The Rev. H. Moseley on certain Elementary Formule. 261 
Therefore (by the method of indeterminate coefficients) equating 
like powers of y in these equations, we have 
Z,+4Z,=2Z, 47, +82Z,=2Z, Z,+12Z,+16Z,=2Z, 
*. Za=—12, “. 8Z,= —22, 14Z,= —12Z,—Z, 
Z,=+4Z1 Z4=—42, 
Substituting in equation (2), 
L=L,{y—3y +5y—iy' t+ ..}e 
Now when y=z, z=1 (see equation (1)) ; 
*, L=Z,{z—}2?+42—72....} 
—ty2+byP—lyAt .... 
oe ae ie, 
which is the formula for determining the logarithm (2) of a given 
number (1+y) to a given base (1+z2). 
Next, to determine the value of y in a series ascending by 
powers of z, assume . 
y=La +L" 2? + 2M a8 + ZIMA t+ ok 
o. L4+ysl14+Zlae+ 2" 224 2M 4 ZoAt oe. 2. (4) 
Squaring both sides, 
(l+y)?=14 2Z!a + (2Z" + Z!*) x? + (2Z" 4+ 2Z'Z")a8 
+ (QZ! 4 27/2" 4 "at... 
But (by equation 1) z becomes 2% when y becomes (1 +y)?, 
. by equation (4), . 
 (L+y)?=14 22a + 42" a? + 82a + L6ZMaA+ .... 
Equating the coefficients of like powers of 2, 
AZ! = 27!" 4 D2) SLM" = 27!" 4 O/T! | LEZ = 7M 4 OZIZIM 4 Zitz 
: Z!2 BYAL — VA 7! VAL 
|| — ee Te ae 
er 2 shih AP pl chee 
1.2.8 ema e 
12.5.4 
Now if the numerator of the second member of equation (3) 
be represented by Y, and its denominator by Z, and if the result- 
ing value of # be substituted in equation*(4), we have 
7! Zii2 2 23 
l+y=1+ Zi tpmaY + 5s mie ola. 
But Y is a multiple of y. No other term than the second in 
the above series contains therefore y in the first power only, so 
