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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. 

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.

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).

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

Sketch of A First Demonstration

1. Divide the interval [1,a] on the s-axis into n ³ 1 segments using the end points s_i = 1+i·[(a-1)/(n)] where 1 £ i £ n. Each segment has length [(a-1)/(n)].

2. On each segment [s_i,s_i+1] construct a rectangle whose top just touches the curve y = [1/(s)] at y = [1/(s_i)]. The sum S_n 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 [s_i,s_i+1] and height [1/(s_i)] has the same area as the rectangle with base [bs_i,bs_i+1] and height [1/(bs_i)]. But the rectangles with base segments [bs_i,bs_i+1] and height [1/(bs_i)] approximate the area S_ba under the curve y = [1/(s)] from s = b to s = ba. So taking the limit as n -> ¥ suggests S_ba = ln(a).

4. Drawing a graph suggests or implies S_ba = ln(ab) -ln(b). Therefore ln(a) = S_ab = ln(ab)-ln(b) as well. So we are done in the first case where a > 1 and b > 0. That is, the area S_ba 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.

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 log_a(x) = [(ln(x))/(ln(a))] when a ¹ 1. The property ln(ax) = ln(a)+ln(x) now implies log_c(ab) = log_c(b) +log_c(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 log_e(x) = ln(x).

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