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.
PS: Search engines also send visitors to the Exponents, Radicals & logs section and to this this earlier single page lesson. Site pages on the natural logarithm altogether provide a full treatment.
The presentation here is to 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. Variants of the exposition given here may be presented less cryptically in other texts.
Site Reviews
* Magellan, the McKinley Internet Directory, 1996: Mathphobics, this site may ease your fears of the subject, perhaps even help you enjoy it. The tone of the little lessons and "appetizers" on math and logic is unintimidating, sometimes funny and very clear. There are a number of different angles offered, and you do not need to follow any linear lesson plan. Just pick and peck. The site also offers some reflections on teaching, so that teachers can not only use the site as part of their lesson, but also learn from it. (Magellan is no longer online)
* The World-Wide Web Virtual Library Education by Country - Canada 1, 2005. Why Slopes: Appetizers and Lessons for Math and Reason. This online classroom offers appetizers and lessons for math from arithmetic to calculus or why slopes; for deductive reason (logic) and critical thinking; and for learning in general. Included here are opinions on the communication of skills and mathematics instruction. The logic appetizers are math free. Each appetizer is different. If one is not to your liking try another. Most are from three books on understanding and explaining math and reason.
may encourage a visit to site entrance
http://www.whyslopes.com.
The natural logarithm ln(a) for a > 0 can be introduced 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
Area(A)
=
(a-1) 1 a = 1 - 1 a
Area(C)
=
(1- 1 a ) 1 = 1 - 1 a
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
ln(ab) = ln(b) +ln(a).
(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
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,
ln(ab)
=
-ln( 1 ab )
=
-ln( 1 a 1 b )
=
-[ln( 1 a )+ln( 1 b )]
=
-ln( 1 a ) + -ln( 1 b ) = ln(a)+ln(b)
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
This suggests that
G¢(x) =
lim
Dx-> 0
G(x+Dx)-G(x) Dx = 1 x
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
is constant. To evaluate the constant, observe that
d = G(1)-F(1) = ln(a)-ln(1) = ln(a)
since ln(1) = 0. Thus we conclude ln(ax)-ln(x) = ln(a) or equivalently
ln(ax) = ln(a)+ln(x)
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
This suggests that
G¢(x) =
lim
Dx-> 0
G(x+Dx)-G(x) Dx = 1 x
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
is constant. To evaluate the constant, observe that
d = G(1)-F(1) = ln(a)-ln(1) = ln(a)
since ln(1) = 0. Thus we conclude ln(ax)-ln(x) = ln(a) or equivalently
ln(ax) = ln(a)+ln(x)
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).
