# 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.

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

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 $$\log_{3}{(x)}$$
\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}