Difference between revisions of "2018 AMC 12B Problems/Problem 7"
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<math>\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 </math> | <math>\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 </math> | ||
− | == Solution 1== | + | == Solutions == |
+ | === Solution 1 === | ||
+ | Change of base makes this <math>\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \log 27}{\log 3 \log 5} = \log_327\cdot \log_525 = \boxed{6}</math> | ||
− | + | === Solution 2 === | |
+ | Using the chain rule for logarithms (<math>\log _{a} b \cdot \log _{b} c = \log _{a} c</math>), we get <math>\log _{3} 7 \cdot \log _{5} 9 \cdot \cdot \cdot \log_{23} 27 = (\log _{3} 7 \cdot \log _{7} 11 \cdot \cdot \cdot \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdot \cdot \cdot \log _{21} 25) = \log _{3} 27 \cdot \log _{5} 25 = 3 \cdot 2 = 6</math>. | ||
− | == Solution | + | == Video Solution == |
+ | https://youtu.be/RdIIEhsbZKw?t=605 | ||
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2018|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2018|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 04:28, 21 January 2021
Problem
What is the value of
Solutions
Solution 1
Change of base makes this
Solution 2
Using the chain rule for logarithms (), we get .
Video Solution
https://youtu.be/RdIIEhsbZKw?t=605
~ pi_is_3.14
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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.