The Natural Logarithm
For real numbers, the following sections describe the area-under-a-curve
definition of the natural logarithm, and how this introduction of the natural
logarithm leads to the definition and properties of all logarithms,
exponentials and powers involving real numbers.
The natural logarithm ln(a) for a > 0 can be introduced
FOOTNOTE: Variants of the exposition given here may be presented less
cryptically in other texts. The presentation here is show briefly the approach
I would like to see favored in schools. Working through the details of this
exposition in its present form could be a subject for discussion in a high
school math club. Understanding this section and the next demands or provides
a sound command of some mathematics beyond arithmetic.
as the (signed) area under the curve y = [1/(s)] from s = 1
to s = a. Equivalently, it may be represented by the signed area
under the curve u = [1/(v)] from v = 1 to v = a.
This definition does not depend on the labelling of the horizontal and vertical
axes. See the next two diagrams.
In the next diagram, the area from s = 1 to s = a > 1
can be approximated by slicing it into n vertical rectangles with the
same base size [(a-1)/(n)], and then
making this base size smaller by letting n®¥
(that is get larger and larger).
FOOTNOTE: The shorthand n®¥ should be read
as n tends to (or goes to) infinity. It is left as an exercise for
advance students to write on paper the Riemann sums whose limit is or should
be the value L.
The sum of the area of the resulting rectangles approximates to a single number L
with greater and greater accuracy, more decimal places say, as n ®¥.
This single limit gives what we call ln(a).
For a ³ 1, the value of ln(a) is
given by the area from s = 1 to s = a under the curve y
= [1/(s)]. Here we take or assume ln(1) = 0. It can be shown that ln(a)
® 0 when when a approaches 1 through values
above or greater than 1.
The natural logarithm ln(b) of a number b when 0 < b
< 1 is defined next.
For 0 < b < 1, the value of ln(b) is given by (-1) times
the area under the curve y = [1/(s)] from s = b to s
= 1.
The above two diagram illustrate the arithmetic or area-based definition of
the natural logarithm ln(a) or ln(b) in the two mutually exclusive
cases a > 1 and 0 < b < 1. These definitions imply that ln(x)
® 0 = ln(1) when x ®
1.
Reading Guide. The rest of this section states and indicates the
proofs of two algebraic properties of the natural logarithm. The first proof is
easy. The second proof is cryptic - material for advanced students. The next
section briefly indicates the relationship between the inverses of the
logarithms and exponential functions - more material for advance students.
Consult another calculus or analysis text for the missing details.
Proof of Property ln([1/(b)]) = -ln(b)
for b > 0.
We will show that 0 = ln(b)+ln([1/(b)]) when b
> 0. For this, first consider the case a > 1. In the
following diagram
By symmetry (or reflection across the line y = s),
ln(a) = Area(B)+Area(A). Therefore ln(a) =
Area(B)+Area(C)
Here A is the rectangle with corners (0,1) and (1/a, 1) while
C is the rectangle with corners (1/a,0) and (1,1)
Now by definition -ln([1/(a)]) =
Area(B)+Area(C).
Therefore -ln([1/(a)]) = ln(a).
This in turn implies ln([1/(a)])+ln(a) = 0.whenever a
> 1.
Finally, we conclude ln([1/(b)])+ln(b) = 0 whenever b
> 0. This follows by putting a = b if b ³
1 and by putting a = [1/(b)] if 0 < b < 1.) The
latter is equivalent to the property ln([1/(b)]) = -ln(b)
which we wanted to show.
Fundamental Property of Logarithms
Next we may derive the fundamental property of logarithms, that is
(This holds when a = 1 and b > 0 since ln(1) = 0 by
definition.) We will now consider the case where a > 1 and b
> 0. For this it suffices to reconsider how the number ln(a) is
computed. Two ways to show this are indicated next.
Sketch of A First Demonstration
1. Divide the interval [1,a] on the s-axis into n
³ 1 segments using the end points si
= 1+i·[(a-1)/(n)] where 1 £
i £ n. Each segment has length [(a-1)/(n)].
2. On each segment [si,si+1]
construct a rectangle whose top just touches the curve y = [1/(s)]
at y = [1/(si)]. The sum Sn
of the areas
| Aj = yj·(si+1-si)
= yi· |
a-1
n |
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of these rectangles provides an approximation to ln(a) which we
assume becomes more accurate as n is made larger.
3. Now the rectangle with base [si,si+1]
and height [1/(si)] has the same area as the
rectangle with base [bsi,bsi+1]
and height [1/(bsi)]. But the rectangles with base
segments [bsi,bsi+1] and
height [1/(bsi)] approximate the area Sba
under the curve y = [1/(s)] from s = b to s
= ba. So taking the limit as n ®¥
suggests Sba = ln(a).
4. Drawing a graph suggests or implies Sba
= ln(ab) -ln(b). Therefore ln(a)
= Sab = ln(ab)-ln(b)
as well. So we are done in the first case where a > 1 and b
> 0. That is, the area Sba under the curve y
= [1/(s)] from s = b to s = ba equals the
area under the curve y = [1/(s)] from s = 1 to s =
ba minus the areas from s = 1 to s = b.
Now the fundamental property of logarithms, that is ln(ab) = ln(b)
+ln(a) holds whenever at least one of the factors a and b
is greater than 1 (since addition and multiplication of real numbers is
commutative.) Now observe for c > 0 that 0 = ln(1) = ln( [1/(c)]
·c) = ln([1/(c)])+ln(c) since c or its reciprocal
must be ³ 1. Hence ln(c) = -ln([1/(c)]).
This was shown before with the aid of some diagrams. The latter equality
prepares us to treat the sole remaining case where both numbers a and b
are between 0 and 1. In this case,
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| -ln( |
1
a |
) + -ln( |
1
b |
) = ln(a)+ln(b) |
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as required. Therefore ln(ab) = ln(a)+ln(b) holds
whenever a and b are both positive.
This indicates a simple demonstration of the fundamental property for the
natural logarithm ln(x) for x > 0. The sketch of an alternative
proof follows.
Sketch of a Second Demonstration. For a > 0, put G(x)
= ln(ax). Then value of G(x) is given by the (signed)
area from s = 1 to s = ax under the curve y = [1/(s)].
Observe G(1) = ln(a). The area of region D in the
following diagram equals G(x+Dx)-G(x).
The height of the region D is approximately [1/(ax)] and its
length is precisely a(x+Dx) -
ax = aDx. Therefore
| G(x+Dx)-G(x)
» Area(D) = |
1
ax |
·aDx
= |
1
x |
·Dx |
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This suggests that
| G¢(x)
= |
lim
Dx®
0
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|
G(x+Dx)-G(x)
Dx |
= |
1
x |
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Similarly F(x) = ln(x) implies that F¢(x)
= [1/(x)]. This implies by the Constant Difference Theorem that
| ln(ax)-ln(x)
= G(x)-F(x)
= d |
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is constant. To evaluate the constant, observe that
| d = G(1)-F(1)
= ln(a)-ln(1) = ln(a) |
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since ln(1) = 0. Thus we conclude ln(ax)-ln(x)
= ln(a) or equivalently
as required.
The height of the region D is approximately [1/(ax)] and its
length is precisely a(x+Dx) -
ax = aDx. Therefore
| G(x+Dx)-G(x)
» Area(D) = |
1
ax |
·aDx
= |
1
x |
·Dx |
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This suggests that
| G¢(x)
= |
lim
Dx®
0
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G(x+Dx)-G(x)
Dx |
= |
1
x |
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Similarly F(x) = ln(x) implies that F¢(x)
= [1/(x)]. This implies by the Constant Difference Theorem that
| ln(ax)-ln(x)
= G(x)-F(x) =
d |
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is constant. To evaluate the constant, observe that
| d = G(1)-F(1)
= ln(a)-ln(1) = ln(a) |
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since ln(1) = 0. Thus we conclude ln(ax)-ln(x)
= ln(a) or equivalently
as required.
Logarithms To Base a > 0
The logarithm to base a > 0 is given by loga(x)
= [(ln(x))/(ln(a))] when a ¹ 1.
The property ln(ax) = ln(a)+ln(x) now implies logc(ab)
= logc(b) +logc(a) holds when a,
b and c are all positive real numbers with c ¹
1. The proof is a simple algebraic exercise. Further note that ln(e) = 1
implies loge(x) = ln(x).
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