Solve the equation $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
The answers are $x_1=\dfrac19$ and $x_2=3$.
For the range we have: $$D_x:\begin{cases}x>0\\x\ne1\\9x^2>0\iff x\ne0\end{cases}\iff x\in(0;1)\cup(1;+\infty)$$
I wrote the equation as follows using $\log_a(b)=\dfrac{1}{\log_b(a)}:$ $$\log_x(9x^2)\cdot\dfrac{1}{\log_x^2(3)}=4\\\iff \log_x(9x^2)=4\log_x^2(3)=\log_x^2(3^4)$$ Is this reasonable and how can one continue?
$$\log_x(9x^2)\cdot\log_3^2(x)=4$$ $$\Rightarrow \dfrac{\log_3(9x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$ $$\Rightarrow \dfrac{\log_3(9)+\log_3(x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$ $$\Rightarrow \dfrac{2+2\log_3(x)}{\log_3(x)}\cdot\log_3^2(x)=4$$ $$\Rightarrow 2\log_3 (x)\left[1+\log_3(x)\right]=4$$ $$\Rightarrow \left[\log_3 (x)\right]^2+\log_3(x) - 2=0$$
Using the quadratic formula, we get $\log_3(x) = \dfrac{-1\pm\sqrt{1+8}}{2}=\dfrac{-1\pm3}{2}=1,-2$
This gives $x=3$ and $x=\dfrac{1}{9}$.
You are on the right track!
Based on the assumption that $x\in(0,1)\cup(1,+\infty)$, we can rewrite the first term from the LHS of the original equation as follows: \begin{align*} \log_{x}(9x^{2}) = \log_{x}(9) + 2\log_{x}(x) = \log_{x}(3^{2}) + 2 = 2\log_{x}(3) + 2 \end{align*} As you have already noticed, the following identity holds: \begin{align*} \log_{x}(3) = \frac{1}{\log_{3}(x)} \end{align*} Gathering all the previous results and letting $y = \log_{3}(x)$, one arrives at the desired solution set: \begin{align*} \left(\frac{2}{y} + 2\right)y^{2} = 4 & \Longleftrightarrow (2 + 2y)y^{2} = 4y \tag{$y\neq 0$}\\ & \Longleftrightarrow y(2y^{2} + 2y - 4) = 0 \tag{$y\neq 0$}\\\\ & \Longleftrightarrow 2y(y^{2} + y - 2) = 0 \tag{$y\neq 0$}\\\\ & \Longleftrightarrow y\in\{-2,1\}\\\\ & \Longleftrightarrow \log_{3}(x)\in\{-2,1\}\\\\ & \Longleftrightarrow x\in\left\{\frac{1}{9},3\right\}. \end{align*}
Hopefully this helps!
$\rm {Hint :}$
$$\begin{align}\log_x9x^2\cdot\log_3^2x&=2\log_x(3x)\cdot\frac {1}{\log^2_x3}\\ &=\frac {2\log_x3+2}{\left(\log_x3\right)^2}\\ &=4\thinspace .\end{align}$$
where, $x>0\wedge x≠1$.
