Question

Помогите решить номер 2

Answer 1

$4^{2x} - 5\cdot 4^{x} + 4\ \textless \ 0$
Замена:$4^x = y$
$y^2 - 5\cdot y + 4 \ \textless \ 0 \\ \\ D = 25 - 16 = 9 \\ \\ y_{12} = \frac{5\pm \sqrt{9}}{2} = \frac{5\pm 3}{2} = \left \{ {{4} \atop {1}} \right. \\ \\(y-4)(y-1)\ \textless \ 0 \\ \\ y\in(1;4) \\ \\ x\in(0;1)$