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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 19:30, 16 February 2018

Problem

What is the value of

\[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Solution 1

Change of base makes this $\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \log 27}{\log 3 \log 5} = \log_327\cdot \log_525 = \boxed{6}$ (mathguy623)

Solution 2

Using the chain rule for logarithms ($\log _{a} b \cdot \log _{b} c = \log _{a} c$), we get $\log _{3} 7 \cdot \log _{5} 9 \cdot \cdot \cdot \log_{23} 27 = (\log _{3} 7 \cdot \log _{7} 11 \cdot \cdot \cdot \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdot \cdot \cdot \log _{21} 25) = \log _{3} 27 \cdot \log _{5} 25 = 3 \cdot 2 = 6$.

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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