If $${\log _3}x + {\log _{9}}{x^2} + {\log _{27}}{x^3}$$     $$ = 9,$$  then x equals to -

Solution (By Examveda Team)

$$\eqalign{ & \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr} $$