204. To find the logarithm of any given number. Let N be any given number whose logarithm is a, in a system whose base is a; then aN and a*2=N1; hence, by the exponential theorem, we have from the last equation 22 1+Axz+A2x2 z2 + =1+A1≈+A ̧2. + 1.2 ... 1.2 and equating the coefficients of z, we get Ax=A1; hence .... because A =(a−1)—}(a −1)2+(a —1)3— . . . in the expansion of aš. A1=(N—1)—↓(N—1)2 +↓(N−1 )3 —- ... in the expansion of N'. and 205. To find the logarithm of a number in a converging series. We have seen that if a=N1, then x= _ (N1—1)—↓ (N1—1)2 + (N1—1)3—↓(N1−−1)1+ . . . Now the reciprocal of the denominator is the modulus of the system; and, representing the modulus by M, we have } x=log. N1=M{(N1—1) —; (N1—1)2 + ↓ (N1— 1 )3— 4 (N1—1)' + . . . } Put N=1+n; then N1-1=n, and we have log. (1+n)=M(+n—¥n2+\n3—¥n1+}n3— . . .) Similarly log. (1—n)=M(—n—\n2 —}n3—\n1—}n3— . . .) 1 1 .. log. (P+1)=log. P+2M {2P+1+3(2P+1)2+5(2P+1)3° ... Hence, if log. P be known, the log. of the next greater number can be found by this rapidly converging series. 206. To find the Napierian logarithms of numbers. In the preceding series, which we have deduced for log. (P+1), we find a number M, called the modulus of the system; and we must assign some value to this number before we can compute the value of the series. Now, as the value of M is arbitrary, we may follow the steps of the celebrated Lord Napier, the inventor of logarithms, and assign to M the simplest possible value. This value will therefore be unity; and we have log. (P+1)=log. P+2{2P+1+ 1 1 {2P+1 + 3(2P+1)3 + 5(2P+1)3 + · Expounding P successively by by 1, 2, 3, 4, &c., we find In this manner the Napierian logarithms of all numbers may be computed. and by means of this equation we can pass from one system of logs. to another, by multiplying, the log. of any number in the system whose base is a, by the reciprocal of log. b in the same system; and thus we shall obtain the log. of the same number in the system whose base is b. Let the two systems be the Napierian and the common, in which the base of the former is €2.718281828... and the base of the latter is b=10, the base of our common system of arithmetic; then we have b=10, and a=e=2.718281828... and consequently if N denote any number, we shall have and the modulus of the common system is, therefore, M= 1 2.3025851 =43429448 .. 2 M=86858896 Hence, to construct a table of common logarithms, we have + 1 1 1 log. (P+1)=log. P+86858896 + 2P+13(2P+1) 5(2P+1)5 + } and thus we have a series for computing the logs. of all numbers, without knowing the log. of the previous number. EXAMPLES IN LOGARITHMS. (1.) Given the log. of 2=0·3010300, to find the logs. of 25 and ‘0125. .. log. 0125=log. 1-log. 10-3 log. 2——1—3 log. 2=2-0969100 (2.) Calculate the common logarithm of 17. Ans. 1.2304489. (3.) Given the logs. of 2 and 3 to find the logarithm of 22.5. Ans. 1+2 log. 3—2 log. 2. (4.) Having given the logs. of 3 and 21, to find the logarithm of 83349. Ans. 6+2 log. 3+3 log. 21. ON EXPONENTIAL EQUATIONS. 209. An exponential equation is an equation in which the unknown appears in the form of an exponent or index; thus, the following are exponential equations: X When the equation is of the form ab, or abc, the value of x is readily obtained by logarithms, as we have already seen in Art. 201. But if the equation be of the form xa, the value of x may be obtained by the rule of double position, as in the following example: Ex. Given æx = 100, to find an approximate value of x. The value of x is evidently between 3 and 4, since 33 27 and 44 = 256; hence, taking the logs. of both sides of the equation, we have x log. x = = log. 100=2* First, let a1 = 3·5; then error =-0957620 Then, as the difference of the results is to the difference of the assumed numbers, so is the least error to a correction of the assumed number corresponding to the least error; that is, 098451 1 :: 002689: 00273; ··00273 = 3.59727, nearly Again, by forming the value of x for x= 3.5972, we find the error to be ―0000841, and for x=3.5973, the error is +·0000149; hence, as 000099 0001 0000149 0000151; therefore x 3.5973-0000151 = 3·5972849, the value nearly. EXAMPLES FOR PRACTICE. (1.) Find a from the equation x* = 5. Ans. 2.129372. Ans. 8-6400268. Ans. 4-8278226. In equations of this kind, the following method may be adopted:-Let x=a; then a log. = log. a; put log. a=y, and log. a=b; then xy=b, and log. x+log. y=log. b; hence y+log. y=log. b. Now, y may be found by double position, as above, and then becomes known. When a is less than unity, put = 1 and a; then we have by=y ... y log. 6=log. y, and if log. b=c, and log. y=z; then cy=x, and log. c+log. y=log. 2, or log. c+2=log. z. Hence z may be found by the preceding method, and then y and a become known. INTEREST AND ANNUITIES. THE solution of all questions connected with interest and annuities may be greatly facilitated by the employment of algebraical formulæ. In treating of this subject we may employ the following notation: Let p pounds denote the principal. interest of £1 for one year. interest of p pounds for t years. amount of p pounds for t years at the rate of interest denoted by r. the number of years that p is put out to interest. SIMPLE INTEREST. PROBLEM I.—To find the interest of a sum p for t years at the rate r. Since the interest of one pound for one year is r, the interest of p pounds for one year must be p times as much, or pr; and for t years t times as much as for one year, consequently, PROBLEM II. To find the amount of a sum p laid out for t years at simple interest at the rate r. The amount must evidently be equal to the principal together with the interest upon that principal for the given time, Hence, S = p + ptr = p(ltr)... .(2) Example 1. Required the interest of £873. 15s. for 23 years at 4 per cent. per annum. |