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 \ \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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<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>
  
 
==Solution 1==
 
==Solution 1==

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 (ProblemsAnswer KeyResources)
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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