2018 AMC 12B Problems/Problem 7

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 $\log _{3} 7 \cdot \log _{5} 9 \cdot \cdot \cdot \log_{23} 27 = (\log _{3} 7 \cdot \log _{7} 11 \cdot \cdot \cdot \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdot \cdot \cdot \log _{21} 25) = \log _{3} 27 \cdot \log _{5} 25 = 3 \cdot 2 = 6$ .

Solution 3

When the entire product is rewritten using change of base, the fractions telescope down to $\frac{\log 25 \log 27}{\log 3 \log 5} = \log_327\cdot \log_525 = \boxed{6}$

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also