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Difference between revisions of "2023 AMC 12A Problems/Problem 19"

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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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==Problem==
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What is the product of all solutions to the equation
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<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
  
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-Jonathan Yu
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<math>\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1\qquad\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2</math>
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==Solution 1==
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For <math>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</math>, transform it into <math>\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}</math>. Replace <math>\ln x</math> with <math>y</math>. Because we want to find the product of all solutions of <math>x</math>, it is equivalent to finding the sum of all solutions of <math>y</math>. Change the equation to standard quadratic equation form, the term with 1 power of <math>y</math> is canceled.  By using Veita, we see that since there does not exist a <math>by</math> term, <math>\sum y=0</math> and <math>\prod x=e^0=\boxed{\textbf{(C)} 1}</math>.
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~plasta
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==See also==
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{{AMC12 box|year=2023|ab=A|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 19:32, 9 November 2023

Problem

What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]

$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1\qquad\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$


Solution 1

For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$, transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$. Replace $\ln x$ with $y$. Because we want to find the product of all solutions of $x$, it is equivalent to finding the sum of all solutions of $y$. Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Veita, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{\textbf{(C)} 1}$.

~plasta

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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