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# (M-18)     Natural Logarithms

## And how to get more accurate Approximate Logs

### Bases of Logarithms

As shown earlier, if log10 x = y (read as "y is the logarithm of x to base 10"),
that means x = 10y.

Numbers other than 10 can also serve as "base" of a set of logarithms, though such logarithms lack an important advantage. This is best seen if x is expressed in scientific notation. In such notation, every number is expressed as product of some number between 1 and 10, multiplied by a power of 10, e.g.

x =1492 = 1.492 103
Then
log x = log 1.492 + 3 = 3.173769..

Then 1.492, the part which gives the "detailed structure" of x, contributes the decimal part of the logarithm ("mantissa"), while the power of 10 which gives the magnitude of x is encoded in the whole-number part of the logarithm only ("the characteristic"). If the base is not 10, this useful feature is lost and logarithms of numbers with the same digits but different positions of the decimal point (e.g. 1492 and 14.92) give completely different logarithms.

Nevertheless, sometimes a different base also has an advantage.

### Switching Bases

Suppose some number "A" is chosen to be the base of our new logarithms. Then if

x = Ay                     (1)
that means
y = logA x                 (2)

What is the connection between these "new" logarithms and "old" log x with base 10? To convert one into the other, we need to know the "old" log of A. Denote it by B

B = log10 A               (3)
which means
10B= A                   (4)

Substitute in (1) and use the formula for raising a power of a number by another power:

x = (10B)y = 10(By)         (5)

Taking logs (to base 10) and using (3) and (2)

log10x   =   By   =   (log10 A)(logA x)         (6)
so
(logA x)   =   (log10x )/ log10 A         (7)

Thus to "convert" an old log of base 10 to a new base A one only has to multiply it by a constant number 1/log10A .
That constant number can also be expressed in a different way. Since (7) holds true for ANY value of x, try x=10
(logA 10) = (log1010 )/ log10 A               (8)

However, log10 10 = 1 (answering the question, to what power do you raise 10 to get 10?). So

(logA10) = 1/ log10 A               (9)

(note in passing that in any base A, logA A = 1. "To what power do you need to raise A to get A?"). Either side of (9) can then provide the constant conversion factor.

### Natural Logarithms and the Binomial Approximation

Logarithms derived to base e are called natural logs and are often denoted "ln" rather than "log" (though if you persist to study calculus, you will find that there, plain "log" usually means a natural logarithm, while those of base 10, rarely used, are the ones needing a subscript). They may not be the most convenient system to use in calculation, since (for instance) the natural logs of 1.492, 14.92, 149.2 and 1492 all have quite different appearances. However, they have a property (used by Henry Briggs in devising his original tables) which is extremely useful: if a number y is very small, then to a good approximation

loge (1 + y)  ~  y                       (10)

We can deduce this approximation using Newton's binomial theorem (Newton lived a bit later, but we won't go into that here):

(1+y) n = 1  +  ny  +  [n(n–1) / (1.2)] y2  +  [n(n–1)(n–2) / (1.2.3)] y3 +...         (11)

The point to note here is that if y is much smaller than 1, y2 is much, much smaller, and higher powers are smaller still. Say y is 1/1000, then y2 is 1/1,000,000, y3 = 1/1,000,000,000 and so forth, and if the power exponent n is relatively modest (say, 5.3), ignoring all higher powers gives the very good approximation
(1 + y) n ~ 1 + ny                 (12)

Now if x is a large number, say 1000, then to a very good approximation

(1 + 1/x) x ~ e                     (13)
and
ey ~ (1 + 1/x) (xy)                 (14)

If y is small--say of order 1/1000, or not to many times larger, xy remains a "modest" number and the binomial approximation still holds

ey   ~  [1 + (1/x)(xy)]   =   1 + y         (15)

Take natural logarithms of both sides and you end up with (10).

### Using     loge(1 + y) ~ y

Calculus shows that equation (10) is just the beginning of a formula which can be extended without limit ("infinite series")

loge(1 + y) = y – y2/2 + y3/3 – y4/4 + ...

If y is a positive number less than 1, the higher powers get smaller and smaller, and the place where we decide to stop the series determines the accuracy of our approximation (if larger than 1, higher powers gradually get bigger and the series cannot be used). Other formulas give logarithms even faster.

With very small values of y, the higher powers are very small, so equation (10)

loge (1 + y)  ~  y                       (10)
may be accurate enough, and then for any n

loge (1 + y)n  ~   ny

Like some early mathematicians, you can assume (say)

loge (1.000 001)   ~   1.000 001
and
loge (1.000 001)n  ~  n (0.000 001)

You easily reach large values of n by squaring the expression inside the logarithm again and again. Each squaring doubles the exponent, so you get numbers whose logarithms are 2, 4, 8, 16, 54, 128, 256, ... up to 1 048 576 times (1.000 001)

loge (1.000 001)1 048 576  ~   (1048576).(0.000001)  =   1.048576...

By expressing n in binary notation, these result help you get the logarithm of (1.000001) to any whole-number power n. For instance, in binary 135=10000111, so

(1.000 001)135  = (1.000 001)(128 + 4 + 2 + 1)  =
(1.000 001)128.(1.000 001)4.(1.000 001)2.(1.000 001)

all numbers whose approximated logarithms are known. Multiply those numbers and get the number whose log is (0.000135)
-------------------
However, we'd rather not impose any long calculation on you. Instead, as iv section (M-16), we enlist simple tricks to speed up the derivation of some logarithms. However all these only give "natural"logarithms to base e : to derive from them the corresponding "common" logarithm to base 10, they must be multiplied by a certain factor, derived below.

### The Conversion Factor

To convert to and from natural logs using equation (7) above, we need to know one of the factors in (9) :

log10e

Here again we allow ourselves a crude approximation. We have found that e = 2.71828.... Approximately, therefore

e  ~   2.7
With that approximation
10 e ~ 27 = 33

Taking common logs
log 10 + log e ~ 3 log 3 ~ 3 (0.475) = 1.425

Since log 10 = 1
log e ~ 0.425         (16)

The accurate figure is 0.434294... , and Henry Briggs (who used something like (10)), derived it to many more decimal points, even though he was not aware of the number "e" and its ties to compound interest. The approximation (13) should be accurate enough to improve our crude logarithms by quite a bit. Have your calculator ready! We had

210= 1024 =(1000)(1.024)
taking logs
10 log 2   =   3 + log (1.024)                 (17)

Isolating log x in equation (7), with x=1.024 in this case and using (10)

log 1.024   =   (loge1.024)(log10e)   ~   (0.024)(0.425)   =   0.0102

So from (17)
10 log 2 ~ 3.0102

log 2   ~   0.30102                 (18)

The tables give 0.3010299.. As shown in section (M-16), corresponding values of the logarithms of 4,8 and 5 follow at once. From (18)

log 8 = log (23) = 3 log 2 ~ 0.90306

Similar steps then give log 3:
34 = 81 = (80)(1.0125)
Take logs
4 log 3   ~   (1 + 0.90306) + (0.0125)(0.425)   =   1.9083725
so
log3   ~   0.477093..                 (19)
while the tables give
log 3 = 0.47712...

Thus in both cases the accuracy of our approximation jumped from 1% to the 5th decimal place, nearly as good as in the printed tables. You can explore log 7 on your own, using the approximation of section M-16. We can also seek a better approximation of the important constant log e. We have

e3   =  20.0855  ~   20                     (20)

Taking the common logarithm of both sides and using (18) above

3 loge   ~   1 + 0.30102   =  1.30102           (21)

So
loge   ~   0.43367                                 (22)

Tables give 0.4342945..., so (22) gives not as great an improvement as (18) or (19), but is still much better than (16). However, you can now "lift yourselves by your bootstraps":

20.00855 = 20.(1.004275)                 (23)

Take again logs to base 10 of (20)

3 loge   ~   1 + 0.30102   + log 1.004275           (24)

By eq. (7) and using (22)

log101.004275  =  (loge1.004275) . (log10e)   ~ (0.004275).(0.43367)   ~   0.001854

Hence
log10e   ~   (1.30102 + 0.001845) / 3   =   0.4342913...           (25)

which is accurate to 5 figures. And one can proceed even further!

### Further Exploring

If you have access to a good university library, look up "Logarithms!" in the American Journal of Physics, vol. 46, p. 101, 1978.

Author and Curator:   Dr. David P. Stern
Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .

Last updated 27 October 2008