a) \(y = {\log _a}M \Leftrightarrow M = {a^y}\)

b) Lấy loogarit theo cơ số b cả hai vế của \(M = {a^y}\) ta được

\({\log _b}M = {\log _b}{a^y} \Leftrightarrow {\log _b}M = y{\log _b}a \Leftrightarrow y = \frac{{{{\log }_b}M}}{{{{\log }_b}a}}\)

a: \(log_2\left(mn\right)=log_2\left(2^7\cdot2^3\right)=7+3=10\)

 \(log_2m+log_2n=log_22^7+log_22^3=7+3=10\)

=>\(log_2\left(mn\right)=log_2m+log_2n\)

b: \(log_2\left(\dfrac{m}{n}\right)=log_2\left(\dfrac{2^7}{2^3}\right)=7-3=4\)

\(log_2m-log_2n=log_22^7-log_22^3=7-3=4\)

=>\(log_2\left(\dfrac{m}{n}\right)=log_2m-log_2n\)

a) \(\log_2\left(mn\right)=\log_2\left(2^7.2^3\right)=\log_22^{7+3}=\log_22^{10}=10.\log_22=10.1=10\)

\(\log_2m+\log_2n=\log_22^7+\log_22^3=7\log_22+3\log_22=7.1+3.1=7+3=10\)

b) \(\log_2\left(\dfrac{m}{n}\right)=\log_2\dfrac{2^7}{2^3}=\log_22^4=4.\log_22=4.1=4\)

\(\log_2m-\log_2n=\log_22^7-\log_22^3=7.\log_22-3\log_22=7.1-3.1=4\)

a: \(log_2\left(M\cdot N\right)=log_2\left(2^5\cdot2^3\right)=log_2\left(2^8\right)=8\)

\(log_2M+log_2N=log_22^5+log_22^3=5+3=8\)

=>\(log_2\left(MN\right)=log_2M+log_2N\)

b: \(log_2\left(\dfrac{M}{N}\right)=log_2\left(\dfrac{2^5}{2^3}\right)=log_2\left(2^2\right)=2\)

\(log_2M-log_2N=log_22^5-log_22^3=5-3=2\)

=>\(log_2\left(\dfrac{M}{N}\right)=log_2M-log_2N\)

Đáp án D.

Ta có

log   6125 7 = log   6125 + log 7 = log 7 2 . 125 + 1 2 log   7

= 5 2 log   7 + log   5 3 = 5 2 n + 3 log   5 = 5 2 n + 3 1 - log   2

= 5 2 n + 3 - 3 m .

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

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

\(log_65=\dfrac{1}{log_56}=\dfrac{1}{log_52+log_53}=\dfrac{1}{a+b}\)

=>Chọn B

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

ĐK \(0< x< 20\)

\(logx\left(20-x\right)=log100\Rightarrow x\left(20-x\right)=100\)

\(\Rightarrow x^2-20x+100=0\Rightarrow x=10\)