TABLE II.-LOGARITHMS OF LEADING NUMBERS WITHOUT INDICES. N.1 0.1 1. 2. 3. I 4. 5. 6. 7. 8. 9. 100 000000000434|000868001301001734 0021660025980030291003461 003891 101004321 004750005181005609 0060380064660068940073210077481008174 1102008600 009026009451 009876010300 010724011147011570011993012415 1103||01283701325901308010141000145210149401015360015779 016197016616 104 017033017451017868|018284018700019116019532 019947020361020775 105021189021603022016022428022841023252023664024075 024486024896 106 025306025715 026125/026533026942 027350027757028164028571028978 107 029384 029789030195030600 031004031408031812032216032619033021 108 0334240338260342271034628035029035430035830036230036629037028 109037426/037825038223038620039017039414039811 040207040602040998 In table I, the logarithms are given, with indices, in columns adjacent to the columns of numbers. In table II, each figure in the row at the top may be annexed to any number in the left-hand column; the logarithm of any number thus formed, will be found at the right of the number in the column, and beneath the figure at the top. The proper index may be supplied in any case, according to the theory of logarithms. Thus, to obtain the logarithm of 1023 by this table, we find 102 in the left-hand column, and 3 in the top row; and opposite the former, and under the latter, we find 009876, the decimal part of the logarithm. Hence, log. 1023 = 3.009876. ln like manner, we find log. 104.2 = 2.017868, log. .1078 = -1.032619. CASE I. 416. To find the logarithms of numbers when their factors are in the tables. RULE.-- Take out from the tables the logarithms of the factors, und find their sum ; the result will be the logarithm required. EXAMPLES FOR PRACTICE. 1. Required the logarithm of 533.5. Observe that 533,5 =106.7X5; hence, log. 106.7 = 2.028164 .698970 2.727134, Ans. log. 5 417. To find the logarithms of numbers intermediate between the numbers on the table. Since the logarithms in any table form a regular series, we may interpolate for intermediate logarithms, by the usual formula, n(n-1) 2 If the logarithm of the given number is intermediate between the logarithms of table I, it will be necessary to take account of the first and second differences. But we may always employ table II, where the logarithms increase so slowly that two terms of the formula will give the result accurately. The first four figures of a number, counting from the left, will be called the four superior figures; and the others, the inferior figures. To apply the formula, a will represent that logarithm of the table which is next less than the required logarithm, and n will denote the inferior figures of the number, regarded as decimal. Hence the following RULE. Take out the logarithm of the four superior figures of the given number; multiply the difference between this logarithm and the next greuter in the table, by the inferior places of the number, considered as a decimal; add this product to the former result, rend the sum will be the logarithm required. EXAMPLES FOR PRACTICE. 1. Required the logarithm of 1.07632. log. 1.077-log. 1.076 = 404 = d. And putting n = .32, we have log. 1.076 = .031812 nd, = 404 X.32 = 129 .031941, Ans. 3579 : 35 = 102.25714+. 2.009451 2.009694 = log. 102.25714 log. 35 = 1.544068 3.553762, Ans. NOTE.—It is obvious that if we divide any number by its first two figures, we may obtain the logarithm of the quotient by means of table II; then we may add the logarithm of the divisor, found by Table I, to obtain the required logarithm. 3. Find the logarithm of 10724. Ans. 4.030357. 4. Find the logarithm of 10.8539. Ans. 1.035586. 5. Find the logarithm of 1021.56. Ans. 3.009264. 6. Find the logarithm of 568.53. Ans. 2.754753. 7. Find the logarithm of 3244. Ans. 3.511081. 8. Find the logarithm of 365.25638. Ans. 2.562598. 9. Find the logarithm of 132.57. Ans. 2.122415. 10. Find the logarithm of 567521. Ans. 5.753982. 11. Find the logarithm of 258.7. Ans. 2.412796. 12. Find the logarithm of 1.296. Ans. .112605. 13. Find the logarithm of 5784. Ans. 3.762228. EXPONENTIAL EQUATIONS. 418. We will now illustrate the application of logarithms to the solution of exponential equations. 1. Given 2* = 10, to find the value of x. Suppose the logarithms of both members of the equation to be taken. We shall have, by (404, 5), x log. 2=log. 10; 1 or, 3.3219+, Ans. .301030 log. 10 2 25 = 2. Given 5* = 4, to find the value of x. Raising both members of the given equation to the power denoted by x, we have 3* 72 Taking the logarithms of both members, log. 25 = x log. 3- log. 7; whence, 1.397940 -3.79899+, Ans. log.3—log. 7 477121.845098 3. Given ra* = b'c, to find the value of x. Taking the logarithms of both members of the equation ,we have, by (404, 3 and 5). log. rtx log. a = 2 log. b + log. c; 2 log. 6 + log. c-log. r whence, log. a log. 25 = EXAMPLES FOR PRACTICE. 1. Given 7* = 8, to find the value of x. Ans. x = 1.06862. 2 2. Given 5* = 30, to find the value of x. Ans. x = -.94640. 3. Given a* = 6°c, to find the value of x. 2 log. 6+3 log. c Ans. = log. a 4. Given aba d =m, to find the value of x. log.(md+c)—log. a Ans. x = log. 6 5. Given ma* = b, to find the value of x. log. a Ans. x = log. 6—log. m C. Given a® + 1 = 2c and a' —b"=2d, to find x and y. log. (c+d) log. (c-d) Ans. x = Py= log. a log. 7 7. Given 729 7292 9 log. 6 8. Given 216* = 12, to find the value of x. Ans. x= 3 log. 12 3 9. Given 516* = 12, to find the value of x. Ans, x= 3 log. 43 +3. log. 12 |