Cho a = log 2 3 , b = log 3 5 , c = log 7 2 . Hãy tính log 140 63 theo a, b, c.
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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:
\(\log_{30}8=\frac{\log 8}{\log 30}=\frac{\log (2^3)}{\log (10.3)}=\frac{3\log2}{\log 10+\log 3}\)
\(=\frac{3\log (\frac{10}{5})}{1+\log 3}=\frac{3(\log 10-\log 5)}{1+\log 3}=\frac{3(1-b)}{1+a}\)
\(a,A=log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78\\ =log_28\\ =log_22^3\\ =3\\ b,B=log_22\cdot log_24...log_22^n\\ =log_22\cdot log_22^2...log_22^n\\ =1\cdot2\cdot...\cdot n\\ =n!\)
a: \(log_49=\dfrac{log9}{log4}=\dfrac{log3^2}{log2^2}=\dfrac{2\cdot log3}{2\cdot log2}=\dfrac{log3}{log2}=\dfrac{b}{a}\)
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: \(log_56=\dfrac{log6}{log5}=\dfrac{log\left(2\cdot3\right)}{log\left(\dfrac{10}{2}\right)}=\dfrac{log2+log3}{log10-log2}\)
\(=\dfrac{a+b}{1-a}\)
\(log_65=\dfrac{1}{log_56}=\dfrac{1}{log_52+log_53}=\dfrac{1}{a+b}\)
=>Chọn B
\(log_a\left(a^3b^2\right)=log_aa^3+log_ab^2=3+2\cdot log_ab\)
=>B
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: \(log_a\left(2b\right)+log_a\left(\dfrac{b^2}{2}\right)\)
\(=log_a\left(2b\cdot\dfrac{b^2}{2}\right)=log_a\left(b^3\right)=3\cdot2=6\)
a,Ta có: \(a^6=\left(a^{\dfrac{3}{5}}\right)^{10}=b^{10}\\ a^3b=\left(a^{\dfrac{3}{5}}\right)^5\cdot b=b^5\cdot b=b^6\\ \dfrac{a^9}{b^9}=\dfrac{\left(a^{\dfrac{3}{5}}\right)^{15}}{b^9}=\dfrac{b^{15}}{b^9}=b^6\)
b, \(log_ab=log_aa^{\dfrac{3}{5}}=\dfrac{3}{5}\\ log_a\left(a^2b^5\right)=log_a\left(a^2\cdot a^3\right)=log_a\left(a^5\right)=5\\ log_{\sqrt[5]{a}}\left(\dfrac{a}{b}\right)=5log_a\left(\dfrac{a}{a^{\dfrac{3}{5}}}\right)=5log_a\left(a^{\dfrac{2}{5}}\right)=2\)
a) \(log_29\cdot log_34=4\)
b) \(log_{25}\cdot\dfrac{1}{\sqrt{5}}=-\dfrac{1}{4}\)
c) \(log_23\cdot log_9\sqrt{5}\cdot log_54=\dfrac{1}{2}\)
Ta có:
log 140 63 = log 140 3 2 . 7 = 2 log 140 3 + log 140 7
Từ đề bài suy ra:
loh 0 , 5 π . log 7 5 = log 7 2 . log 2 3 . log 35 = cab
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