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

(→‎Solution 2)
(→‎Solution 2)
Line 13: Line 13:
  
 
==Solution 2==
 
==Solution 2==
−
<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
+
<math>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</math>
 
Rearranging it give us:
 
Rearranging it give us:
−
<cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</cmath>
+
<math>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</math>
−
<cmath>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</cmath>
+
<math>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</math>
 
let <math>\log_{2023}x</math> be <math>a</math>, we get
 
let <math>\log_{2023}x</math> be <math>a</math>, we get
−
<cmath>(\log_{2023}7+a)(\log_{2023}289+a)=1+a</cmath>
+
<math>(\log_{2023}7+a)(\log_{2023}289+a)=1+a</math>
−
<cmath>a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a</cmath>
+
<math>a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a</math>
−
<cmath>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</cmath>
+
<math>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</math>
 
by veita's formula,
 
by veita's formula,
−
<cmath>a_1+a_2=0</cmath>
+
<math>a_1+a_2=0</math>
−
<cmath>\log_{2023}{x_1}+\log_{2023}{x_2}=0</cmath>
+
<math>\log_{2023}{x_1}+\log_{2023}{x_2}=0</math>
−
<cmath>\log_{2023}{x_1x_2}=0</cmath>
+
<math>\log_{2023}{x_1x_2}=0</math>
−
<cmath>x_1x_2=\boxed{\textbf{(C)} 1}</cmath>
+
<math>x_1x_2=\boxed{\textbf{(C)} 1}</math>
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2023|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2023|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:37, 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

Solution 2

$\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$ Rearranging it give us: $\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x$ $(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)$ let $\log_{2023}x$ be $a$, we get $(\log_{2023}7+a)(\log_{2023}289+a)=1+a$ $a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a$ $a^2+\log_{2023}7 \cdot \log_{2023}289-1=0$ by veita's formula, $a_1+a_2=0$ $\log_{2023}{x_1}+\log_{2023}{x_2}=0$ $\log_{2023}{x_1x_2}=0$ $x_1x_2=\boxed{\textbf{(C)} 1}$

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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