2021 AMC 12B Spring Problem 9

Attempt Problem 9 of the 2021 AMC 12B Spring below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

$0$

$1$

$\dfrac{5}{4}$

$2$

$\log_2 5$

Answer: D

Concepts:logarithm algebraic manipulation

Difficulty rating: 1520

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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