Difference between revisions of "2018 AMC 12B Problems/Problem 7"
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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>. | 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>. | ||
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| + | == Solution 3 == | ||
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| + | When the entire product is rewritten using change of base, the fractions telescope down to <math>\frac{\log 25 \log 27}{\log 3 \log 5} = \log_327\cdot \log_525 = \boxed{6}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 19:42, 16 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 
(mathguy623)
Solution 2
Using the chain rule for logarithms (
), we get 
.
Solution 3
When the entire product is rewritten using change of base, the fractions telescope down to 
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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