2021 AMC 12B Problem 9

Using $\dfrac{1}{\log_{40}2}=\log_2 40$ and $\dfrac{1}{\log_{20}2}=\log_2 20,$ the expression becomes $(\log_2 80)(\log_2 40)$ $-(\log_2 160)(\log_2 20).$ Let $t=\log_2 5.$ Then $\log_2 80=4+t,$ $\log_2 40=3+t,$ $\log_2 160=5+t,$ $\log_2 20=2+t.$ The value is $(4+t)(3+t)$ $-(5+t)(2+t)$ $=(12+7t+t^2)$ $-(10+7t+t^2)$ $=2.$ Thus, the correct answer is D.