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

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

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) \(log_216=4\)

b) \(log_3\dfrac{1}{27}=-3\)

c) \(log1000=3\)

d) \(9^{log_312}=144\)

\({a^{\frac{1}{2}}} = b \Leftrightarrow {\log _a}b = \frac{1}{2} \Leftrightarrow 2{\log _a}b = 1\)

Chọn B.

a) \(log_69+log_64=log_636=2\)

b) \(log_52-log_550=log_5\left(2:50\right)=-2\)

c) \(log_3\sqrt{5}-\dfrac{1}{2}log_550=-1,0479\)

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

a, Hàm số \(y=log_{\dfrac{1}{2}}x\) có cơ số \(\dfrac{1}{2}< 1\) nên hàm số nghịch biến trên \(\left(0;+\infty\right)\)

Mà \(4,8< 5,2\Rightarrow log_{\dfrac{1}{2}}4,8>log_{\dfrac{1}{2}}5,2\)

b, Ta có: \(log_{\sqrt{5}}2=2log_52=log_54\)

Hàm số \(y=log_5x\) có cơ số 5 > 1 nên hàm số đồng biến trên \(\left(0;+\infty\right)\)

Do \(4>2\sqrt{2}\Rightarrow log_54>log_52\sqrt{2}\Rightarrow log_{\sqrt{5}}2>log_52\sqrt{2}\)

c, Ta có: \(-log_{\dfrac{1}{4}}2=-\dfrac{1}{2}log_{\dfrac{1}{2}}2=log_{\dfrac{1}{2}}\dfrac{1}{\sqrt{2}}\)

Hàm số \(y=log_{\dfrac{1}{2}}x\) có cơ số \(\dfrac{1}{2}< 1\) nên nghịch biến trên \(\left(0;+\infty\right)\)

Do \(\dfrac{1}{\sqrt{2}}>0,4\Rightarrow log_{\dfrac{1}{2}}\dfrac{1}{\sqrt{2}}< log_{\dfrac{1}{2}}0,4\Rightarrow-log_{\dfrac{1}{4}}2< log_{\dfrac{1}{2}}0,4\)

\(log_{12}21=\dfrac{log_321}{log_312}=\dfrac{log_3\left(7\cdot3\right)}{log_3\left(2^2\cdot3\right)}=\dfrac{log_37+log_33}{log_34+log_33}\)

\(=\dfrac{log_37+1}{log_32^2+1}=\dfrac{log_37+1}{2\cdot log_32+1}=\dfrac{b+1}{2a+1}\)

ĐK: \(x>0\)

\(logx=2log5-log2\\ \Leftrightarrow logx=log25-log2\\ \Leftrightarrow logx=log\dfrac{25}{2}\Leftrightarrow x=12,5\)

Chọn C.