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