# Difference between revisions of "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

From the Change of Base Formula, we have $$\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.$$

## Solution 2

Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get \begin{align*} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{\textbf{(C) } 6}. \end{align*}

~ pi_is_3.14

## See Also

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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