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$$x^{ 5-\log _{ 3 }{ x } }=9x^2$$

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.

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    $\begingroup$ $-\log_3 9x^2=-\log_3 9 - \log_3 x^2=-2-\log_3 x^2=-2-2\log_3 x$. Last equality holds because we know $x>0$. $\endgroup$
    – user236182
    Oct 5 '15 at 4:16
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    $\begingroup$ we know $x>0$ because the term $\log_3 x$ occurs in the first line of the Q $\endgroup$ Oct 5 '15 at 4:28
  • $\begingroup$ @user236182 Feel free to add that comment as an actual answer. I'll accept it and +1 $\endgroup$ Oct 5 '15 at 8:36
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Something to try:

Convert your right side to $$-\log _{ 3 }{ 9x^{ 2 } }=-\log _{ 3 }{( 3x)^{ 2 } }=-2\log_{3}{(3x)}$$

Then you convert your left-side terms to $$\log_{3}{(3x)}$$ instead of $$\log_{3}{(x)}$$

See where that takes you.

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

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