Ta có : 

\(a=\log_{12}18=\frac{\log_218}{\log_212}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+2\log_23}{2+\log_23}\)

\(\Rightarrow a\left(a+\log_23\right)=1+2\log_23\Leftrightarrow\log_23=\frac{1-2a}{a-2}\left(1\right)\)

\(b=\log_{24}54=\frac{\log_254}{\log_224}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+3\log_23}{3+\log_23}\)

\(\Rightarrow b\left(3+\log_23\right)=1+3\log_23\Leftrightarrow\log_23=\frac{1-3b}{b-3}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{1-2a}{a-2}=\frac{1-3b}{b-3}\Leftrightarrow\left(1-2a\right)\left(b-3\right)=\left(1-3b\right)\left(a-2\right)\)

                                           \(\Leftrightarrow ab+5\left(a-b\right)=1\Rightarrow\) Điều phải chứng minh

Ta có : \(\log_a^2\frac{b}{c}=\left(\log_a\frac{b}{c}\right)^2=\left[\log_a\left(\frac{c}{b}\right)^{-1}\right]^2=\left(-\log_a\frac{c}{b}\right)^2=\left(\log_a\frac{c}{b}\right)^2=\log_a^2\frac{c}{b}\)

=> Điều phải chứng minh

Ta có : \(\log_{\frac{a}{b}}^2\frac{c}{b}=\log_{\frac{a}{b}}^2\frac{b}{c};\log_{\frac{b}{c}}^2\frac{a}{c}=\log_{\frac{b}{c}}^2\frac{c}{a};\log_{\frac{c}{a}}^2\frac{b}{a}=\log_{\frac{c}{a}}^2\frac{a}{b}\)

\(\Rightarrow\log_{\frac{a}{b}}^2\frac{c}{b}.\log_{\frac{b}{c}}^2\frac{a}{c}.\log_{\frac{c}{a}}^2\frac{b}{c}=\log_{\frac{a}{b}}^2\frac{c}{b}.\log^2_{\frac{b}{c}}\frac{c}{a}\log_{\frac{c}{a}}^2\frac{a}{b}=\left(\log_{\frac{a}{b}}\frac{c}{b}.\log_{\frac{b}{c}}\frac{c}{a}\log_{\frac{c}{a}}\frac{a}{b}\right)^2=1^2=1\)

\(\Rightarrow\) Trong 3 số không âm \(\log_{\frac{a}{b}}^2\frac{c}{b};\log^2_{\frac{b}{c}}\frac{c}{a};\log_{\frac{c}{a}}^2\frac{a}{b}\) luôn có ít nhất 1 số lớn hơn 1

\(A=\log_3\left(\log_{2\sqrt{2}}\sqrt[3]{\sqrt{2}}\right)=\log_3\left(\log_{2^{\frac{3}{2}}}2^{\frac{1}{6}}\right)=\log_3\left(\frac{1}{6}.\frac{2}{3}\right)=\log_33^{-2}=-2\)

\(B=\log_{\sqrt{6}}3\log_336=\log_{\sqrt{6}}36=\log_{6^{\frac{1}{2}}}6^2=4\)

\(N=\log_{\frac{1}{3}}5\log_{25}\frac{1}{7}=\log_{3^{-1}}5\log_{5^5}3^{-3}=\left(-5\right)\left(-\frac{3}{2}\right).\log_35\log_53=\frac{15}{2}\)

\(D=\left(\sqrt[3]{9}\right)^{\frac{2}{2\log_53}}=\left(3^{\frac{2}{3}}\right)^{\frac{3\log_35}{2}}=3^{\log_35}=5\)

\(B=25^{\frac{1}{2}+\frac{1}{9}\log_{\frac{1}{2}}27+\log_{125}81}=\left(5^2\right)^{\frac{1}{2}+\frac{1}{9}\log_{5^{-1}}3^3+\log_{5^3}3^4}\)

   \(=5^{1-\frac{2}{3}\log_53+\frac{8}{3}\log_53}=5^{1+2\log_53}=5.5^{\log_53^2}=5.9=45\)

\(F=\log_{3-2\sqrt{2}}\left(27^{\log_92}+2^{\log_827}\right)=\log_{3-2\sqrt{2}}\left[\left(3^3\right)^{^{\log_92^2}}+2^{\log_{2^3}3^3}\right]\)

   \(=\log_{3-2\sqrt{2}}\left(3^{\frac{3}{2}\log_32}+2^{\log_23}\right)\)

   \(=\log_{3-2\sqrt{2}}\left(3^{\log_32^{\frac{3}{2}}}+2^{\log_23}\right)\)

   \(=\log_{3-2\sqrt{2}}\left(2^{\frac{3}{2}}+3\right)=\log_{\left(3-2\sqrt{2}\right)^{-1}}\left(3-2\sqrt{2}\right)=-1\)

Trong hệ SI đơn vị đo cường độ âm là

A. Jun trên mét vuông J/m².

B. Đêxiben dB.

C. Ben B.

D. Oát trên mét vuông W/m².