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

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== Video Solution by OmegaLearn (Logarithmic Manipulation) ==
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https://youtu.be/4_dUQu0a9ZA
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~ pi_is_3.14
  
 
==See Also==
 
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Revision as of 22:21, 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


Video Solution by OmegaLearn (Logarithmic Manipulation)

https://youtu.be/4_dUQu0a9ZA

~ pi_is_3.14

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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