Property 7. Given any real number ξ

Property 7. Given any real number ξ there exists a positive real number x
such that logx < ξ.
4516 A. H. Salas
Indeed, choose a positive integer m for which ξ > −m. Then 1+ξ/m > 0. We
define x = 1
2(1+ξ/m)m. Taking into account (2.5) we obtain fn(x) ≤ m( m
√
x−
1) for any n ≥ m. Letting n→∞in this inequality gives log x ≤ m( m
√
x−1).
Then
log x ≤ m( m
√
x − 1) < m( m

(1 + ξ/m)m − 1) = ξ. (3.11)
Property 8. The range of logarithmic function is the set of real numbers.
Indeed, let y be any real number. By Property 7, we may find a number a > 0
and a number b such that loga < y and log b < −y. Then
loga < y < −log b = log b, where b =
1
b .
It is clear that a < b. Let H = {t > 0 such that logt < y }. This set is
not empty since a ∈ H . If t ∈ H then logt < y < log b and this gives t < b.
Thus, H is bounded from above by b. Let x = supH. We claim that log x = y.
Indeed, suppose that logx < y. Let ε = y−log x. There exists δ > 0 such that
|log x − log t| < ε if |x − t| < δ. Let t = x+δ/2 > x. Then logt < log x+ε = y
and t ∈ H so that t ≤ x. But t > x. Contradiction.
On the other hand, suppose that logx > y. Let ε = log x−y. There exists
δ > 0 such that |log x − log s| < ε if |x − s| < δ. Choose some s such that
0 < s < x and |x − s| < δ. Then logs > log x − ε = y > log t so that t < s for
any t ∈ H. We conclude that s is an upper bound of H and then x ≤ s. But
x > s. Contradiction.
We have proved that log x = y. Since y was arbitrarily chosen, we conclude
that the range of log is the set of real numbers.

ผลลัพธ์ (อังกฤษ) 2:

Property 7. Real Number CTnew given any positive there exists A Real Number x.
Such that logx <CTnew.
4516 AH Salas
Indeed, for which CTnew Choose A positive Integer M>-M. Then 1 + ξ / m> 0. We
define x = 1
2 (1 + CTnew / M) M. Taking Into Account (2.5) We obtain FN (x) ≤ M (M.
√
x-
1) for any n ≥ M. Letting n → ∞ in this Inequality gives log x ≤ M (M.
√
x-1).
Then
log x ≤ M (M
√
x - 1) <M (M.
?
(1 + CTnew / M) M - 1) = CTnew. . (Three eleven)
Property 8. The Range of Logarithmic Function is The Set of Real Numbers.
Indeed, Let Y be any Real Number. By Property 7, We may Find A Number A> 0.
B and A Number? such that loga <y and log b? <-Y. Then
Loga <Y <-log B? = Log B, Where B =.
1
B? .
It is Clear that A <B. Let H = {t> 0 such that logt <y}. Set this is
Not empty since A ∈ H. If T ∈ H then Logt <Y <log B and this gives T <B.
Thus, H is bounded by B from Above. Let x = supH. We Claim that log x = Y.
Indeed, suppose that logx <Y. Let ε = y-log x. There exists N> 0 Such that.
| log x - log T | <lotus if | x - T | <N. Let t = x + δ / 2> x. Then Logt <= Y log x + lotus.
∈ H and T so that T ≤ x. But t> x. Contradiction.
On The Other hand, suppose that logx> Y. Let ε = log x-y. There exists
N> 0 Such that | log x - log S | <lotus if | x - S | <N. Choose some S Such that
0 <S <x and | x - S | <N. Then logs> log x - lotus = Y> log T so that T <S for.
any T ∈ H. We conclude that S is an Upper Bound of H and then x ≤ S. But
x> S. Contradiction.
We have proved that log x = Y. Since Y was arbitrarily Chosen, We conclude.
that The Range of log is The Set of Real Numbers.

ผลลัพธ์ (อังกฤษ) 3:

Property 7. Given any real number ξ there exists a positive real number x
such that logx < ξ.
4516 A. H. Salas
Indeed, Choose a positive integer m for which ξ > − m. Then 1 ξ / M > 0. We
define x = 1
2 (1 ξ / M) m. Taking into account (2.5) we obtain FN (x) < = m (m


√ x − 1) for any n > = m. Letting n - > ∞ in this inequality gives log x < = m (m

√ x − 1).

log Then x < = m (m

√ x − 1) < m (m 

(1 ξ / M) m − 1) = ξ. (3.11)
Property 8. The range of logarithmic function is the set of real numbers.
Indeed let y, be Any real number. By Property 7 we may, find a number a > 0
and a number B  such that loga < y and log B  < − y. Then
loga < y < − log B  = log b where B, 1

.
B  It is clear that a < B. Let H = {T > 0 such that logt < y}. This set is
not empty Since a ∈ HIf t ∈ H then logt < y < log B and this gives T < B.
Thus H is, bounded from above by B. Let x = supH. We claim that log X = y.
Indeed suppose that, logx < y. Let ε = y − log X. There exists δ > 0 such that
| log x − log t | < ε if | x − t | < δ.? Let t = x δ / 2 > x. Then logt < log x ε = y
and t ∈ H so that T < = x. But T > x. Contradiction.
On the, other hand suppose That logx > yLet ε = log x − y. There exists
δ > 0 such that | log x − log s | < ε if | x − s | < δ. Choose some s such that
0 < s < x and | x − s | < δ. Then logs > log x − ε. = Y > log t so that T < s for
any t ∈ H. We conclude that s is an upper bound of H and Then x < = s. But
x > s. Contradiction.
We have proved that log x = y. Since y was, arbitrarily chosen we conclude
That the range of log is the set of real numbers