Equations & Inequations

$\log_{\frac{1}{3}}(2^{x+2}-4^x)\geq -2$

We can now evaluate $\left(\frac{1}{3}\right)^{LHS}$ and $\left(\frac{1}{3}\right)^{RHS}$ , but we also have to change the $\geq$ to a $\leq$ since $f(x)=\left(\frac{1}{3}\right)^x$ is monotonically decreasing

$2^{x+2}-4^x\leq 9$

$4\cdot 2^x-(2^x)^2\leq 9$

Let $y=2^x$

$4y-y^2\leq 9$

$y^2-4y+9\geq 0$

$(y-2)^2+5\geq 0$

Which is trivially true for all values of $y$ and therefore for all values of $x$

Now we have to consider the original inequality again, since the domain of $\log(x)$ is $(0,\infty)$ we have to make sure that

$2^{x+2}-4^x> 0$

$2^x(4-2^x)> 0$

$4-2^x> 0\phantom{cccccccccc}\color{#3D99F6}\text{since } 2^x>0\phantom{c}\forall x\in\mathbb{R}$

$2^x<4$

$\boxed{x<2}$

In other words, the solutions set is $\boxed{(-\infty,2)}$