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

Revision as of 07:59, 12 November 2023

Problem

What is the product of all solutions to the equation $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$

$\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1$ $\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2$

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 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

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.

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

https://youtu.be/OcNU62SMh4o

Video Solution

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

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

@@ Line 50: / Line 50: @@
 <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 <math>\ln{x}</math> is 0, thus the product of all solutions of <math>x</math> is <math>\boxed{\textbf{(C)} 1}</math>
+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

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

Page

Toolbox

Search