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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 $\alpha < -3, \frac{1}{3} \le \alpha \le 2, \alpha > 8$?

Right answer is $−3 < α ≤ \text{ or } 2 ≤ α < 8 $

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1  
$\alpha = 2^x$. – user17762 Apr 15 '12 at 7:24
up vote 4 down vote accepted

$$\frac{f(\alpha)}{g(\alpha)}=\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)} \leq 0$$

$1.$ $f(\alpha) \leq 0$ 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 $g(\alpha) <0$

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 :

$\alpha \in \left(-3, \frac{-1}{4}\right] \cup[2,8)$

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I don't get the first part: $f(\alpha) \leq 0$ and $g(\alpha) \gt 0$. It could be $f(\alpha) \ge 0$ and $g(\alpha) \lt 0$ too? – Jiew Meng Apr 16 '12 at 0:47
    
@JiewMeng Yes , you have two cases . A final solution is union of these two particular solutions.. – pedja Apr 16 '12 at 2:48

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