<s>[INST] Solve the equation $\log_x(9x^2)\cdot\log_3^2(x)=4$ 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?
 [/INST] $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9)+\log_3(x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{2+2\log_3(x)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow 2\log_3 (x)\left[1+\log_3(x)\right]=4$$
$$\Rightarrow \left[\log_3 (x)\right]^2+\log_3(x) - 2=0$$
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 $x=\dfrac{1}{9}$.
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