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

(The "Solution 3" that was added was identical to Solution 1 . . .)
(Sol 1 already talked about Change of Base, although it used complicated notation. I hope it is ok to delete a repetitive solution. Sorry ....)
 
(9 intermediate revisions by 6 users not shown)
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== Problem ==
 
== Problem ==
What is the value of
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What is the value of <cmath> \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? </cmath>
 
 
<cmath> \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? </cmath>
 
 
<math>\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 </math>
 
<math>\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 </math>
  
== Solution 1==
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== Solution 1 ==
 +
From the Change of Base Formula, we have <cmath>\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}.</cmath>
  
Change of base makes this <math>\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}</math> (mathguy623)
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== Solution 2 ==
 +
Using the chain rule of logarithms <math>\log _{a} b \cdot \log _{b} c = \log _{a} c,</math> we get
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<cmath>\begin{align*}
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\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 \\
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&= 3 \cdot 2 \\
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&= \boxed{\textbf{(C) } 6}.
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\end{align*}</cmath>
  
== Solution 2 ==
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== Video Solution ==
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https://youtu.be/RdIIEhsbZKw?t=605
  
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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~ pi_is_3.14
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2018|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2018|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 00:57, 3 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*}

Video Solution

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

~ pi_is_3.14

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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