Nếu $\log _{12}6 = a; \log _{12} 7 = b$ thì:

${\log _a}b$  

${\log _b}a$

${\ln _a}b$

${\ln _b}a$

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

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

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

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

$a < 0,b > 0$

$0 < a \ne 1,b < 0$

$0 < a \ne 1,b > 0$

$0 < a \ne 1,0 < b \ne 1$

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

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

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

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

${a^b} = N$  

${\log _a}N = b$       

${a^N} = b$

${b^N} = a$

$P = \dfrac{{\sqrt 3 }}{3}$ 

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

$P = 3\sqrt 3 $

$P = 27$

${\log _a}1 = 1$        

${\log _a}a = a$

${\log _a}1 = a$        

${\log _a}a = 1$