An algebra problem by Hobart Pao

$log_x {9}+log_ {(log_x {27})} 81=6$ $2log_x {3}+4log_ {(3log_x {3})} 3=6$ ${\boxed{log_x {3}=y}}$ $2y+4log_ {(3y)} 3=6$ $y+2log_ {(3y)} 3=3$ $2log_ {(3y)} 3=3-y$ $(3y)^{(3-y)}=3^2$ ${\boxed{y=1}}\rightarrow log_x {3}=1$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{x=3}}}}$