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

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\end{align*}</cmath>
 
\end{align*}</cmath>
  
== Solution 3 ==
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== Video Solution by OmegaLearn ==
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)}
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https://youtu.be/RdIIEhsbZKw?t=605
  
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).  
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~ pi_is_3.14
  
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}
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/FmBU-BT89H4
  
== Video Solution ==
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~Education, the Study of Everything
https://youtu.be/RdIIEhsbZKw?t=605
 
 
 
~ pi_is_3.14
 
  
 
==See Also==
 
==See Also==

Latest revision as of 01:35, 28 May 2023

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*}

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/FmBU-BT89H4

~Education, the Study of Everything

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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