2021 AMC 12B Spring Problem 9

Below is the professionally curated solution for Problem 9 of the 2021 AMC 12B Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

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Concepts:logarithm algebraic manipulation

Difficulty rating: 1520

What is the value of $\dfrac{\log_2 80}{\log_{40}2}-\dfrac{\log_2 160}{\log_{20}2}?$

$0$

$1$

$\dfrac{5}{4}$

$2$

$\log_2 5$

Solution:

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.

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