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Theo công thức biến đổi có số ta có : \(\log_{a^n}x=\frac{\log_ax}{\log_aa^n}=\frac{1}{n}\log_ax\)
Từ đó ta có :
\(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}\)
\(=\frac{1}{\log_ax}+\frac{2}{\log_ax}+\frac{4}{\log_ax}+...+\frac{n}{\log_ax}\)
\(=\frac{1+2+3+...+n}{\log_ax}=\frac{n\left(n+1\right)}{\log_ax}\)
Vậy \(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}=\frac{n\left(n+1\right)}{\log_ax}\)
1.\(\dfrac{log_ac}{log_{ab}c}=log_ac.log_c\left(ab\right)=log_ac.\left(log_ca+log_cb\right)=log_ac.log_ca+log_ac.log_cb=\dfrac{log_ac}{log_ac}+\dfrac{log_cb}{log_ca}=1+log_ab\)
2. \(log_{ax}bx=\dfrac{log_abx}{log_aax}=\dfrac{log_ab+log_ax}{log_aa+log_ax}=\dfrac{log_ab+log_ax}{1+log_ax}\)
3. \(\dfrac{1}{log_ax}+\dfrac{1}{log_{a^2}x}+...+\dfrac{1}{log_{a^n}x}=log_xa+log_xa^2+...+log_xa^n\)
\(=log_xa+2log_xa+...+n.log_xa=log_xa+2log_xa+...+n.log_xa\)
\(=log_xa.\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2}log_xa=\dfrac{n\left(n+1\right)}{2.log_ax}\)
\(log_xa=\dfrac{1}{p};log_xb=\dfrac{1}{q}\)
\(log_xabc=\dfrac{1}{r}\Rightarrow log_xa+log_xb+log_xc=\dfrac{1}{r}\)
\(\Rightarrow\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{log_cx}=\dfrac{1}{r}\)
\(\Rightarrow\dfrac{1}{log_cx}=\dfrac{1}{r}-\dfrac{1}{p}-\dfrac{1}{q}=\dfrac{pq-pr-qr}{pqr}\)
\(\Rightarrow log_cx=\dfrac{pqr}{pq-pr-qr}\)
\(\Leftrightarrow log_{\frac{1}{3}}xy\le log_{\frac{1}{3}}\left(x+y^2\right)\)
\(\Rightarrow xy\ge x+y^2\) (do \(\frac{1}{3}< 1\))
\(\Rightarrow x\left(y-1\right)\ge y^2\) (\(y-1>0\) do
Nếu \(y\le1\Rightarrow\left\{{}\begin{matrix}VT\le0\\VP>0\end{matrix}\right.\) (vô lý)
\(\Rightarrow y>1\Rightarrow x\ge\frac{y^2}{y-1}\)
\(\Rightarrow P=2x+3y\ge\frac{2y^2}{y-1}+3y=5y+2+\frac{2}{y-1}\)
\(\Rightarrow P\ge5\left(y-1\right)+\frac{2}{y-1}+7\ge2\sqrt{\frac{10\left(y-1\right)}{y-1}}+7=7+2\sqrt{10}\)
\(P_{min}=7+2\sqrt{10}\) khi \(\left\{{}\begin{matrix}y=1+\frac{\sqrt{10}}{5}\\x=\frac{y^2}{y-1}=...\end{matrix}\right.\)
\(P=3log_{a^2b}a-\dfrac{3}{4}log_a2.log_2\left(\dfrac{a}{b}\right)\)
\(=\dfrac{3}{log_a\left(a^2b\right)}-\dfrac{3}{4.log_2a}.\left(log_2a-log_2b\right)\)
\(=\dfrac{3}{log_aa^2+log_ab}-\dfrac{3}{4.log_2a}.log_2a+\dfrac{3}{4}.\dfrac{log_2b}{log_2a}\)
\(=\dfrac{3}{2+3}-\dfrac{3}{4}+\dfrac{3}{4}.log_ab=\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{9}{4}=\dfrac{21}{10}\)
\(a;b>0\Rightarrow3a+2b+1>1\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\) đồng biến
Mà \(9a^2+b^2\ge2\sqrt{9a^2b^2}=6ab\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge2\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}log_{6ab+1}\left(3a+2b+1\right)=1\\3a=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6ab+1=3a+2b+1\\b=3a\end{matrix}\right.\)
\(\Rightarrow18a^2+1=3a+6a+1\)
\(\Leftrightarrow18a^2-9a=0\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\)
\(P=log_{\dfrac{\sqrt{a}}{b}}a+log_{\dfrac{\sqrt{a}}{b}}\sqrt[3]{b}=log_{\dfrac{\sqrt{a}}{b}}a+\dfrac{1}{3}log_{\dfrac{\sqrt{a}}{b}}b\)
\(=\dfrac{1}{log_a\dfrac{\sqrt{a}}{b}}+\dfrac{1}{3.log_b\dfrac{\sqrt{a}}{b}}=\dfrac{1}{log_a\sqrt{a}-log_ab}+\dfrac{1}{3\left(log_b\sqrt{a}-log_bb\right)}\)
\(=\dfrac{1}{\dfrac{1}{2}-2}+\dfrac{1}{3\left(\dfrac{1}{4}-1\right)}=-\dfrac{10}{9}\)
\(log_{a^2}\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}log_a\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}\left[log_aa^3-log_a\sqrt[5]{b^3}\right]=\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)\)
\(\Rightarrow\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)=3\)
\(\Rightarrow log_ab=-5\)
a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)
b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)
c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)
d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)
\(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)
\(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)
Từ giả thiết ta thấy tất cả các biểu thức đều xác định :
Ta có : \(\log_ax=1+\log_ax.\log_az\Leftrightarrow\log_ax=\frac{1}{1-\log_az}=\frac{1}{1-\log_a\frac{a}{z}}=\log_{\frac{a}{z}}z\)
Do đó \(\log_xa.\log_{\frac{a}{z}}z=1\)
Tương tự \(\log_ya.\log_{\frac{a}{x}}x=1\)
Hơn nữa, thay \(\log_ax=\frac{1}{1-\log_az}\) vào \(\log_ay=1+\log_ay.\log_ax\), ta được :
\(\log_ay=1+\frac{\log_ay}{1-\log_az}\Leftrightarrow1-\log_az=\frac{\log_ay}{\log_ay-1}\)
\(\Leftrightarrow\log_za=1+\log_ay.\log_az\)
Tương tự như trên ta cũng có :
\(\log_za.\log_{\frac{a}{y}}y=1\)
Từ đó suy ra :
\(A=\left(\log_{\frac{a}{x}}a.\log_ya\right)\left(\log_{\frac{a}{y}}a.\log_za\right)\left(\log_{\frac{a}{z}}a.\log_xa\right)=1\)