Exponential inequality: $4^x-9\cdot2^x+8>0$

Question

I am facing difficulties in solving the following equation:

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

Thanks in advance!

Answer 1

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

Answer 2

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)$