Question

log6(3!)-log3 1-2log2 1 pe 4

Answer 1

 

$\displaystyle\bf\\log_6(3!)-log_3(1)-2log_2\bigg(\frac{1}{4}\bigg)=\\\\=log_6(1\times2\times3)-log_3(1)-2log_2\bigg(\frac{1}{2^2}\bigg)=\\\\=log_6(6)-log_3(1)-2log_2(2^{-2})=\\\\=1-0-2\times(-2)=\\\\=1+4=\boxed{\bf5}$