Difference between revisions of "2021 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | < | + | <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> | ||
| − | < | + | <cmath>= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}</cmath> |
| − | < | + | <cmath>=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}</cmath> |
| − | < | + | <cmath>=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}</cmath> |
| − | Expanding, < | + | 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>. | ||
| Line 19: | Line 19: | ||
~ SoySoy4444 | ~ SoySoy4444 | ||
| + | |||
| + | ==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) == | == Video Solution by OmegaLearn (Logarithmic Manipulation) == | ||
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~ pi_is_3.14 | ~ 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== | ==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
Contents
Problem
What is the value of![\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\]](http://latex.artofproblemsolving.com/8/a/9/8a93eb4b50cbbb0b051fa1401e2e58d5ab02d33e.png)
Solution
![\[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\]](http://latex.artofproblemsolving.com/e/c/a/ecaa739cc3d887ba1766dc6a6e1c4f7ff7eba881.png)
Note that 
, and similarly 
![\[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\]](http://latex.artofproblemsolving.com/8/e/a/8eae9612f64d6e8ddf97686a8e74c8309b6d6ef7.png)
![\[=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}{20})\log_{2}{20}\]](http://latex.artofproblemsolving.com/7/3/6/736771a0eba460514a82ed669ee1b1d9855e7047.png)
![\[=(2+\log_{2}{20})(1+\log_{2}{20})-(3+\log_{2}{20})\log_{2}{20}\]](http://latex.artofproblemsolving.com/4/b/2/4b21abc78fc7647612ff5c9ecfd62e29c9773171.png)
Expanding, ![\[2+2\log_{2}{20}+\log_{2}{20}+(\log_{2}{20})^2-3\log_{2}{20}-(\log_{2}{20})^2\]](http://latex.artofproblemsolving.com/b/e/2/be239702516f13b4dd8831ce1f08e672926ba2ca.png)
All the log terms cancel, so the answer is 
.
~ SoySoy4444
Video Solution (Just 2 min!)
~Education, the Study of Everything
Video Solution by Punxsutawney Phil
https://youtu.be/yxt8-rUUosI&t=157s
Video Solution by OmegaLearn (Logarithmic Manipulation)
~ 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 (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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
