2023 AMC 12A Problems/Problem 19


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 $\prod x=e^0=\boxed{\textbf{(C)} 1}$.


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:

\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x\]


let $\log_{2023}x$ be $a$, we get


\[a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a\]

\[a^2+\log_{2023}7 \cdot \log_{2023}289-1=0\]

by Vieta's Formulas,




\[x_1x_2=\boxed{\textbf{(C)} 1}\]


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


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


Video Solution


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

See also

2023 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png