The inequality for x is:
$(\log_3 x)^2 \lt \log_9( x^4)$
I plotted the graphs to find as answer: $1\lt x \lt 9$
My approach:
In my attempt to solve it, I used the logarithm properties to equal the bases:
$\log_3x \cdot \log_3x \lt 4\log_9x$
$\log_3x \cdot \log_3x \lt 4\log_3x^{1/2}$
$\log_3x \cdot \log_3x \lt 2\log_3x$
Then I changed variables, letting $u = \log_3x$, thus:
$u^2 \lt 2u$
$u^2 - 2u \lt0$
$u(u-2) \lt 0$
$u < 0$ or $u < 2$
Replacing the prior variable:
$\log_3x\lt0$ and $\log_3x\lt2$
Which gives: $x\lt 3^0$ and $x\lt3^2$:
$x<0$ and $x\lt9$.
Thats not the answer. I would appreciate if someone could point out my mistake.
Edit: I made a mistake when typing the first equations. The right Log was supposed to have the base 9, not 3, as initially was there for some of you to solve. Now it's fixed.