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\(N=\log_{\frac{1}{3}}5\log_{25}\frac{1}{7}=\log_{3^{-1}}5\log_{5^5}3^{-3}=\left(-5\right)\left(-\frac{3}{2}\right).\log_35\log_53=\frac{15}{2}\)
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 \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)
\(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)
\(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)
\(D=\log_{5^{-1}}\left(5^2\right)-3\log_{3^2}\left(3^{-1}\right)+4.\log_{2^{\frac{3}{2}}}2^6=-2+\frac{3}{2}+16=\frac{31}{2}\)
\(E=16\left[\log_{3^{-2}}3^{\frac{3}{2}}\right]^2+23\log_{2^{\frac{9}{2}}}2^{\frac{5}{2}}-12\log_55^{-3}=16\left(-\frac{3}{4}\right)^2+9\frac{5}{9}-12\left(-3\right)=50\)
Ta có :
\(\begin{cases}5>1;3>1\Rightarrow\log_53>0\\15>1;4>1\Rightarrow\log_{15}4>0\\0< \frac{1}{3}< 1;\frac{7}{2}>1\Rightarrow\log_{\frac{1}{3}}\frac{14}{5}< 0\\0< 0,3< 1;\frac{7}{2}>1\Rightarrow\log_{0,3}\frac{7}{2}< 0\end{cases}\)
\(\Rightarrow A=\frac{\log_53.\log_{15}4}{\log_{\frac{1}{3}}\frac{14}{5}\log_{0,3}\frac{7}{2}}>0\)
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 \(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}=\frac{n\left(n+1\right)}{\log_ax}\)
\(\Leftrightarrow log_{\frac{1}{3}}xy\le log_{\frac{1}{3}}\left(x+y^2\right)\)
\(\Rightarrow xy\ge x+y^2\) (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 \(\left\{{}\begin{matrix}y=1+\frac{\sqrt{10}}{5}\\x=\frac{y^2}{y-1}=...\end{matrix}\right.\)
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 :
\(A=\left(\log_{\frac{a}{x}}a.\log_ya\right)\left(\log_{\frac{a}{y}}a.\log_za\right)\left(\log_{\frac{a}{z}}a.\log_xa\right)=1\)
Ta có :
\(\log_62-\frac{1}{2}\log_{\sqrt{6}}5=\log_62-\log_65=\log_6\frac{2}{5}\)
\(\Rightarrow\left(\frac{1}{6}\right)^{\log_62-\frac{1}{2}\log_{\sqrt{6}}5}=\left(\frac{1}{6}\right)^{\log_6\frac{2}{5}}=\left(6^{-1}\right)^{\log_6\frac{2}{5}}=6^{\log_6\frac{2}{5}}=\frac{5}{2}=\sqrt[3]{\left(\frac{5}{2}\right)^3}=\sqrt[3]{\frac{125}{8}}\)
Mà :
\(\sqrt[3]{\frac{125}{8}}>\sqrt[3]{\frac{124}{8}}\Rightarrow\left(\frac{1}{6}\right)^{\log_62-\frac{1}{2}\log_{\sqrt{6}}5}>\sqrt[3]{\frac{31}{2}}\)
\(\Rightarrow B=\left(\frac{1}{6}\right)^{\log_62-\frac{1}{2}\log_{\sqrt{6}}5}-\sqrt[3]{\frac{31}{2}}>0^{ }\)
\(B=25^{\frac{1}{2}+\frac{1}{9}\log_{\frac{1}{2}}27+\log_{125}81}=\left(5^2\right)^{\frac{1}{2}+\frac{1}{9}\log_{5^{-1}}3^3+\log_{5^3}3^4}\)
\(=5^{1-\frac{2}{3}\log_53+\frac{8}{3}\log_53}=5^{1+2\log_53}=5.5^{\log_53^2}=5.9=45\)