Difference between revisions of "2018 AMC 12B Problems/Problem 7"

Revision as of 20:03, 2 February 2022

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 $\begin{align*} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{\textbf{(C) } 6}. \end{align*}$

Solution 3

Using the Change of Base Formula, we have \frac{\log(7) \cdot \log(9)\cdot \log(11) ... \log(25) \cdot \log(27)}{\log(3) \cdot \log(5)\cdot \log(7)...\log(21) \cdot \log(23)}

From this, we realize that the top logs from \log(7) - \log(23) cancel with the bottom logs from \log(7) - \log(23). This leaves us with \frac{\log(25) \cdot \log(27)}{\log(3) \cdot \log(5). Going from this step, we realize that we can use the communicative property to achieve \frac{\log(25) \cdot \log(27)}{\log(5) \cdot \log(3).

Therefore, we can use the Change of Base Formula to get back to \log_525 \cdot log_327. This equals to 2 \cdot 5 = \boxed{\textbf{(C) } 6}

Video Solution

https://youtu.be/RdIIEhsbZKw?t=605

~ pi_is_3.14

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