$\log_{10}2 = a$、$\log_{10}3 = b$とするとき、$\log_{24}75$の値を$a$、$b$で表せ。代数学対数底の変換対数計算2025/5/81. 問題の内容log102=a\log_{10}2 = alog102=a、log103=b\log_{10}3 = blog103=bとするとき、log2475\log_{24}75log2475の値をaaa、bbbで表せ。2. 解き方の手順まず、底の変換公式を使って、log2475\log_{24}75log2475の底を10に変換します。log2475=log1075log1024\log_{24}75 = \frac{\log_{10}75}{\log_{10}24}log2475=log1024log1075次に、75と24を素因数分解します。75=3×5275 = 3 \times 5^275=3×5224=23×324 = 2^3 \times 324=23×3よって、log1075=log10(3×52)=log103+2log105\log_{10}75 = \log_{10}(3 \times 5^2) = \log_{10}3 + 2\log_{10}5log1075=log10(3×52)=log103+2log105log1024=log10(23×3)=3log102+log103\log_{10}24 = \log_{10}(2^3 \times 3) = 3\log_{10}2 + \log_{10}3log1024=log10(23×3)=3log102+log103log105\log_{10}5log105はlog105=log10(102)=log1010−log102=1−log102=1−a\log_{10}5 = \log_{10}(\frac{10}{2}) = \log_{10}10 - \log_{10}2 = 1 - \log_{10}2 = 1-alog105=log10(210)=log1010−log102=1−log102=1−aと変形できます。これらを代入して、log1075=log103+2(1−a)=b+2(1−a)=b+2−2a\log_{10}75 = \log_{10}3 + 2(1 - a) = b + 2(1 - a) = b + 2 - 2alog1075=log103+2(1−a)=b+2(1−a)=b+2−2alog1024=3log102+log103=3a+b\log_{10}24 = 3\log_{10}2 + \log_{10}3 = 3a + blog1024=3log102+log103=3a+bしたがって、log2475=log1075log1024=b+2−2a3a+b=2−2a+b3a+b\log_{24}75 = \frac{\log_{10}75}{\log_{10}24} = \frac{b + 2 - 2a}{3a + b} = \frac{2 - 2a + b}{3a + b}log2475=log1024log1075=3a+bb+2−2a=3a+b2−2a+b3. 最終的な答え2−2a+b3a+b\frac{2 - 2a + b}{3a + b}3a+b2−2a+b