Difference between revisions of "2018 AMC 12B Problems/Problem 7"
(The "Solution 3" that was added was identical to Solution 1 . . .) |
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== Solution 1== | == 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> | + | 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 == | == Solution 2 == |
Revision as of 02:08, 19 February 2018
Contents
Problem
What is the value of
Solution 1
Change of base makes this
Solution 2
Using the chain rule for logarithms (), we get .
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 |
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