Bài 1:

\(A=\log_380=\log_3(2^4.5)=\log_3(2^4)+\log_3(5)\)

\(=4\log_32+\log_35=4a+b\)

\(B=\log_3(37,5)=\log_3(2^{-1}.75)=\log_3(2^{-1}.3.5^2)\)

\(=\log_3(2^{-1})+\log_33+\log_3(5^2)=-\log_32+1+2\log_35\)

\(=-a+1+2b\)

Bài 2:

\(\log_{30}8=\frac{\log 8}{\log 30}=\frac{\log (2^3)}{\log (10.3)}=\frac{3\log2}{\log 10+\log 3}\)

\(=\frac{3\log (\frac{10}{5})}{1+\log 3}=\frac{3(\log 10-\log 5)}{1+\log 3}=\frac{3(1-b)}{1+a}\)

Bài 3:

\(\log_{27}5=a; \log_87=b; \log_23=c\)

\(\Leftrightarrow \frac{\ln 5}{\ln 27}=a; \frac{\ln 7}{\ln 8}=b; \frac{\ln 3}{\ln 2}=c\)

\(\Leftrightarrow \frac{\ln 5}{\ln (3^3)}=a; \frac{\ln 7}{\ln (2^3)}=b; \ln 3=c\ln 2\)

\(\Leftrightarrow \frac{\ln 5}{3\ln 3}=a; \frac{\ln 7}{3\ln 2}=b; \ln 3=c\ln 2\)

\(\Rightarrow \frac{\ln 5}{3c\ln 2}=a; \frac{\ln 7}{3\ln 2}=b\)

\(\Rightarrow \ln 35=\ln 5+\ln 7=3ac\ln 2+3b\ln 2\)

Do đó:
\(D=\log_6 35=\frac{\ln 35}{\ln 6}=\frac{\ln 35}{\ln 2+\ln 3}=\frac{\ln 35}{\ln 2+c\ln 2}=\frac{3ac\ln 2+3b\ln 2}{\ln 2+c\ln 2}\)

\(=\frac{3ac+3b}{1+c}\)

\(a,A=log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78\\ =log_28\\ =log_22^3\\ =3\\ b,B=log_22\cdot log_24...log_22^n\\ =log_22\cdot log_22^2...log_22^n\\ =1\cdot2\cdot...\cdot n\\ =n!\)

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a: \(log_49=\dfrac{log9}{log4}=\dfrac{log3^2}{log2^2}=\dfrac{2\cdot log3}{2\cdot log2}=\dfrac{log3}{log2}=\dfrac{b}{a}\)

b: \(log_612=\dfrac{log12}{log6}=\dfrac{log2^2+log3}{log2+log3}=\dfrac{2\cdot log2+log3}{log2+log3}\)

\(=\dfrac{2a+b}{a+b}\)

c: \(log_56=\dfrac{log6}{log5}=\dfrac{log\left(2\cdot3\right)}{log\left(\dfrac{10}{2}\right)}=\dfrac{log2+log3}{log10-log2}\)

\(=\dfrac{a+b}{1-a}\)

a,Ta có: \(a^6=\left(a^{\dfrac{3}{5}}\right)^{10}=b^{10}\\ a^3b=\left(a^{\dfrac{3}{5}}\right)^5\cdot b=b^5\cdot b=b^6\\ \dfrac{a^9}{b^9}=\dfrac{\left(a^{\dfrac{3}{5}}\right)^{15}}{b^9}=\dfrac{b^{15}}{b^9}=b^6\)

b, \(log_ab=log_aa^{\dfrac{3}{5}}=\dfrac{3}{5}\\ log_a\left(a^2b^5\right)=log_a\left(a^2\cdot a^3\right)=log_a\left(a^5\right)=5\\ log_{\sqrt[5]{a}}\left(\dfrac{a}{b}\right)=5log_a\left(\dfrac{a}{a^{\dfrac{3}{5}}}\right)=5log_a\left(a^{\dfrac{2}{5}}\right)=2\)

\(log_3\sqrt{3}=log_33^{\dfrac{1}{2}}=\dfrac{1}{2}\)

\(lne^3=log_ee^3=3\)

\(log_{27}3=log_{3^3}3=\dfrac{1}{3}\)

\(\log_{\sqrt{3}}3=log_{3^{\dfrac{1}{2}}}3=1:\dfrac{1}{2}=2\)

\(\log_{0,125}2=log_{2^{-3}}2=\dfrac{1}{-3}\)

\(\log_{\sqrt[3]{49}}7=\log_{7^{\dfrac{2}{3}}}7=1:\dfrac{2}{3}=\dfrac{3}{2}\)

\(\log_{\dfrac{1}{125}}5=\log_{5^{-3}}5=-\dfrac{1}{3}\)

\(\log_84=log_{2^3}2^2=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)

\(\log_{25}\left(\dfrac{1}{5}\right)=\log_{5^2}5^{-1}=\dfrac{1}{2}\cdot\left(-1\right)=-\dfrac{1}{2}\)

\(\log_{\dfrac{1}{5}}\sqrt{5}=\log_{5^{-1}}5^{\dfrac{1}{2}}=\dfrac{1}{-1}\cdot\dfrac{1}{2}=-\dfrac{1}{2}\)

\(log_{\dfrac{1}{7}}\sqrt[5]{49}=\log_{7^{-1}}7^{\dfrac{2}{5}}=\dfrac{1}{-1}\cdot\dfrac{2}{5}=-\dfrac{2}{5}\)

\(\log_4\left(\dfrac{1}{\sqrt{2}}\right)=\log_{2^2}\left(\sqrt{2}\right)^{-1}\)

\(=\log_{2^{-2}}\left(\sqrt{2}\right)^{-\dfrac{1}{2}}=\dfrac{1}{-2}\cdot\dfrac{-1}{2}=\dfrac{1}{4}\)

\(\log_{27}3\sqrt{3}=\log_{3^3}3^{\dfrac{3}{2}}=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)

\(log_a\left(a^3b^2\right)=log_aa^3+log_ab^2=3+2\cdot log_ab\)

=>B