2021 AMC 12B Problems/Problem 9

Problem

What is the value of\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\]$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5$

Solution

\[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\]

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 $2\implies\boxed{\text{(D)}}$.

~ SoySoy4444


Video Solution (Just 2 min!)

https://youtu.be/jQaB8k9lZNc

~Education, the Study of Everything

Video Solution by Punxsutawney Phil

https://youtu.be/yxt8-rUUosI&t=157s

Video Solution by OmegaLearn (Logarithmic Manipulation)

https://youtu.be/4_dUQu0a9ZA

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by TheBeautyofMath

https://youtu.be/kuZXQYHycdk?t=925

~IceMatrix

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AMC 12 Problems and Solutions

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