NAPIER COMMEMORATIVE LECTURE. 5D 
_We have logy sz — logws; 
=r log " _— Pr log ah 
es aaa 
— r log, 2 
r log ss 
= log. (1+ = *) 
= — rlog. fips 1 92)); 
$1 
Then it follows as in Theorem I. that 
Pe ee Iooxss — ‘loaner so! ¢ (Ee), 
82 $1 
We can now grasp Napier’s method. He takes, to begin 
with, his First Table. The logarithm of the first term is 
zero. Theorem I., above, shows that the logarithm of the 
second term lies between r—s and (r — s)-5 that is, between 
1°000,000,0 and 1°000,000,1. If this is taken as the arith- 
metic mean of these two numbers, we have for the 
logarithm of this term 1°000,000,05. The remaining terms 
in the table form with these twoa G.P. Therefore, the 
logarithms have a common difference 1°000,000,05. Hence 
the logarithm of the last term is 100°000,005. 
We pass from the logarithm of the last term in the First 
Table, to that of the second term in the Second Table, by 
the help of Theorem II. The logarithm of this term is 
found to lie between two limits, close to each other, and 
approximately as above, it is to be taken as 100°000,500,0. 
As the first term is radius, its logarithm is zero; and as 
the terms in the table form a G.P., their logarithms are 
obtained by adding step by step the common difference. 
In this way the logarithm of the last term is given as 
3,000°025,000,1. | 
By a process exactly similar, it is easy to pass from the 
last term of the Second Table to the second term in the 
first column of the Third Table. The first term in that 
