Let y=log2023x. Then (y+log20237)(y+log2023289)=1+y. Expand and simplify to obtain y2+(log20237+log2023289−1)y+log20237⋅log2023289−1=0. Since log2023289=2log202317 and log20237+log2023289=log2023(7⋅289)=log20232023=1, this becomes y2+log20237⋅log2023289−1=0.
By Vieta's Formula, y1+y2=0. Therefore log2023x1+log2023x2=0, so log2023(x1x2)=0. Hence x1x2=1.