2018 AMC 12B Problems/Problem 7: Difference between revisions
The "Solution 3" that was added was identical to Solution 1 . . . |
|||
| Line 7: | Line 7: | ||
== 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 01:08, 19 February 2018
Problem
What is the value of
![\[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\]](http://latex.artofproblemsolving.com/1/0/8/108141fcccb63d051a626897ded91e10882ab0db.png)
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
