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Difference between revisions of "2023 AMC 12A Problems/Problem 19"
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==Solution 1==
 
==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>.
+
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 Vieta, 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>.
  
 
~plasta
 
~plasta

Revision as of 20:44, 9 November 2023

Problem

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

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


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 Vieta, 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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