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

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

Lấy logarit cơ số c hai vế:

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

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

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

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

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

Chọn B.

Lời giải:

Ta có \(\left\{\begin{matrix} \log_ab=\frac{b}{4}\\ \log_2a=\frac{16}{b}\end{matrix}\right.\Rightarrow 4=\log_2a.\log_ab=\log_2b\)

\(\Rightarrow b=16\).

\(\log_2a=\frac{16}{b}=1\Rightarrow a=2\)

Do đó \(a+b=18\). Đáp án D.

em chưa có học

\(a,a^{log_ab^{\alpha}}=c\Leftrightarrow log_ac=log_ab^{\alpha}\Leftrightarrow c=b^{\alpha}\Rightarrow a^{log_ab^{\alpha}}=b^{\alpha}\\ a^{\alpha log_ab}=c\Leftrightarrow\alpha log_ab=log_ac\Leftrightarrow log_ab^{\alpha}=log_ac\Leftrightarrow b^{\alpha}=c\Rightarrow a^{\alpha log_ab}=b^{\alpha}\\ \Rightarrow a^{log_ab^{\alpha}}=a^{\alpha log_ab}\)

\(b,a^{log_ab^{\alpha}}=a^{\alpha log_ab}\\ \Rightarrow log_ab^{\alpha}=\alpha log_ab\)

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

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

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: \(log_aa^{-3}=-3\)