468. We have explained in Arts. 459-461 the advantages of the Common System of Logarithms, which may be stated in a more general form thus :

Let A be any sequence of figures (such as 2.35916), having one digit in the integral part.

Then any number N having the same sequence of figures (such as 235 916 or '00235916) is of the form A × 10", where n is an integer, positive or negative.

Therefore log10 N=log10 (4 × 10′′)=log10 A+n.

Now A lies between 10° and 10', and therefore log A lies between 0 and 1, and is therefore a proper fraction.

But log10 N and log10 A differ only by the integer n;

logo A is the fractional part of log10 N.

Hence the logarithms of all numbers having THE SAME SEQUENCE OF FIGURES have the same mantissa.

Therefore one register serves for the mantissa of logarithms of all such numbers. This renders the tables more comprehensive.

Again, considering all numbers which have the same sequence of figures, the number containing two digits in the integral part=10. A, and therefore the characteristic of its logarithm is 1.

Similarly the number containing m digits in the integral part=10m. 4, and therefore the characteristic of its logarithm is m.

Also numbers which have no digit in the integral part and one cypher after the decimal point are equal to A. 10-1 and A. 10-2 respectively, and therefore the characteristics of their logarithms are -1 and -2 respectively.

Similarly the number having m cyphers following the decimal point =A.10−(m+1);

.. the characteristic of its logarithm is −(m+1).

Hence we see that the characteristics of the logarithms of all numbers can be determined by inspection and there

fore need not be registered. This renders the tables less bulky.

469. The method of using Tables of Logarithms does not fall within the scope of this treatise, but an account of it may be found in the Author's work on ELEMENTARY TRIGONOMETRY.

470. We proceed to give a short explanation of the way in which Logarithms are applied to the solution of questions relating to Compound Interest.

471. Supposer to represent the interest on £1 for a year, then the interest on P pounds for a year is represented by Pr, and the amount of P pounds for a year is represented by P+Pr.

472.

To find the amount of a given sum for any time

at compound interest.

Let P be the original principal,

r the interest on £1 for a year,

Then if P1, P2, P3... P2 be the amounts at the end of 1, 2, 3 ... n years,

P1=P+Pr=P (1+r),

P=P+Pr=P1 (1+r) = P(1+r)2,

P=P2+P2r=P2(1+r)=P(1+r)3,

P=P(1+r)".

473. Now suppose Pn, P and r to be given: then by the aid of Logarithms we can find n, for

474. If the interest be payable at intervals other than a year, the formula P2=P (1+r)" is applicable to the solution of the question, it being observed that r represents the interest on £1 for the period on which the interest is calculated, halfyearly, quarterly or for any other period, and n represents the number of such periods.

For example, to find the interest on P pounds for 4 years at compound interest, reckoned quarterly at 5 per cent. per

N.B.-The Logarithms required may be found from the extracts from the Tables given in pages 329, 330.

1. In how many years will a sum of money double itself at 4 per cent. Compound interest?

2. In how many years will a sum of money double itself at 3 per cent. Compound interest?

3. In how many years will a sum of money double itself at 10 per cent. Compound interest?

4. In how many years will a sum of money treble itself at 5 per cent. Compound interest?

5. If £P at Compound interest, rate r, double itself in n years, and at rate 2r in m years: shew that m:n is greater than 1:2.

6. In how many years will £1000 amount to £1800 at 5 per cent. Compound interest?

7. In how many years will £P double itself at 6 per cent. per ann. Compound interest payable half-yearly?

APPENDIX.

475. The following is another method of proving the principal theorem in Permutations, to which reference is made in the note on page 289.

To prove that the number of permutations of n things taken r at a time is n. (n-1)......(n−r+1).

Let there be n things a, b, c, d ............

......

If n things be taken 1 at a time, the number of permutations is of course n.

Now take any one of them, as a, then n-1 are left, and any one of these may be put after a to form a permutation, 2 at a time, in which a stands first: and hence since there are n things which may begin and each of these n may have n- -1 put after it, there are altogether n (n-1) permutations of n things taken 2 at a time.

Take any one of these, as ab, then there are n − 2 left, and any one of these may be put after ab, to form a permutation, 3 at a time, in which ab stands first: and hence since there are n (n − 1) things which may begin, and each of these n (n − 1) may have n- -2 put after it, there are altogether n (n-1) (n-2) permutations of n things taken 3 at a time.

If we take any one of these as abc, there are n-3 left, and so the number of permutations of n things taken 4 at a time is n. (n-1) (n − 2) (n − 3).

So we see that to find the number of permutations, taken r at a time, we must multiply the number of permutations, taken r-1 at a time, by the number formed by subtracting r-1 from n, since this will be the number of endings any one of these permutations may have.

Hence the number of permutations of n things taken 5 at a time is

n (n-1) (n-2) (n−3) × (n−4), or n (n − 1) (n-2) (n−3) (n−4); and since each time we multiply by an additional factor the number of factors is equal to the number of things taken at a time, it follows that the number of permutations of n things taken r at a time is the product of the factors

n. (n-1) (n-2).............. (n−r+1).