Ответ:
а) 19·4x – 5·2x+2 + 1=0
19·22x – 5·22·2x + 1=0
19·22x – 20·2x + 1=0
Замена: 2x = t, t > 0
19t2–20t+1=0
D=400–76=324, > 0 = > 2 корня
t1=(20–18)/38=1/19
t2=1
1)2x=1/19
x=log2(1/19)=log2(19–1)=–log2(19)
2)2x=1
x=0
б)24=16, 25=32, = > –log2(19)≈–4,...
= > –log2(19)∈[–5;–4].
0∉[–5;–4]