Difference between revisions of "2018 AMC 12B Problems/Problem 7"
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== Problem == | == Problem == | ||
| − | What is the value of | + | 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> |
| + | <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 == | ||
| + | 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> | ||
| + | |||
| + | == Solution 2 == | ||
| + | Using the chain rule of logarithms <math>\log _{a} b \cdot \log _{b} c = \log _{a} c,</math> we get | ||
| + | <cmath>\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*}</cmath> | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/RdIIEhsbZKw?t=605 | ||
| − | + | ~ pi_is_3.14 | |
| − | |||
| − | == Solution == | + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
| + | https://youtu.be/FmBU-BT89H4 | ||
| − | + | ~Education, the Study of Everything | |
==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 01:35, 28 May 2023
Contents
Problem
What is the value of ![\[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\]](http://latex.artofproblemsolving.com/1/0/8/108141fcccb63d051a626897ded91e10882ab0db.png)
Solution 1
From the Change of Base Formula, we have ![\[\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.\]](http://latex.artofproblemsolving.com/3/d/0/3d0aba10cc169dfcc3b353ffe370b5a527574723.png)
Solution 2
Using the chain rule of logarithms 
we get
Video Solution by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=605
~ pi_is_3.14
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
