Đặt $a = \log_{2}3, b = \log_{5}3$. Hãy biểu diễn $\log_{6}45$ theo $a$ và $b$:

$a < 0,b > 0$

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

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

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

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

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

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

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

$P = \dfrac{{\sqrt 3 }}{3}$ 

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

$P = 3\sqrt 3$

$P = 27$

 ${\log _2}16 = {\log _3}81$

${\log _3}9 = 3$

${\log _4}16 = {\log _2}8$    

${\log _2}4 = {\log _3}6$

${2^{{{\log }_2}3}} = {5^{{{\log }_3}5}}$  

${2^{{{\log }_2}3}} = {5^{{{\log }_5}3}}$

${5^{{{\log }_5}3}} = {\log _2}3$

${2^{{{\log }_2}4}} = 2$

${\log _a}\left( {bc} \right) = {\log _a}b + {\log _b}c$

${\log _a}\dfrac{b}{c} = {\log _a}b + {\log _a}c$

${\log _a}\dfrac{b}{c} = \dfrac{{{{\log }_a}b}}{{{{\log }_a}c}}$      

${\log _a}\left( {bc} \right) = {\log _a}b + {\log _a}c$

$\log_{a}b < 1 < \log_{b}a$    

$1 < \log_{a}b < \log_{b}a$  

$\log_{b}a < \log_{a}b < 1$

$\log_{b}a < 1 < \log_{a}b$