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I am facing difficulties in solving the following equation:

$$4^x-9\cdot2^x+8>0.$$

Thanks in advance!

Tianlalu
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Love.math
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2 Answers2

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HINT Let $y=2^x$ and it becomes the quadratic inequality of $y$, because $4^x=(2^x)^2=y^2$.

Tianlalu
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    With the limitation that $y>0$ when we consider the possible solutions. – user Nov 10 '18 at 02:41
  • Could someone explain why there is a downvote here? I've already changed my attitude toward the questions with no attempts. See here. – Tianlalu Nov 12 '18 at 07:21
  • I don't know, I've only given a suggestion to improve the hint but originally I've upvoted that. – user Nov 12 '18 at 07:41
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Note that $$4^x = (2^x)^2$$

Thus your inequality is $$ (2^x-1)(2^x-8)>0$$

The solution is $ 2^x<1 $ or $2^x>8$

That is $x<0$ or $x>3$ which could be expressed as $ (-\infty, 0)\cup (3,\infty)$

Mohammad Riazi-Kermani
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