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\(B=\left(\log b_a+\log_ba+2\right)\left(\log b_a-\log b_{ab}\right)-1=\left(\log b_a+\frac{1}{\log b_a}+2\right)\left(\log b_a.\log_ba-\left(\log_{ab}b.\log_ba\right)\right)-1\)
\(=\frac{\log^2_ab+2\log_ab+1}{\log_ab}\left(1-\log_{ab}a\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{\log_aab}\right)-1\)
\(=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{1+\log_ab}\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}.\frac{\log_ab}{1+\log_ab}-1=\log_ab+1-1=\log_ab\)
Rút gọn biểu thức sau :
\(A=\left(\log_ab+\log_ba+2\right)\left(\log_ab-\log_{ab}b\right)\log_ba-1\)
\(=\left(\log_ab+\log_ba+2\right)\left(1-\log_{ab}a\right)-1\)
\(=\left(\log_ab+\log_ba+2\right)\left(1-\frac{1}{1+\log_ab}\right)-1\)
\(=\frac{1}{1+\log_ab}\left(\log_ab+\log_ba+2\right)-1\)
\(=\frac{1}{1+\log_ab}\left[\left(\log_ab+\log_ba+2\right)-1-\log_ab\right]\)
\(=\frac{1}{1+\log_ab}\left(\log_ab+\log^2_ba\right)=\log_ab\)
Ta có \(A=\left(\log^3_ba+2\log^2_ba+\log_ba\right)\left(\log_ab-\log_{ab}b\right)-\log_ba\)
\(=\left(\log_ba+1\right)^2\left(1-\frac{1}{\log_aab}\right)-\log_ba\)
\(=\left(\log_ba+1\right)^2\left(1-\frac{1}{1+\log_ab}\right)-\log_ba\)
\(=\left(\log_ba+1\right)^2\left(1-\frac{\log_ba}{\log_ba+1}\right)-\log_ba\)
\(=\log_ba+1-\log_ba=1\)
Chọn 2 làm cơ số, ta có :
\(A=\log_616=\frac{\log_216}{\log_26}=\frac{4}{1=\log_23}\)
Mặt khác :
\(x=\log_{12}27=\frac{\log_227}{\log_212}=\frac{3\log_23}{2+\log_23}\)
Do đó : \(\log_23=\frac{2x}{3-x}\) suy ra \(A=\frac{4\left(3-x\right)}{3+x}\)
b) Ta có :
\(B=\frac{lg30}{lg125}=\frac{lg10+lg3}{3lg\frac{10}{2}}=\frac{1+lg3}{3\left(1-lg2\right)}=\frac{1+a}{3\left(1-b\right)}\)
c) Ta có :
\(C=\log_65+\log_67=\frac{1}{\frac{1}{\log_25}+\frac{1}{\log_35}}+\frac{1}{\frac{1}{\log_27}+\frac{1}{\log_37}}\)
Ta tính \(\log_25,\log_35,\log_27,\log_37\) theo a, b, c .
Từ : \(a=\log_{27}5=\log_{3^3}5=\frac{1}{3}\log_35\)
Suy ra \(\log_35=3a\) do đó :
\(\log_25=\log_23.\log35=3ac\)
Mặt khác : \(b=\log_87=\log_{2^3}7=\frac{1}{3}\log_27\) nên \(\log_27=3b\)
Do đó : \(\log_37=\frac{\log_27}{\log_23}=\frac{3b}{c}\)
Vậy : \(C=\frac{1}{\frac{1}{3ac}+\frac{1}{3a}}+\frac{1}{\frac{1}{3b}+\frac{c}{3b}}=\frac{3\left(ac+b\right)}{1+c}\)
d) Điều kiện : \(a>0;a\ne0;b>0\)
Từ giả thiết \(\log_ab=\sqrt{3}\) suy ra \(b=a^{\sqrt{3}}\). Do đó :
\(\frac{\sqrt{b}}{a}=a^{\frac{\sqrt{3}}{2}-1};\frac{\sqrt[3]{b}}{\sqrt{a}}=a^{\frac{\sqrt{3}}{3}-\frac{1}{2}}=a^{\frac{\sqrt{3}}{3}\left(\frac{\sqrt{3}}{2}-1\right)}\)
Từ đó ta tính được :
\(A=\log_{a^{\alpha}}a^{\frac{-\sqrt{3}}{3}\alpha}=\log_{a^{\alpha}}\left(a^{\alpha}\right)^{\frac{-\sqrt{3}}{3}}=\frac{-\sqrt{3}}{3}\) với \(\alpha=\frac{\sqrt{3}}{2}-1\)
Vì \(a,b>1\) và \(c\ge0\Rightarrow0< \log_ba\le\log_b\left(a+c\right)\)
\(\Rightarrow\frac{1}{\log_ba}\ge\frac{1}{\log_b\left(a+c\right)}\Leftrightarrow\log_ab\ge\log_{a+c}b\)
\(\Rightarrow\) điều phải chứng minh
\(B=\log_{25}15\) biết \(\log_{25}3=a\)
Ta có : \(a=\log_{15}3=\frac{1}{\log_3\left(3.5\right)}=\frac{1}{1+\log_35}\)
\(\Rightarrow\log_35=\frac{1}{a}-1=\frac{1-a}{a}\)
\(\Rightarrow B=\log_{25}15=\frac{\log_315}{\log_325}=\frac{\log_3\left(3.5\right)}{\log_35^2}=\frac{1+\frac{1-a}{a}}{2.\log_35}=\frac{1}{2\left(1-a\right)}\)
Ta có :
\(a=\log_615=\frac{\log_215}{\log_26}=\frac{\log_23+\log_25}{1+\log_23}\left(1\right)\)
\(b=\log_{12}18=\frac{\log_118}{\log_212}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+2\log_23}{2+\log_23}\left(2\right)\)
Từ \(\left(2\right)\Rightarrow b\left(2+\log_23\right)=1+2\log_23\Leftrightarrow\left(b-2\right)\log_23=1-2b\Leftrightarrow\log_23=\frac{1-2b}{b-2}\)
Từ \(\left(1\right)\Rightarrow\log_25=a\left(a+\log_23\right)-\log_23=\left(a-1\right)\log_23+a=\left(a-1\right)\frac{1-2b}{b-2}+a=\frac{b-5}{4b-2a-2ab-2}\)
\(\Rightarrow F=\log_{25}24=\frac{\log_224}{\log_225}=\frac{\log_2\left(2^3.3\right)}{\log_25^2}=\frac{3+\log_23}{2\log_25}=\frac{3+\frac{1-2b}{b-2}}{2.\frac{2b-a-ab-1}{b-2}}=\frac{b-5}{4b-2a-2ab-2}\)
Ta có \(a=\log_{25}7=\frac{\log_27}{\log_225}=\frac{\log_27}{2\log_25}=\frac{\log_27}{2b}\Rightarrow\log_27=2ab\)
\(\Rightarrow H=\log_{\sqrt[3]{5}}\frac{49}{8}=\frac{\log_2\frac{49}{8}}{\log_2\sqrt[3]{5}}=\frac{\log_2\frac{7^2}{2^2}}{\log_25^{\frac{1}{3}}}=\frac{2\log_27-3}{\frac{1}{3}\log_25}=\frac{12ab-9}{b}\)
a) Áp dụng công thức: \(\log_ab.\log_bc=\log_ac\)
b) Vì \(\dfrac{1}{\log_{a^k}b}=\dfrac{1}{\dfrac{1}{k}\log_ab}=\dfrac{k}{\log_ab}\) nên biểu thức vế trái bằng:
\(VT=\dfrac{1}{\log_ab}\left(1+2+...+n\right)\)
\(=\dfrac{1}{\log_ab}.\dfrac{n\left(n+1\right)}{2}=VP\)
a) Từ công thức đổi cơ số suy ra ∀a,b,c > 0 (a,b ≠ 1), logab. logbc = logac.
Do đó log36. log89. log62 = ( log36. Log62).
= log32. 
log23 = 
.
b) logab2+
= logab2 + logab2 =2logab2 = 4 loga|b|.