Math Gold Medalist

Lor

2023 AMC 12A 

Problem 19

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

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

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   Solution