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

(Created page with "<math>\frac{log_{2}{80}}{log_{40}{2}}-\frac{log_{2}{160}}{log_{20}{2}}</math> Note that <math>log_{40}{2}=\frac{1}{log_{2}{40}}</math>, and similarly <math>log_{20}{2}=\frac{...")
 
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<math>\frac{log_{2}{80}}{log_{40}{2}}-\frac{log_{2}{160}}{log_{20}{2}}</math>
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==Problem==
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What is the value of<cmath>\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?</cmath><math>\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5</math>
  
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Note that <math>log_{40}{2}=\frac{1}{log_{2}{40}}</math>, and similarly <math>log_{20}{2}=\frac{1}{log_{2}{20}}</math>
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==Solution==
 +
<math>\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}</math>
  
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<math>= log_{2}{80}\cdot log_{2}{40}-log_{2}{160}\cdot log_{2}{20}</math>
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Note that <math>\log_{40}{2}=\frac{1}{\log_{2}{40}}</math>, and similarly <math>\log_{20}{2}=\frac{1}{\log_{2}{20}}</math>
  
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<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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<math>= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot log_{2}{20}</math>
  
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<math>=(2+log_{2}{20})(1+log_{2}{20})-(3+log_{2}{20})log_{2}{20}</math>
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<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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Expanding, <math>2+2log_{2}{20}+log_{2}{20}+(log_{2}{80})^2-3log_{2}{20}-(log_{2}{20})^2</math>
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<math>=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}</math>
  
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All the log terms cancel, so answer <math>=2</math>.
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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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All the log terms cancel, so the answer is <math>2\implies\boxed{\text{(D)}}</math>.
  
 
~ SoySoy4444
 
~ SoySoy4444
 +
 +
==See Also==
 +
{{AMC12 box|year=2021|ab=B|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Revision as of 20:43, 11 February 2021

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, $2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{80})^2-3\log_{2}{20}-(\log_{2}{20})^2$

All the log terms cancel, so the answer is $2\implies\boxed{\text{(D)}}$.

~ SoySoy4444

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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