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

27

Conversely,
log 27 =3 log 3, and log 3 = log 27
log 4- -4 log 3- l-log 2, and log2=4 log 27-1-lo-o 4-.
Then log 16=4 log2=4{ log27-1-log 4-}
and log 15 =1-+logr 3-log 2=2+log 4 1 —log 27.
8. As the preceding.
1V.
These equivalents are shown to be true bly Arts. 2, 3, 4.
V.
1. 2'0629957. 2. *1661035. 3. 1-2419383. 4. 19796043. 5. 19599946.
6. -4507275. 7. '2544670.
YI.
If the logarithms of the two melmbelrs of each equation be taken, the values of x
may be found by help of the logarithms of the prime numbers given in the note to
Art. 5, p. 4.
VII.
By taking the logarithms of the members of these equations the values of the
unknown quantities may be found in terms of the quantities supposed to be of
known values.
log a +x logc=log b+x log d: mx log (1 -+c-l)==log1 b: z xlog a -+nx log b=log C:
imx log (,Se - mC) = log n: x log a+ —nx. log c = (mzx- z) log 1
(3:x —2) loga+(5x+3)logb=(x-1)logc+(4x-5)logd: x =2:.alog (c+b)=log (C-b): ~x(log b-logc.)O=loga+logb: x=-.
izx log, a = lx log b+-x log c+ log (1 +c c).
VIII.
HIere xloga-+ylogb=logp, xlogc+?ylogd=logq, are two equations of the first
degree, with two unknown quantities. The same remark applies to the next two
equations: Here cx+Y = ax a= ax bx =b. By equating values of x, (x-y)2 ==n, and
=y=F —l. By equating values of x or y, (x++y)2 =4q2 and ==y2, from which x and
p may be found. Divide the first by the second equation. Taking the logarithms
of each equation, (loga)2q-loga.logz=(logc)2+logc.logy and logx log =loga logy,
i)y substitution logx and logy may be determined.
n m
X.= x(n d- andy= t.)..-.
IX.
1. Take the logarithms of each equation, and there results three equations Nwith
three unknown quantities.
2. The same remark applies to the sets 2 and 3.
X.
From the first equation ax may te found as from an ordinary quadratic. The
next six equations when reduced are ar- cax = -, ax - a'' =1, ax — 2c6x =8,
~m' - 2c-' log a = (log a)2, x2 (log a- log b-9 log c) + x logb = r log a,
logex(m logeao-log - logb) = 0.
4x =7andl. 3 =2187 and-2188. 4=-64 and-80. 10=10 and - 6. The
reduced equation is 55<-5' = O0, 3.3 = 2 and 1. 3x-27 and-21. 2x =256 and- 56.
The equation loge [-7/a-j_- =, which expressed in an exponential form
Va-1 -— aJ