2023 AMC 12A Problems/Problem 19

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

$\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$ 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 veita's formula, $a_1+a_2=0$ $\log_{2023}{x_1}+\log_{2023}{x_2}=0$ $\log_{2023}{x_1x_2}=0$ $x_1x_2=\boxed{\textbf{(C)} 1}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2023 AMC 12A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See also