Solve $\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$ Solve $$\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$$
So letting $\alpha = 2^x$,
$$\frac{4\alpha^2-9\alpha+2}{\alpha^2-5\alpha-24}\le 0$$
$$\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)}\le 0$$
Multiply both sides by denominator squared: 
$$(4\alpha - 1)(\alpha-2)(\alpha+3)(\alpha-8)\le 0$$
Is this step right? 
If I do this, then I will get $\alpha &lt -3, \frac{1}{3} \le \alpha \le 2, \alpha > 8$? 
Right answer is $−3 &lt α ≤ \text{ or } 2 ≤ α &lt 8
$

$$\frac{f(\alpha)}{g(\alpha)}=\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)} \leq 0$$
$1.$ $f(\alpha) \leq 0$ and $g(\alpha)>0$
Hence :
$\left(\alpha \in \left[\frac{-1}{4},2\right]\right) \cap (\alpha \in (-\infty,-3) \cup (8,+\infty)) \Rightarrow \alpha \in \emptyset$
$2.$ $f(\alpha) \geq 0$ and $g(\alpha) &lt0$
Hence :
$\left(\alpha \in \left(-\infty,\frac{-1}{4}\right] \cup [2,+\infty)\right)\cap (\alpha \in (-3,8)) \Rightarrow \alpha \in \left(-3, \frac{-1}{4}\right] \cup[2,8)$
So , final solution is :
$\alpha \in \left(-3, \frac{-1}{4}\right] \cup[2,8)$
