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).

9.

What is the value of the following expression? log⁡280log⁡402−log⁡2160log⁡202\dfrac{\log_2 80}{\log_{40}2}-\dfrac{\log_2 160}{\log_{20}2}
00
11
54\dfrac{5}{4}
22
log⁡25\log_2 5
Answer: D
Concepts:logarithmalgebraic manipulation
Small hint:
1log⁡402=log⁡240,\dfrac{1}{\log_{40}2}=\log_2 40, and similarly for the other reciprocal
Big hint:
Write everything in terms of t=log⁡25,t=\log_2 5, since log⁡280=4+t,\log_2 80=4+t, etc.
Solution:
Using 1log⁡402=log⁡240\dfrac{1}{\log_{40}2}=\log_2 40 and 1log⁡202=log⁡220,\dfrac{1}{\log_{20}2}=\log_2 20, the expression becomes (log⁡280)(log⁡240)(\log_2 80)(\log_2 40) −(log⁡2160)(log⁡220).-(\log_2 160)(\log_2 20).
Let t=log⁡25.t=\log_2 5. Then log⁡280=4+t,\log_2 80=4+t, log⁡240=3+t,\log_2 40=3+t, log⁡2160=5+t,\log_2 160=5+t, log⁡220=2+t.\log_2 20=2+t.
The value is (4+t)(3+t)(4+t)(3+t) −(5+t)(2+t)-(5+t)(2+t) =(12+7t+t2)=(12+7t+t^2) −(10+7t+t2)-(10+7t+t^2) =2.=2.
Thus, the correct answer is D.
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