Elementary arithmetic, with brief notices of its history... by Robert Potts.

13.. log,2 =loge1 + 2 - + +. + &. =693147174...
Let u = 5, v - 4,
(111 11
log,5 =log e4+ 2 + + + &c. 1-609437892.
whence log 2 + og5 log 10 = 2-302585066....
and hence 21= I 1 '43429448... the modulus.
log 10 2-302585066
16. PROP. The diference between the locarithlms of any two successive
numbers diminishes as the numbers increase.
Lot u and u + 1 be any two successive numbers,
mn and m + d their logarithms respectively.
Then log10 (?4 + 1)= m + ci, and logot = n,
and (m + c) - 2 = log, (ut + 1) - loglo t
-log, {i -1
01.. =loglo{ 1 + }
IIence as uz increases, 1 + - diminishes,
and.. also log1o 1 + - } diminishes: consequently d also diminishes.
17. PROP. To find the mantissa of the logaritfhm of a num ber consisting of
six dijits, having gqien the mantissa of the logarithm of the nutmber conmposed
of the first five digits, and the mantissa the logarithm of the next greater
nimvber.'
1 Ex. 1. Given log0o43712=4-6406007, and loglo43713=4 6406106; to find
loglo43712 8.
Generally m1 = 1-t- (m.7 ---2) 1a10
Here a=8, m, = '6406106, m — '6406007;.. m 2-m1 = 0000099;
and (=a-2m,) = 0000099 X 8 -0000079, the increment for 8;
10 10.'. m =?2I +(m2z- }ml) 1= 6406007+ 0000079 = '6406086;
and since the numnber of integral places is 5, the characteristic is 4,
and consequently loguo 43712-8=4'6406086 the logarithm required.
Ex. 2. Conversely. Given 4'6406007 =logo43712, and 4 6406106 =logo43713,
to find the number corresponding to the logarithm 4 '606086.
Generally 00,- 71 = a
I0.2 -0M1 10
Here m7 = '640086 w,= '6406106
n = '6406007 n 6 —6406007
nl -- i = '0000079 Mq,- 0 = — '0000099
- -m '0000079 8.o. -n 0000079'- = 8 8 nearly, the sixth digit.
z2- - l 'lC000)99 10
Hence 43712+ 8 =43712 8 is the number required.