Hàm số \(y=log_cx\) nghịch biến

\(\Rightarrow0< c< 1\) và các hàm \(y=log_ax,y=log_bx\) đồng biến nên \(a,b>1\)

Ta chọn \(x=100\Rightarrow log_a>log_b100\Rightarrow a< b\Rightarrow b>a>c\)

\(\Rightarrow B\)

B

\(log_cx\) nghịch biến biến nên 0<c<1

\(log_ax;log_bx\) đồng  biến nên a>1; b>1

=>Loại D

\(log_ax>log_bx\left(x>1\right)\)

=>\(\dfrac{1}{log_xa}< \dfrac{1}{log_xb}\)

=>a<b

=>Chọn B

Đáp án D

⇔ log   z - 1 log   z = 1 1 - log   x

⇔ 1 - log   x = log   z log   z - 1

⇔ log   x = - 1 log   z - 1 ⇔ x = 10 1 1 - log   z .

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

=>B

Vì \(\dfrac{1}{e}\simeq0,368< 1\)

\(\Rightarrow y=log_{\dfrac{1}{e}}\left(x\right)\) nghịch biến trên D = \(\left(0;+\infty\right)\)

Chọn C.

0<1/e<1

=>\(log_{\dfrac{1}{e}}\left(x\right)\) nghịch biến 

=>C

Đáp án C.

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) \(\log_a\left(a^2b\right)=\log_aa^2+\log_ab=2.\log_aa+\log_ab=2.1+2=4\)

b) \(\log_a\dfrac{a\sqrt{a}}{b\sqrt[3]{a}}=\log_a\left(a\sqrt{a}\right)-\log_a\left(b\sqrt[3]{b}\right)=\log_aa^{\dfrac{3}{2}}-\log_ab^{\dfrac{4}{3}}=\dfrac{3}{2}.\log_aa-\dfrac{4}{3}\log_ab=\dfrac{3}{2}.1-\dfrac{4}{3}.2=-\dfrac{7}{6}\)

c) \(\log_a\left(2b\right)+\log_a\left(\dfrac{b^2}{2}\right)=\log_a2+\log_ab+\log_ab^2-\log_a2=\log_ab+2\log_ab=3\log_ab=3.2=6\)

a: \(=log_aa^2+log_ab=2+2=4\)

b: \(log_a\left(\dfrac{a\sqrt{a}}{b\sqrt[3]{b}}\right)=log_aa^{\dfrac{3}{2}}-log_ab^{\dfrac{4}{3}}\)

=3/2-4/3*2

=3/2-8/3

=9/6-16/6=-7/6

c: \(log_a\left(2b\right)+log_a\left(\dfrac{b^2}{2}\right)\)

\(=log_a\left(2b\cdot\dfrac{b^2}{2}\right)=log_a\left(b^3\right)=3\cdot2=6\)

Chọn D

Đáp án A