## 2.3 The Base of the Natural Logarithm

As discussed in the previous chapter, a logarithm is the power to which a base number must be raised in order to produce a desired numerical value. However, we’ve just described a function – the specific area under a certain hyperbola – that exhibits the properties of logarithms. But before proceeding to find actual values of the natural logarithms of specific numbers using this process, an immediate question arises. If $$\ln x$$ as just defined is the natural logarithm of the number $$x$$, what is the base being used for this logarithm?

We give this special base number the symbol “$$e$$” recognizing that it will be used to describe the natural exponential operation. For any real number $$x>0$$, the base number $$e$$ and the natural logarithm $$\ln x$$ must satisfy

$\ln x = \log_e x = p, ~~~~~{\rm if}~~~~~ x = e^p.$

If $$p=1$$, then for $$x=e$$ we find $$\ln e$$ = 1, another property of any logarithm and its base. So geometrically we can see that, equivalently, if we find the area under the curve $$v(u)=1/u$$ between the limits of 1 and $$e$$, we should arrive at a value of 1:

$\ln e = \int_1^e \frac{1}{u}\;du \equiv 1.$ So we could, in principle, find the area under our curve and vary the value of the upper limit of the integration until a value of 1 is achieved for the integral, to an acceptable accuracy. The value of $$e$$ that makes the integral equal to 1 would be the base number, and this indeed can be done using iterative numerical methods.

However, let’s take a slightly different approach. Let’s start with the fact that

$\ln e^x = \log_e e^x= x$

which, again, is just the basic relationship between a logarithm and its base. We now perform another calculus operation and take the derivative with respect to $$x$$ on both sides of the above equation. Since our logarithm is defined as an integral of the function $$1/u$$, then the derivative of $$\ln u$$ will be $$1/u$$. This in itself is an interesting result:

$\frac{d}{du} \ln u = \frac{1}{u}$

and if $$u=u(x)$$, then we can also use the chain rule to note that

$\frac{d}{dx} \ln u(x) = \frac{1}{u(x)}\cdot \frac{du}{dx}.$

But go back to our previous relationship, $$\ln e^x = x,$$ and let’s take the derivative of both sides:

$\frac{d}{dx} \ln e^x = \frac{d}{dx} x.$

Through the chain rule for derivatives, the left-hand side of this equation reduces to

$\frac{d}{dx}\;\ln e^x = \frac{1}{e^x} \;\frac{d}{dx}e^x.$

But the right-hand side of our equation has the derivative $$\frac{d}{dx}x$$ which is just equal to 1. So, re-equating the two sides of our original equation we get

$\frac{1}{e^x} \;\frac{d}{dx}e^x = 1$

or,

$\frac{d}{dx}e^x = e^x.$

The derivative of the function $$e^x~~$$ is $$~~e^x~!~~~~$$ That is to say, for our special, natural base $$e$$, the instantaneous rate at which the function $$e^x$$ is changing is equal to the instantaneous value of the function itself, and is true for all values of $$x$$. This is truly a uniquely special feature.

This remarkable result gives us all we need to easily compute $$e$$ (and $$e^x$$) to any desired accuracy. To do so, we use another result from calculus, namely a Taylor Series expansion. If the values of the derivatives of a function $$f(x)$$ are known at a point $$x=a$$, then it can be shown that

$f(x) = f(a) + f'(a) (x-a) + \frac{1}{2!}f''(a)\; (x-a)^2 + \frac{1}{3!}f'''(a) \;(x-a)^3 + ...$

where $$f'(x)$$, $$f''(x)$$, etc., are the first, second, etc., derivatives of $$f(x)$$, while

$n! = n\cdot(n-1)\cdot(n-2)\cdot \ldots \cdot 3\cdot 2\cdot 1$

is the factorial of $$n$$. Specific examples include

$0! \equiv 1 \\ 1! = 1 \\ 2! = 2\cdot 1 = 2 \\ 3! = 3\cdot 2\cdot 1 = 6 \\ 4! = 4\cdot 3\cdot 2\cdot 1 = 24$

and so on.

Using the Taylor formula, we take $$a = 0$$ and $$f(x) =e^x$$ for our case. And note that since $$f'(x) = df/dx = d(e^x)/dx = e^x$$, then $$f''(x) = df'/dx = d(e^x)/dx = e^x$$, and so on. Then, since all other derivatives will be $$e^x$$ as well, we find that $$f(0) = f'(0) = f''(0) =... = e^0 = 1$$. We thus arrive at a straightforward expression for the exponential function in terms of an infinite series containing terms of $$x$$ raised to integer powers:

$e^x = 1 + x + \frac{1}{2}x^2 +\frac{1}{6}x^3+\frac{1}{24}x^4 +\frac{1}{120}x^5+ ... \\ ~~~~~~~~~~~~~~~~= \frac{1}{0!}x^0 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 +\frac{1}{3!}x^3+\frac{1}{4!}x^4 +\frac{1}{5!}x^5+ ... \\$

And finally, for $$x=1$$, we have a way to compute the value of our natural base, $$e=e^1$$, namely

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} +\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+ ...$

Let’s compute $$e$$ using a computer program. We’ll look at the results when using a variable number of the first $$n$$ terms, nTerms, in the above infinite series, and look for when the series converges to within 6 decimal places by looking at the difference between successive estimates:

nTerms = seq(1,13)
eEst   = cumsum( 1/factorial( nTerms-1 ) )
eDiff  = c(0,diff(eEst))
eCalc  = eEst[eDiff !=0 & eDiff<1e-6][1]
nTerms eEst eDiff
1 1.000000 0.0000000
2 2.000000 1.0000000
3 2.500000 0.5000000
4 2.666667 0.1666667
5 2.708333 0.0416667
6 2.716667 0.0083333
7 2.718056 0.0013889
8 2.718254 0.0001984
9 2.718279 0.0000248
10 2.718281 0.0000028
11 2.718282 0.0000003
12 2.718282 0.0000000
13 2.718282 0.0000000

We see that the series converges to within 6 decimal places after the use of 11 terms, with a final estimated value of $$e \approx$$ eCalc= 2.718282.

We have just computed a fairly accurate approximation of the base number for the natural logarithm.6 As a check, if we integrate $$1/x$$ from 1 to $$e$$ we should get 1, since $$\ln e = 1$$ if it is the base number. We can do that with our computer programming language which has an integration function built in:

fx = function(x){ 1/x }
lne = integrate(fx, 1, eCalc)
paste("ln e = ",round(lne\$value,6), ", to 6 significant digits.")
## [1] "ln e =  1 , to 6 significant digits."

The base number, $$e$$, for the natural logarithm is called Euler’s number, so-named after the Swiss mathematician Leonhard Euler who implemented the use of this symbol.7 It is also sometimes called Napier’s constant as John Napier actually arrived at its value (actually, its inverse, $$1/e$$) during his pioneering studies into logarithms about 100 years earlier.8 His colleague Henry Briggs, in 1615, convinced Napier to revise his system of logarithms to use Base 10, for which $$\log 10 = 1$$ and thus simplifies the use of logarithms in everyday computations.9 The properties of the number $$e$$, only a few of which we have mentioned, make it a special number indeed, and it is regarded as one of the most important numbers in all of mathematics.

1. We should note that $$e$$ is an irrational number, as is $$\pi$$, and hence its decimal representation will never repeat.↩︎

2. Euler was indeed the first to use this symbol, but it isn’t clear if he meant for the “e” to stand for his name, or if it stood for “exponential”, or if it was just the next vowel to use after “a”, which was the other symbol he used in his work at the time. He is credited for using the symbols “f(x)” to define a function, for instance, as well as “i” for $$\sqrt{-1}$$, among many other standard notations.↩︎

3. For further discussion on this topic, see the section Comments on Napier’s Logarithms.↩︎

4. See The Development of Mathematics, Dover 1992, (second edition, McGraw-Hill, 1945), pp. 161-162.↩︎