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Ta có : \(A=\log_{20}0,04=\log_{20}\frac{2}{5^3}=\frac{\log_2\frac{2}{5^3}}{\log_2\left(2^2.5\right)}=\frac{1-3\log_25}{2+\log_25}=\frac{1-3a}{2+a}\)
Ta có :
\(a=\log_{14}7=\frac{1}{\log_7\left(2.7\right)}=\frac{1}{1+\log_72}\Rightarrow\log_72=\frac{1}{a}-1=\frac{1-1}{a}\)
\(b=\log_{15}5=\frac{\log_75}{\log_7\left(7.2\right)}=\frac{\log_72}{1+\log_72}\Rightarrow\log_75=b\left(1+\log_72\right)=b\left(1+\frac{1-a}{a}\right)=\frac{b}{a}\)
\(\Rightarrow E=\log_{35}28=\frac{\log_727}{\log_735}=\frac{\log_7\left(7.2^2\right)}{\log_7\left(7.5\right)}=\frac{1+\log_72}{1+\log_75}=\frac{1+2.\frac{1-a}{a}}{1+\frac{b}{a}}=\frac{2-a}{a+b}\)
\(B=\log_{25}15\) biết \(\log_{25}3=a\)
Ta có : \(a=\log_{15}3=\frac{1}{\log_3\left(3.5\right)}=\frac{1}{1+\log_35}\)
\(\Rightarrow\log_35=\frac{1}{a}-1=\frac{1-a}{a}\)
\(\Rightarrow B=\log_{25}15=\frac{\log_315}{\log_325}=\frac{\log_3\left(3.5\right)}{\log_35^2}=\frac{1+\frac{1-a}{a}}{2.\log_35}=\frac{1}{2\left(1-a\right)}\)
Ta có \(a=\log_{\sqrt{2}}\left(\frac{1}{\sqrt[3]{5}}\right)=\log_{2^{\frac{1}{2}}}5^{-\frac{1}{3}}=-\frac{2}{3}\log_25\)
\(\Rightarrow\log_25=-\frac{3a}{2}\)
\(\Rightarrow C=\log40=\frac{\log_240}{\log_210}=\frac{\log_2\left(2^3.5\right)}{\log_2\left(2.5\right)}=\frac{3+\log_25}{1+\log_25}=\frac{6-3a}{2-3a}\)
Ta có :
\(a=\log_615=\frac{\log_215}{\log_26}=\frac{\log_23+\log_25}{1+\log_23}\left(1\right)\)
\(b=\log_{12}18=\frac{\log_118}{\log_212}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+2\log_23}{2+\log_23}\left(2\right)\)
Từ \(\left(2\right)\Rightarrow b\left(2+\log_23\right)=1+2\log_23\Leftrightarrow\left(b-2\right)\log_23=1-2b\Leftrightarrow\log_23=\frac{1-2b}{b-2}\)
Từ \(\left(1\right)\Rightarrow\log_25=a\left(a+\log_23\right)-\log_23=\left(a-1\right)\log_23+a=\left(a-1\right)\frac{1-2b}{b-2}+a=\frac{b-5}{4b-2a-2ab-2}\)
\(\Rightarrow F=\log_{25}24=\frac{\log_224}{\log_225}=\frac{\log_2\left(2^3.3\right)}{\log_25^2}=\frac{3+\log_23}{2\log_25}=\frac{3+\frac{1-2b}{b-2}}{2.\frac{2b-a-ab-1}{b-2}}=\frac{b-5}{4b-2a-2ab-2}\)
Ta có : \(b=lg2=lg\left(\frac{10}{5}\right)=1-lg5\Rightarrow lg5=1-b\)
\(\Rightarrow G=\log_{125b}30=\frac{lg30}{lg125}=\frac{lg\left(3.10\right)}{lg\left(5^3\right)}=\frac{1+lg3}{3lg5}=\frac{1+a}{3\left(1-b\right)}\)
Ta có :
\(\begin{cases}a=\log_{27}5=\frac{\log_25}{\log_227}=\frac{\log_25}{3\log_23}=\frac{\log_25}{3c}\Rightarrow\log_25=3ac\\b=\log_87=\frac{\log_27}{\log_28}=\frac{\log_27}{3}\Rightarrow\log_27=3b\end{cases}\)
\(\Rightarrow J=\log_635=\frac{\log_235}{\log_26}=\frac{\log_25+\log_27}{1+\log_23}=\frac{3ac+3b}{1+c}\)
Ta có : \(\log_25=\log_23.\log_35=ab\)
\(\Rightarrow I=\log_{140}63=\frac{\log_263}{\log_2140}=\frac{\log_2\left(3^2.7\right)}{\log_2\left(2^2.5.7\right)}=\frac{2\log_23+\log_27}{2+\log_25+\log_27}=\frac{2a+c}{2+ab+c}\)
Ta có : \(D=\log_6\left(21,6\right)=\frac{\log_2\left(21,6\right)}{\log_26}=\frac{\log_2\frac{2^2.3^3}{5}}{\log_2\left(2.3\right)}=\frac{2+3\log_23-\log_25}{1+\log_23}=\frac{2+3a-b}{1+a}\)
Ta có \(a=\log_{25}7=\frac{\log_27}{\log_225}=\frac{\log_27}{2\log_25}=\frac{\log_27}{2b}\Rightarrow\log_27=2ab\)
\(\Rightarrow H=\log_{\sqrt[3]{5}}\frac{49}{8}=\frac{\log_2\frac{49}{8}}{\log_2\sqrt[3]{5}}=\frac{\log_2\frac{7^2}{2^2}}{\log_25^{\frac{1}{3}}}=\frac{2\log_27-3}{\frac{1}{3}\log_25}=\frac{12ab-9}{b}\)