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

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)

@@ Line 3: / Line 3: @@
 <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==
-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
@@ Line 55: / Line 53: @@
 ==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>
-Maa is a troll, trying to exploit your cleverness to waste your time. Just select the simplest answer. C. It’s just 1 rather than some complicated expression full of logs and huge numbers.
+~MLiang2018
-There are plenty of AMC problems that have simple answers which are not correct. Similar to what numerophile said on problem 17. ~andliu766
 ==Video Solution 1 by OmegaLearn==

2023 AMC 12A (Problems • Answer Key • Resources)
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