smc

$$\frac{{\log }_{2}80}{{\log }_{40}2}-\frac{{\log }_{2}160}{{\log }_{20}2}$$ Note that ${\log }_{40}2=\frac{1}{{\log }_{2}40}$, and similarly ${\log }_{20}2=\frac{1}{{\log }_{2}20}$ $$={\log }_{2}80\cdot {\log }_{2}40-{\log }_{2}160\cdot {\log }_{2}20$$ $$=({\log }_{2}4+{\log }_{2}20)({\log }_{2}2+{\log }_{2}20)-({\log }_{2}8+{\log }_{2}20){\log }_{2}20$$ $$=(2+{\log }_{2}20)(1+{\log }_{2}20)-(3+{\log }_{2}20){\log }_{2}20$$ Expanding, $$2+2{\log }_{2}20+{\log }_{2}20+({\log }_{2}20{)}^2-3{\log }_{2}20-({\log }_{2}20{)}^2$$ All the log terms cancel, so the answer is $2⟹\boxed{2}$.