Difference between revisions of "2021 AMC 12B Problems/Problem 9"

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(Video Solution (Just 2 min!))
 
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==Solution==
 
==Solution==
<math>\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}</math>
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<cmath>\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}</cmath>
  
 
Note that <math>\log_{40}{2}=\frac{1}{\log_{2}{40}}</math>, and similarly <math>\log_{20}{2}=\frac{1}{\log_{2}{20}}</math>
 
Note that <math>\log_{40}{2}=\frac{1}{\log_{2}{40}}</math>, and similarly <math>\log_{20}{2}=\frac{1}{\log_{2}{20}}</math>
  
<math>= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot log_{2}{20}</math>
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<cmath>= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}</cmath>
  
<math>=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}</math>
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<cmath>=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}</cmath>
  
<math>=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}</math>
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<cmath>=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}</cmath>
  
Expanding, <math>2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{80})^2-3\log_{2}{20}-(\log_{2}{20})^2</math>
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Expanding, <cmath>2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2</cmath>
  
 
All the log terms cancel, so the answer is <math>2\implies\boxed{\text{(D)}}</math>.
 
All the log terms cancel, so the answer is <math>2\implies\boxed{\text{(D)}}</math>.
  
 
~ SoySoy4444
 
~ SoySoy4444
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 +
 +
==Video Solution (Just 2 min!)==
 +
https://youtu.be/jQaB8k9lZNc
 +
 +
<i>~Education, the Study of Everything</i>
 +
 +
==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
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 +
==Video Solution by Hawk Math==
 +
https://www.youtube.com/watch?v=VzwxbsuSQ80
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==Video Solution by TheBeautyofMath==
 +
https://youtu.be/kuZXQYHycdk?t=925
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2021|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:41, 19 July 2023

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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