i cannot remember how to solve for this:
9^x^2<=27
Since the expoential is positive and the log is an increasing function we can take the log of both sides of the equation.
If you use the base 3 log you get
$\displaystyle \log_3(9)^{x^2}=\log_3(27) \implies 2x^2 \le 3 \iff 2\left(x^2-\frac{3}{2}\right) \le 0$
$\displaystyle \left(x-\sqrt{\frac{3}{2}} \right)\left(x+\sqrt{\frac{3}{2}} \right)\le 0 \implies x \in \left[ -\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}\right]$