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 (Same idea with Solution 1)

$\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$

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,

$a_1+a_2=0$

$\log_{2023}{x_1}+\log_{2023}{x_2}=0$

$\log_{2023}{x_1x_2}=0$

$x_1x_2=\boxed{\textbf{(C)} 1}$

~lptoggled

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2023 AMC 12A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2 (Same idea with Solution 1)

See also