Question

raspunsul este C

Multumesc

Answer 1

 $a= \log_{6}2 \\ \\ \\ \log_{3}12 = \dfrac{\log_{6}12}{\log_{6}3} = \dfrac{\log_{6}(6\cdot 2)}{\log_{6}\frac{6}{2}} = \dfrac{\log_6 6+\log_{6}2}{\log_6{6}-\log_{6}2} = \boxed{\dfrac{1+a}{1-a}}$