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

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<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
 
<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
  
−
<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</math>
+
<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>
−
<math>\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2</math>
 
−
 
 
  
 
==Solution 1==
 
==Solution 1==
−
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>.
+
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 exponential of 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
  
−
==Solution 2==
+
==Solution 2 (Same idea as Solution 1 with easily understand steps)==
  
−
<math>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</math>
+
<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
  
 
Rearranging it give us:
 
Rearranging it give us:
  
−
<math>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</math>
+
<cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</cmath>
  
−
<math>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</math>
+
<cmath>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</cmath>
  
 
let <math>\log_{2023}x</math> be <math>a</math>, we get
 
let <math>\log_{2023}x</math> be <math>a</math>, we get
  
−
<math>(\log_{2023}7+a)(\log_{2023}289+a)=1+a</math>
+
<cmath>(\log_{2023}7+a)(\log_{2023}289+a)=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+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a</cmath>
  
−
<math>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</math>
+
<cmath>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</cmath>
  
−
by veita's formula,
+
by Vieta's Formulas,
  
−
<math>a_1+a_2=0</math>
+
<cmath>a_1+a_2=0</cmath>
  
−
<math>\log_{2023}{x_1}+\log_{2023}{x_2}=0</math>
+
<cmath>\log_{2023}{x_1}+\log_{2023}{x_2}=0</cmath>
  
−
<math>\log_{2023}{x_1x_2}=0</math>
+
<cmath>\log_{2023}{x_1x_2}=0</cmath>
  
−
<math>x_1x_2=\boxed{\textbf{(C)} 1}</math>
+
<cmath>x_1x_2=\boxed{\textbf{(C)} 1}</cmath>
  
 
~lptoggled
 
~lptoggled
 +
 +
==Solution 3==
 +
Similar to solution 1, change the bases first
 +
<cmath>\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}</cmath>
 +
Cancel and cross multiply to get
 +
<cmath>(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})</cmath>
 +
Simplify to get
 +
<cmath>(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2</cmath>
 +
<cmath>\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}</cmath>
 +
The sum of all possible <math>\ln{x}</math> is 0, thus the product of all solutions of <math>x</math> is <math>\boxed{\textbf{(C)} 1}</math>
 +
 +
~dwarf_marshmallow
 +
 +
==Solution 4==
 +
We take the reciprocal of both sides:
 +
<cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath> Using logarithm properties, we have <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath> Simplify to obtain <cmath>2023x^2=2023x,</cmath> from which we have <math>x=\boxed{\textbf{(C)} 1}</math>
 +
 +
~MLiang2018
 +
 +
==Video Solution 1 by OmegaLearn==
 +
https://youtu.be/OcNU62SMh4o
 +
 +
==Video Solution==
 +
 +
https://youtu.be/-CZkFE-wriQ
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==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 17:17, 23 June 2024

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 exponential of 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 (Same idea as Solution 1 with easily understand steps)

\[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]

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 Vieta's Formulas,

\[a_1+a_2=0\]

\[\log_{2023}{x_1}+\log_{2023}{x_2}=0\]

\[\log_{2023}{x_1x_2}=0\]

\[x_1x_2=\boxed{\textbf{(C)} 1}\]

~lptoggled

Solution 3

Similar to solution 1, change the bases first \[\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}\] Cancel and cross multiply to get \[(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\] Simplify to get \[(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2\] \[\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}\] The sum of all possible $\ln{x}$ is 0, thus the product of all solutions of $x$ is $\boxed{\textbf{(C)} 1}$

~dwarf_marshmallow

Solution 4

We take the reciprocal of both sides: \[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\] Using logarithm properties, we have \[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\] Simplify to obtain \[2023x^2=2023x,\] from which we have $x=\boxed{\textbf{(C)} 1}$

~MLiang2018

Video Solution 1 by OmegaLearn

https://youtu.be/OcNU62SMh4o

Video Solution

https://youtu.be/-CZkFE-wriQ

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

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

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