Resposta :
a)
[tex]\log_518=\\\\\log_5(2\cdot3^2)=\\\\\log_52+\log_53^2=\\\\\log_52+2\cdot\log_53=\\\\0,43+2\cdot0,68=\\\\0,43+1,36=\\\\\boxed{1,79}[/tex]
b) Passemos para a base 5;
[tex]\log_29=\\\\\frac{\log_59}{\log_52}=\\\\\frac{\log_53^2}{\log_52}=\\\\\frac{2\cdot\log_53}{\log_52}=\\\\\frac{2\cdot0,68}{0,43}=\\\\\frac{1,36}{0,43}=\\\\\boxed{3,16}[/tex]
[tex]\log_518=\\\\\log_5(2\cdot3^2)=\\\\\log_52+\log_53^2=\\\\\log_52+2\cdot\log_53=\\\\0,43+2\cdot0,68=\\\\0,43+1,36=\\\\\boxed{1,79}[/tex]
b) Passemos para a base 5;
[tex]\log_29=\\\\\frac{\log_59}{\log_52}=\\\\\frac{\log_53^2}{\log_52}=\\\\\frac{2\cdot\log_53}{\log_52}=\\\\\frac{2\cdot0,68}{0,43}=\\\\\frac{1,36}{0,43}=\\\\\boxed{3,16}[/tex]