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

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

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.

@@ Line 1: / Line 1: @@
 == 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>
-<cmath> \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27? </cmath>
+== Solution 1 ==
-<math>\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10 </math>
+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>
-== Solution 1==
+== Video Solution by OmegaLearn ==
+https://youtu.be/RdIIEhsbZKw?t=605
-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)
+~ pi_is_3.14
-== Solution 2 ==
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/FmBU-BT89H4
-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>.
+~Education, the Study of Everything
 ==See Also==
@@ Line 17: / Line 29: @@
 {{AMC12 box|year=2018|ab=B|num-b=6|num-a=8}}
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

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