(2^(x-3))^2 -71/8*2^(x-3)+7< = 0 , Пусть 2^(x-3) = y ,то получим y^2 -71/8y+7<=0 , D/4= (71/16)^2-7 = 5041/256-7 =(5041-1792)/256= 3249/256 =(57/16/)2>0 , y1= (71/16+57/16) =128/16=8 , y2 =(71/16-57/16) = 14/16 =7/8 , (y-8) *(y - 7/8)<=0, 7/8<=y<=8 , 7/8<=2^x-3 <= 2^3 log по 2 оснавани 7/8<=x-3<=3 , log по оснавани2 7- 3+3<= x<=6 , log по оснавани2 7 <=x<=6 .