2.5. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 71
For the claim about L, and Lemma 2.5.1, L(
xy
)≡ A
(1, x
y
)= A
(yy ,
xy
)= A(y,x) =
A(1,x)−A(1,y) = L(x)−L(y). Finally, from the above,
L(xy) = L(
xy−1
)= L(x)−L
(y−1)
Now L(y−1)≡ A
(1,y−1
)= A
(yy ,
1y
)= A(y,1) = −A(1,y) = −L(y) and so the above
yields L(xy) = L(x)+L(y). In particular, L(1) = L(12)= L(1)+L(1) so L(1) = 0 and
L(y) =−L(y−1).
Theorem 2.5.4 The function of Definition 2.5.2 for x,y > 0 has the following prop-erties.
1. L(xy) = L(x)+L(y) , L(1) = 0, L(
xy
)= L(x)−L(y) , L(x) =−L
( 1x
)2. x → L(x) is strictly increasing so L is one to one and L(x) = −L
( 1x
)< 0 if x < 1
while L(x)> 0 if x > 1.
3. If x > 0, then if y > x/2, |L(y)−L(x)| ≤ 2|x−y||x| .
4. L : (0,∞)→ R is onto.
Proof: The first claim is in the above lemma.Consider the second claim. To see this, first note that if x < y, then L(y)− L(x) =
L(y/x) by the first claim, and L(y/x)> 0 so it follows that L is strictly increasing. Hencethe function L is also one to one. From the definition and the first part, L(x) =−L
( 1x
)< 0
if x < 1 and L(x)> 0 if x > 1.Now consider the third claim. If a ≤ b, then L(b/a)+L(a) = L(b) and so, from the
definition in terms of area,
0 ≤ L(b)−L(a) = L(b/a) = A(1,b/a)≤(
ba−1)
1 =b−a
a.
1 b/a
If x/2 < y ≤ x. Then, from the above,
|L(x)−L(y)|= L(x)−L(y)≤ x− yy
≤ |x− y|x/2
=2 |x− y|
x
If y > x, then since 2/x > 1/x,
|L(x)−L(y)|= L(y)−L(x)≤ y− xx
≤ 2 |x− y|x
Finally consider the last claim. From the definition, it follows that L(2)> 1/2. (Drawa picture.) Therefore, from the first claim,
L(2n) = L
n times︷ ︸︸ ︷2×2×·· ·×2
= nL(2)> n/2, L(
12n
)= nL(1/2)<−n/2.