Question

−2,9⋅(lg16+lg81)/2lg2+2lg3

Answer 1

$\frac{-2,9*(lg16+lg81)}{2lg2+2lg3}=\frac{-2,9*lg(16*81)}{lg2^{2}+lg3^{2}}=\frac{-2,9*lg36^{2}}{lg(4*9)}=\frac{-2,9*2lg36}{lg36} =-2,9*2=-5,8$