Решите неравенство: 2^x+3*2^-x<=4

$2^x+3*2^{-x} \leq 4$
$2^x+ \frac{3}{2^{x}} -4 \leq 0$
Замена: $2^x=a,$ $a\ \textgreater \ 0$
$a+ \frac{3}{a} -4 \leq 0$
$\frac{a^2+3-4a}{a} \leq 0$
$\frac{a^2-4a+3}{a} \leq 0$
$a^2-4a+3=0$
$D=(-4)^2-4*1*3=4$
$a_1= \frac{4+2}{2} =3$
$a_2= \frac{4-2}{2} =1$

$\frac{(a-3)(a-1)}{a} \leq 0$

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$1 \leq a \leq 3$
$1 \leq 2^x \leq 3$
$2^0 \leq 2^x \leq 2^{log_23}$
$0 \leq x \leq {log_23}$

Ответ: [0; log₂3]