Hitting a roadblock while solving a logarithmic equation $$x^{ 5-\log _{ 3 }{ x }  }=9x^2$$
Steps I took:
$$\log _{ 3 }{ x^{ 5-\log _{ 3 }{ x }  } } =\log _{ 3 }{ 9x^{ 2 } } $$
$$(5-\log _{ 3 }{ x } )(\log _{ 3 }{ x) } =\log _{ 3 }{ 9x^{ 2 } } $$
$$5\log _{ 3 }{ x } -(\log _{ 3 }{ x } )^{ 2 }=\log _{ 3 }{ 9x^{ 2 } } $$
$$(\log _{ 3 }{ x } )^{ 2 }-5\log _{ 3 }{ x } =-\log _{ 3 }{ 9x^{ 2 } } $$
I am trying to turn this into a quadratic equation to then solve with substitution, but I can't seem to manipulate the right hand side of this equation in any way that will allow me to do this. 
Hints are much better appreciated than the actual answer.  

$$\begin{align}
x^{ 5-\log _{ 3 }x }&=9x^2
\\x^{3-\log_{3}x}&=9\qquad\text{as $x\neq0$}
\\&=x^{\log_x 9}
\\&=x^{2\log_x 3}
\\&=x^{\frac 2{\log_3 x}}
\\3-u&=\frac 2u\qquad\text{(putting $u=\log_3x$)}
\\u^2-3u+2&=0
\\(u-2)(u-1)&=0
\\u=\log_3x&=1,2
\\x&=3,9
\qquad\blacksquare
\end{align}$$
