Find the product of root of the equation (log₃)²-2(log₃x)-5=0 .

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$$\begin{gathered} \mathrm{Let} {\mathrm{x}}_1 \mathrm{and} {\mathrm{x}}_2 \mathrm{be} \mathrm{roots} \mathrm{of}\mathbf{ }{\left({\log }_{3}x\right)}^2-2\left({\log }_{3}x\right)-5=0. \mathrm{So} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{find} x_1{x}_2 \\ \mathrm{Now} \\ {\left({\log }_{3}x\right)}^2-2\left({\log }_{3}x\right)-5=0 ; \mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{Let} {\log }_{3}x=t \\ \Rightarrow {\log }_3{x}_1=t_1 and {\log }_3{x}_2=t_2 \\ \mathrm{Now} \mathrm{equation}\left(\mathrm{i}\right) \mathrm{becomes}, \\ {t}^2-2t-5=0 \\ \mathrm{Its} \mathrm{roots} \mathrm{are} t_1 \mathrm{and} t_2 \\ \mathrm{Sum} \mathrm{of} \mathrm{roots}=-\frac{\left(\mathrm{Coefficient} \mathrm{of} \mathrm{t}\right)}{\left(\mathrm{Coefficient} \mathrm{of} {\mathrm{t}}^2\right)} \\ \Rightarrow t_1+t_2=-\frac{\left(-2\right)}{1} \\ \Rightarrow {\log }_3{x}_1+{\log }_3{x}_2=2 \\ \mathrm{We} \mathrm{know} \mathrm{that} {\log }_a\left(m\right)+{\log }_a\left(n\right)={\log }_a\left(mn\right) \\ \Rightarrow {\log }_3\left(x_1{x}_2\right)=2 \\ \mathrm{We} \mathrm{know} \mathrm{that} {\log }_a\left(x\right)=y\Rightarrow x=a^y \\ \Rightarrow x_1{x}_2=3^2 \\ \Rightarrow x_1{x}_2=9 \\ \mathbf{Hence}\mathbf{ }\mathbf{product}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{roots}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{equation}\mathbf{ }\mathbf{ }{\left(\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{3}}\mathbf{x}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{2}\left(\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{3}}\mathbf{x}\right)\mathbf{-}\mathbf{5}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{9}\mathbf{.} \end{gathered}$$

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