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

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Rearranging it give us:
 
Rearranging it give us:
  
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<cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</cmath>
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<cmath>3}289x=\log_{2023}2023x</cmath>
  
 
<cmath>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</cmath>
 
<cmath>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</cmath>
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~dwarf_marshmallow
 
~dwarf_marshmallow
  
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==Solution 4==
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==Solution 4(Fakesolve)==
 
We take the reciprocal of both sides:
 
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>
 
<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>
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~silk-hyacinth
 
~silk-hyacinth
  
−
==Solution 5(Similar to solution 4)==
+
==Solution 5(Similar to solution 4, Fakesolve)==
 
First, we take the reciprocal of both sides. We get <cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath>
 
First, we take the reciprocal of both sides. We get <cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath>
 
Flip the logarithms to get <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath>
 
Flip the logarithms to get <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath>
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~Yrock
 
~Yrock
 +
 +
==Solution 6==
 +
We can begin by changing the logarithms' base to <math>x</math> and simplifying:
 +
<math>\newline</math>
 +
<math>\frac{(\log_{x} 2023)^2}{\log_{x} 7 +1} \cdot \frac{1}{\log_{x} 289 + 1} = 1</math>
 +
<math>\newline</math>
 +
After that, we can multiply the denominator on the <math>\text{LHS}</math> on both sides, resulting in:
 +
<math>\newline</math>
 +
<math>(\log_{x} 2023)^2 = (\log_{x} 7 +1)(\log_{x} 289 +1)</math>
 +
<math>\newline</math>
 +
<math>(\log_{x} 2023)^2 = \log_{x} 2023 + \log_{x} 7 + \log_{x} 289 + 1</math>
 +
<math>\newline</math>
 +
<math>(\log_{x} 2023)^2 = 2\log_{x} 2023 + 1</math>
 +
<math>\newline</math>
 +
Now, we can substitute some variable <math>a</math> as <math>\log_{x} 2023</math>, giving the quadratic <math>a^2-2a-1=0</math>. By Vieta's, the product of the roots is <math>1</math>, thus the solution is <math>\boxed{\textbf{(C)} 1}</math>.
 +
 +
~cyberhacker
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==

Latest revision as of 08:18, 12 August 2025

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 understood steps)

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

Rearranging it give us: 

\[3}289x=\log_{2023}2023x\] (Error compiling LaTeX. Unknown error_msg)

\[(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)\]

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(Fakesolve)

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

This solution works for this problem by chance but do note that the simplification step to get \[2023x^2=2023x\]is not how log properties work and that the actual solutions for x are \[x = e^{ \pm \sqrt{ln^2(7)+ln(17)ln(2092529)}}\] (as shown in solution 3) which multiply to 1

~silk-hyacinth

Solution 5(Similar to solution 4, Fakesolve)

First, we take the reciprocal of both sides. We get \[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\] Flip the logarithms to get \[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\]

Now we can use $\log_{a}bc=\log_{a}b+\log_{a}c$. We get \[\log_{2023}7+\log_{2023}x+\log_{2023}289+\log_{2023}x=\log_{2023}2023+\log_{2023}x.\] The $\log_{2023}7+\log_{2023}289$ and $\log_{2023}2023$ terms cancel, giving \[2\cdot\log_{2023}x=\log_{2023}x\] so now we are sure that $\log_{2023}x=0$, so the only solution is $x=\boxed{\textbf{(C)} 1}$.

~Yrock

Solution 6

We can begin by changing the logarithms' base to $x$ and simplifying: $\newline$ $\frac{(\log_{x} 2023)^2}{\log_{x} 7 +1} \cdot \frac{1}{\log_{x} 289 + 1} = 1$ $\newline$ After that, we can multiply the denominator on the $\text{LHS}$ on both sides, resulting in: $\newline$ $(\log_{x} 2023)^2 = (\log_{x} 7 +1)(\log_{x} 289 +1)$ $\newline$ $(\log_{x} 2023)^2 = \log_{x} 2023 + \log_{x} 7 + \log_{x} 289 + 1$ $\newline$ $(\log_{x} 2023)^2 = 2\log_{x} 2023 + 1$ $\newline$ Now, we can substitute some variable $a$ as $\log_{x} 2023$, giving the quadratic $a^2-2a-1=0$. By Vieta's, the product of the roots is $1$, thus the solution is $\boxed{\textbf{(C)} 1}$.

~cyberhacker

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
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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