Đặt ${\log _2}60 = a;{\log _5}15 = b.$ Tính $P = {\log _2}12$ theo $a$ và $b$.

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{2} + \dfrac{1}{2}{\log _a}b$

${\log _{{a^2}}}\left( {ab} \right) = 2 + {\log _a}b$

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{4}{\log _a}b$

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{2}{\log _a}b$

$P = {\log _2}{\left( {\dfrac{a}{b}} \right)^2}$       

$P = {\log _2}\left( {\dfrac{{2a}}{{{b^2}}}} \right)$        

$P = {\log _2}\left( {2a{b^2}} \right)$           

$P = {\log _2}{\left( {ab} \right)^2}$

$P = \dfrac{{\sqrt 3 }}{3}$ 

$P = \dfrac{1}{3}$

$P = 3\sqrt 3 $

$P = 27$

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{2}{\log _a}b$           

${\log _{{a^2}}}\left( {ab} \right) = 2 + 2{\log _a}b$

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{4}{\log _a}b$                       

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{2} + \dfrac{1}{2}{\log _a}b$

\(2\log a + \log b\)

$\log a + 2\log b$

$2\left( {\log a + \log b} \right)$

$\log a + \dfrac{1}{2}\log b$

$0$

$1$

$\ln (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a})$

$\ln (abcd)$

\({\log _{0,5}}a > {\log _{0,5}}b \Leftrightarrow a > b > 0\)

\(\log x < 0 \Leftrightarrow 0 < x < 1\)

\({\log _2}x > 0 \Leftrightarrow x > 1\)

\({\log _{\dfrac{1}{3}}}a = {\log _{\dfrac{1}{3}}}b \Leftrightarrow a = b > 0\)