presses the constant relation wbich the logarithms of n have to each other in the systems to which they belong.

But the only system of these numbers, deserving of notice, except that above described is the one that furdishes what have been usually called hyperbolic or Neperian logarithms, the base é of which is 2.718281828 459.

Hence, in comparing these with the common or tabular logarithms, we shall have, by putting a in the latter of the above formula = 10, the expression

1
log. n=l.nx or I. n=log. n Xl. 10.

1.10' Where log in this case, denotes the common tabular logarithm of the number n, and I. its hyperbolic logarithm ; the constant factor, or multiplier,

which is

1.10 1

or its equal .4342944819, 2.3025850929' being what is usually called the modulus of the common system of logarithms. (f)

PROBLEM 1.

To compute the logarithm of any of the natural numbers 1, 2, 3, 4, 5, &c.

(f) It may here be remarked, that, although the common loga. rithms have superseded the use of hyperbolic or Neperian loga. rithms, in all the ordinary operations to which these numbers are generally applied, yet the laiter are not without some advantages peculiar to themselves ; heing of trequent occurrence in the application of the fluxionary Calcnlus, to many analytical and phy. sical probiems, where they are required for the firding of certain Avents, which could not be so readily determined without theirassistance; on which account, great pains have been isken to calcu. lats tables of lypurbolic logarithms, to a consists Xiunt, chief for this purpose Mr Barlow, in a Cute Sinu v 1.the. matical Tables lately published, has given them tor the first 10000 numbers.

1. Take the geometrical series, 1, 10, 100, 1000, 10000, &c. and apply to it the arithmetical series, 0, 1, 2, 3, 4, &c. as logarithms.

2. Find a geometric mean between 1 and 10, 10 and 100, or any other two adjacent terms of the series, betwixt which the number proposed lies.

3. Also, between the mean, thus found, and the nearest extreme, find another geometrical mean, in the same manner ; and so on, till you are arrived within the proposed limit of the number whose logarithm is sought.

4. Find, likewise, as many arithmetical means between the corresponding terms of the other series, 0, 1, 2, 3, 4, &c. in the same order as you found the geometrical ones, and the last of these will be the logarithm answering to the number required.

EXAMPLES.

1. Let it be required to find the logarithm of 9.
Here the proposed number lies between 1 and 10.
First, then, the log. of 10 is 1, and the log of 1 is 0.
Therefore (10x1)=v10=3.1622777 is the geome-
trical mean;

And y(i+0)=1=.5 is the arithmetical mean ;-
Hence the log. of 3.1622777 is .5. .

Secondly, the log. of 10 is 1, and the log of 3.1622777 is .5.

Therefore (10 X3.1622777)=5.6234132 is the geometrical mean ;

And 1(1+.5)=.75 is the arithmetical mean ;
Hence the log. of 5 6234132 is .75.

Thirdly, the log. of 10 is 1, and the log of 5.6234132 is .75 ;

Therefore (10 X 5.6234132)=7.4989422 is the geometrical mean ;

And ](1+.75)=.875 is the arithmetical mean ;

Hence the log. of 7.4989422 is .875.

Fourthly, the log. of 10 is 1, and the log. of 7.4989422 is :875 ;

Therefore 7/10X 7.4989422) = 8.6596431 is the geometrical mean,

And 1(1+.875)=.9375 is the arithmetical mean;
Hence the log. of 8.6596431 is .9375.

Fifthly, the log. of 10 is 1, and the log. of 8.6596431 is .9375.

Therefore v (10 X 8.6596431)=9.3057204 is the geometrical mean,

Aod (1+.9375)=.96875 is the arithmetical mean;
Hence the log. of 9.3057204 is .96875.

Sixthly, the log. of 8.6596431 is .9375, and the log. of 9.3057204 is .96875 ;

Therefore ✓(8.6596431X9.3057204)=-8.9768713 is the geometrical mean,

And (.9375+.96875)=.953125 is the arithmetical mean ;

Hence the log. of 8.9768713 is .953125. And, by proceeding in this manner, it will be found, after 25 extractions, that the logarithm of 8.9999998 is .9542425 ; which may be taken for the logarithm of 9, as it differs from it so little, that it may be considered as sufficiently exact for all practical purposes.

And in this manner were the logarithms of all the prime numbers at first computed.

RULE II.

When the logarithm of any number (n) is known, the logarithm of the next greater number may be readily found from the following series, by calculating a sufficient

+

3

2+1 +1x++;&c.

1

1

}

number of its terms, and then adding the given logarithm to their sum.

1

1 Log. (n+1)=log.n+m'{

-+
2n+1 '3(2n+1)3''5(2n+1)5
1
+
+7(2n+1); *9(2n+1)o +11(2n+1)'

Or
M'

Зв
Log. (n+1)=log. n + { -

2n+1 3(2+nl)2' 5(2n+1)3 50 70

SE -to

-&e. 7(2n+1)

29(2n+1)? 11(2n+1)2 Where A, B, C, &c. represent the terms immediately preceding those in which they are first used, and n'= twice the modulus=.8685889638 (8)

А

#

+5(2+n1)?

+

+

EXAMPLES.

1. Let it be required to find the common logarithm of the number 2.

Here, because n+1=2, and consequently n=1 and 2n+1=3, we shall have .8685889638

=.289529654 (A) 2n + 1

3 .289529654

=.010723321 (B) 3(2n+1)2 3.32

M

A

(8) It may here be remarked, that the difference between the logarithms of any two consecutive numbers, is so much the less as the numbers are greater ; and consequently the series which comprises the latter part of the above expression, will in that case converge so much the faster. Thus log. n and log. (n+1), or its equal log. n + log. (1+), will, obviously, differ but little from

, each other when n is a large number.

Which logarithm is true to the last figure inclusively.

2. Let it be required to compute the logarithm of the number 3.

Here, since n+1=3, and consequently n=2, and 2n+1=5, we shall have .868588964

=.173717793 (A) 2n+1

5

M

A

.173717793

3.52

=.002316237 (B)

3(2n+1)2

3B
5(2n+1)2

3+.002316237

=.000055590 (c) 5.52