1/log2 x < 1+1/(log2 x-1)
Dziedzina :
1. x > 0 (z def. logarytmu )
2. log2 x≠ 0 ⇔ ( x≠2^0 ) ⇔ x≠1
3. log2 x-1 ≠ 0 ⇔ log2 x≠1 ⇔ x≠2
Stąd : x∈(0,1)∪(1,2)∪(2,∞) ( * )
1/log2 x < 1+1/(log2 x-1)
log2 x=t
1/t < 1+ 1/(t-1)
1/t < (t-1+1)/(t-1)
1/t < t/(t-1)
1/t-1/(t-1) < 0
(t-1-1)/(t(t-1) < 0
(t-2)/t(t-1) < 0
t(t-1)(t-2) < 0
t=0 ∨t-1=0∨ t-2=0
t=0 ∨ t=1 ∨ t=2
log2 x=0 ∨ log2 x=1 ∨ log2 x=2
x=1 ∨ x=2 ∨ x=4
x∈(0,1)∪(2,∞)
Po uwzględnieniu dziedziny ( * ) , mamy : x∈(0,1)∪(2,∞) .