First post
Sabendo que log6^{11} =1,34 e log6^{2} =0,37 20210416_181112.jpg ,calcule:
A) log22^{3}
B) log6^{4\sqrt{11}}
C) log\sqrt{2}^{22}
20210416_181600.jpg
Última msg
Harison ,
a) \log_{22}3
=\frac{\log_63}{\log_622}
=\frac{\log_6(\frac{6}{2})}{\log_6(2.11)}
=\frac{\log_66-\log_62}{\log_62+\log_611}
=\frac{1-0,37}{0,37+1,34}
{\color{red}\boxed{≈0,37}}...