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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\qquad\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>
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==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 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 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 (Same idea as Solution 1 with easily understood steps)==
 +
 +
<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath>
 +
 +
Rearranging it give us:
 +
 +
<cmath>\log_{2023}7x\cdot \log_{2023}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>
 +
 +
let <math>\log_{2023}x</math> be <math>a</math>, we get
 +
 +
<cmath>(\log_{2023}7+a)(\log_{2023}289+a)=1+a</cmath>
 +
 +
<cmath>a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a</cmath>
 +
 +
<cmath>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</cmath>
 +
 +
by Vieta's Formulas,
 +
 +
<cmath>a_1+a_2=0</cmath>
 +
 +
<cmath>\log_{2023}{x_1}+\log_{2023}{x_2}=0</cmath>
 +
 +
<cmath>\log_{2023}{x_1x_2}=0</cmath>
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 +
<cmath>x_1x_2=\boxed{\textbf{(C)} 1}</cmath>
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 +
~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
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<cmath>(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2</cmath>
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<cmath>\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}</cmath>
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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>
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 +
~dwarf_marshmallow
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==Solution 4(Fakesolve)==
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We take the reciprocal of both sides:
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<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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 +
~MLiang2018
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 +
This solution works for this problem by chance but do note that the simplification step to get <cmath>2023x^2=2023x</cmath>is not how log properties work and that the actual solutions for x are <cmath> x = e^{ \pm \sqrt{ln^2(7)+ln(17)ln(2092529)}}</cmath> (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 <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>
 +
 +
Now we can use <math>\log_{a}bc=\log_{a}b+\log_{a}c</math>. We get <cmath>\log_{2023}7+\log_{2023}x+\log_{2023}289+\log_{2023}x=\log_{2023}2023+\log_{2023}x.</cmath> The <math>\log_{2023}7+\log_{2023}289</math> and <math>\log_{2023}2023</math> terms cancel, giving <cmath>2\cdot\log_{2023}x=\log_{2023}x</cmath> so now we are sure that <math>\log_{2023}x=0</math>, so the only solution is <math>x=\boxed{\textbf{(C)} 1}</math>.
 +
 +
~Yrock
 +
 +
==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}}

Latest revision as of 19:00, 30 September 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 understood 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(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

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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