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>. |
Revision as of 02:43, 13 February 2021
Contents
[hide]Problem
What is the value of
Solution
Note that , and similarly
Expanding,
All the log terms cancel, so the answer is .
~ SoySoy4444
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
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.