(31g2-lg24):(lg3+lg27) помогите

Question

(31g2-lg24):(lg3+lg27)
помогите

Answer 1

(3lg2-lg24)/(lg3+lg27)=(lg2³-g24)/lg(3×27)=lg(8/24)/lg81=
lg(1/3)/lg81=lg₈₁3⁻¹=-(1/4)lg₃3=-1/4