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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

  • 1
    $\alpha = 2^x$.2012-04-15

1 Answers 1

5

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

$1.$ $f(\alpha) \leq 0$ 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 :

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

  • 0
    @JiewMeng Yes , you have two cases . A final solution is union of these two particular solutions..2012-04-16