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

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

---- - -----(0)-----+------[1]---- - -----[3]------+-------
////////////// ///////////////
-----------(0)-----------------------------------------------
/////////////////////////////////////////////////////
$1 leq a leq 3$
$1 leq 2^x leq 3$
$2^0 leq 2^x leq 2^{log_23}$
$0 leq x leq {log_23}$

Ответ: [0; log₂3]