smc

For ${\log }_{7x}2023\cdot {\log }_{289x}2023={\log }_{2023x}2023$, transform it into $\frac{\ln 289+\ln 7}{\ln 7+\ln x}\cdot \frac{\ln 289+\ln 7}{\ln 289+\ln x}=\frac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$. 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 $\prod x=e^0=\boxed{1}$.