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]