Question

$Log_{2}16$ - $Log_{4}32$ e igual a :

Deixem calculos bem explicativos por favor 

Answer 1

$log_2{16}-log_4{32}=log_2{2^4}-\frac{log_2{32}}{log_24}=\\\\ =log_2{2^4}-\frac{log_2{2^5}}{log_2{2^2}}=\\\\ =4log_22-\frac{5log_22}{2log_22}=\\\\ =4.1-\frac{5.1}{2.1}=4-\frac{5}{2}=\frac{2.4-5}{2}=\frac{8-5}{2}=\frac{3}{2}$