Treść:

Niech $L={\log }_{\sqrt{2}}2\cdot {\log }_2\sqrt{3}\cdot {\log }_{\sqrt{3}}4$   . Wtedy:

A. L = 1 B. L = 2 C. L = 3 D. L = 4

 

Rozwiązanie:

$L={\log }_{\sqrt{2}}2\cdot {\log }_2\sqrt{3}\cdot {\log }_{\sqrt{3}}4$  

 

Skorzystajmy ze wzoru na zamianę podstawy logarytmu:

${\log }_{\sqrt{2}}2=\frac{{\log }_{2}2}{{\log }_2\sqrt{2}}=\frac{1}{{\log }_2\sqrt{2}}=\frac{1}{{\log }_2{2}^{\frac{1}{2}}}=\frac{1}{\frac{1}{2}{\log }_{2}2}=\frac{1}{\frac{1}{2}}=2$  

 

${\log }_{\sqrt{3}}4=\frac{{\log }_{2}4}{{\log }_2\sqrt{3}}=\frac{{\log }_2{2}^2}{{\log }_2\sqrt{3}}=\frac{2{\log }_{2}2}{{\log }_2\sqrt{3}}=\frac{2}{{\log }_2\sqrt{3}}$       

 

$L={\log }_{\sqrt{2}}2\cdot {\log }_2\sqrt{3}\cdot {\log }_{\sqrt{3}}4=2\cdot \sout{{\log }_2\sqrt{3}}{}^1\cdot \frac{2}{\sout{{\log }_2\sqrt{3}}{}_1}=4$  

 

Odp.: D