Question

 РЕШИТЕ УРАВНЕНИЕ 4^x-10*2^x+16<0

Answer 1

$(2^x)^2-10*2^x+16<0 \ 2^x=t \ t^2-10t+16<0 \ D=100-4*16=36 \ t_1=8;t_2=2 \ (t-2)(t-8)<0 \ t in (2;8) \ left { {{t>2} atop {t<8}} right \ left { {{2^x>2} atop {2^x<8}} right \ left { {{2^x>2^1} atop {2^x<2^3}} right \ left { {{x>1} atop {x<3}} right \ x in(1;3)$

Answer 2

$4^x-10cdot2^x+16<0, \ a=2^x, a^2-10a+16<0, \ a^2-10a+16=0, \ a_1=2, a_2=8, \ (a-2)(a-8)<0, \ 2<a<8, \  2<2^x<2^3, \ 1<x<3. \ xin(1;3).$