35=\frac{a+ab}{1+a}\)

\(\log_{5}7\)
\(\displaystyle =\frac{\log_{3}7}{\log_{3}5}\)
\(\displaystyle =\frac{\log_{3}7}{a}\)
\(b=\log_{7}9\)より、
\(\displaystyle b=\frac{\log_{3}9}{\log_{3}7}\)
\(\displaystyle \log_{3}7=\frac{2}{b}\)
よって、
\(\displaystyle \log_{5}7=\frac{2}{ab}\)

\(=\log_{10}(2^2×3)\)
\(=2\log_{10}2+\log_{10}3\)
\(=2a+b\)

\(=\log_{10}8-\log_{10}9\)
\(=\log_{10}2^3-\log_{10}3^2\)
\(=3\log_{10}2-2\log_{10}3\)
\(=3a-2b\)

\(=\displaystyle \log_{10}\frac{10}{2}\)
\(=\log_{10}10-\log_{10}2\)
\(=1-\log_{10}2\)
\(=1-a\)

\(\displaystyle =\frac{\log_{10}3}{\log_{10}4}\)
\(\displaystyle =\frac{\log_{10}3}{\log_{10}2^2}\)
\(\displaystyle =\frac{b}{2a}\)