Logarithmic equation with logarithm in power.

Question

$$x^{\log_{\,3}(3x)}=9$$ I tried to turn the exponential to logarithm form $- \log_{\,x}(9) = \log_{\,x}(3x)$. I also tried using the property $a=\log_{\,b}(b^a)$, but it didn't get me anywhere. I never encounter these kind of problems so I would like a bit of help.

Answer 1

$$x^{\log_3(3x)}=9$$ $$\log_3x^{\log_3(3x)}=\log_33^2$$ $$\log_3x\log_3(3x)=2$$ $$\log_3x(1+\log_3x)=2$$ $$\log^2_3x-1+\log_3x-1=0$$ $$(\log_3x-1)(\log_3x+2)=0$$ $x=3$ and $x=3^{-2}$ are real solutions

Answer 2

Hint: $$x^{\log_3(3x)}=3^{\log_3(x)\log_3(3x)}=3^{\log_3(x)(\log_3(x)+1)}$$

Now take $\log_3$ of both sides.

Answer 3

$$ x^{\log_3 (3x)} = 9 $$

Taking $\log_3$ of both sides:

$$\log_3 3x\times \log_3 x = \log_3 9 $$

$$ (1 + \log_3 x) \log_3 x = 2$$

$$ (\log_3 x-1)(\log_3 x+2) = 0$$

$$ \log_3 x = 1 = \log_3 3$$

$$ x = 3$$

$$ \log_3 x = -2$$

$$ x = 3^{-2}$$