TEI

498

TBI

In VI. substitute — for z, and we obtain,x

_ . x + x (x a? x* \

X ' L x M Car 2® ,+ 3:e» 4x i+,& )

,X X ’2 x 3 X* -v

XI. L* (x + x) — l* x = m ( - - + tp -3 — -j—. +, &c.) and, by transposition, we obtain,

v • ' V x 2 x 2 o x' 4 x* J

XII. l* (*+?) = 1’* + M (-f-—Is- + T?-’ &C - )

X

In IX. substitute — for z. and we obtain,x

x + x „ /x x 3 , a s x 7 \

x — x \x 3 x 3 5 X? 7 x 1 /

V - 1

In IX. for z substitute-and we obtain,

XIII. ^ = 2 M (^+^ + ^ + ^+,&c.)

ps le#®'

This series will always converge, whatever be the value of v ; but as v is greater, the degree of convergency ^ e ? 0B ^ a icul a 'and, therefore, the logarithms of small numbers may be easily found; but, in the computation of large numbers, t > e neC estion would extend to too great a length to be of any practical utility. In order, therefore, to save trouble, it becomesary to derive the logarithm of one number from that of another.

In XIII. for v substitute v + 1, and we obtain,

xnr. -• (i + .) = s «( TJ ^ ) + +• te )

In IX. substitute ^— for z, and we obtain,

2 —x

XT. „• j-Lj = 2. + 5^. +. to.)

In XIII. let v = - and we obtain,

z

+ 5 (2z > + 7\ > + ’ &C ~) thereflM " i >

XVL z^ ~ 2m \1(2z + z) + 3 (2z + zf ^ 5(2 z+z)‘

XVII. !■(« + ») -i * = 2 M (+ 3(2 - + ? ) + to.) therefore,

XVIII (.+ .)= . + 2 K + + , fa.)

Practical Example.

The logarithm of 18 being 1.255273, to find the logarithm 19 by the series XIX.Here * = 18, z = 1;

therefore, r.’19 = ..' 18 + 2>t (i + +, to.)

= r.’ 19 + 2 X .4342945 ^ +. to.)

2 X .4342945 = .868589 (37)’ = 1369.