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

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

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

=>B

\(a^2+4b^2=23ab\Rightarrow a^2+4ab+4b^2=27ab\Rightarrow\left(a+2b\right)^2=27ab\)

\(\Rightarrow\dfrac{\left(a+2b\right)^2}{9}=3ab\)\(\Rightarrow\left(\dfrac{a+2b}{3}\right)^2=3ab\)

Lấy logarit cơ số c hai vế:

\(log_c\left(\dfrac{a+2b}{3}\right)^2=log_c\left(3ab\right)\)

\(\Rightarrow2log_c\dfrac{a+2b}{3}=log_c3+log_ca+log_cb\)

\(\Rightarrow log_c\dfrac{a+2b}{3}=\dfrac{1}{2}\left(log_ca+log_cb+log_c3\right)\)

1) X=log1-log2+log2-log3+...+log99-log100

=log1-log100

=0-2

=-2

Đáp án C

2)X=-log₃100=-log₃10²=-2log₃(2.5)=-2log₃2-2log₃5=-2a-2b

Đáp án A

a) \(\log_{12}12^3=3.\log_{12}12=3.1=3\)

b) \(\log_{0,5}0,25=\log_{2^{-1}}2^{-2}=\dfrac{-2}{-1}\log_22=2.1=2\)

c) \(\log_aa^{-3}=-3.\log_aa=-3.1=-3\)

a: \(log_{12}12^3=3\)

b: \(=log_{0.5}0.5^2=2\)

c: \(log_aa^{-3}=-3\)

Đáp án C.

a)    \({\log _c}b = {\log _a}b.{\log _c}a \Leftrightarrow {a^{{{\log }_c}b}} = {a^{{{\log }_a}b.{{\log }_c}a}} \Leftrightarrow {c^{{{\log }_c}b}} = {\left( {{c^{{{\log }_c}a}}} \right)^{{{\log }_a}b}} \Leftrightarrow b = {a^{{{\log }_a}b}} \Leftrightarrow b = b\) (luôn đúng)

Vậy \({\log _c}b = {\log _a}b.{\log _c}a\)

b)    Từ \({\log _c}b = {\log _a}b.{\log _c}a \Leftrightarrow {\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}\)

\(a,log_a1=c\Leftrightarrow a^c=1\Leftrightarrow c=0\Rightarrow log_a1=0\\ b,log_aa=c\Leftrightarrow a^c=a\Leftrightarrow c=1\Rightarrow log_aa=1\\ c,log_aa^c=b\Leftrightarrow a^b=a^c\Leftrightarrow b=c\Rightarrow log_aa^c=c\\ d,a^{log_ab}=c\Leftrightarrow log_ab=log_ac\Leftrightarrow b=c\Rightarrow a^{log_ab}=b\)

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

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