TABLE II.-LOGARITHMS OF LEADING NUMBERS WITHOUT INDICES. N. 0. 1. 2. 3. 4. | 5. 6. 7. 8. 9. 100 000000000434 000868 001301 001734 002166 002598 003029003461 003891 101 004321 004750 005181 005609006038 006466 006894007321 007748008174 102 008600 009026 009451 009876 010300 010724011147011570011993 012415 103012837013259013680 014100 014521 014940015360 015779 016197 016616 104 017033 017451 017868 018284018700 019116019532 019947 020361 020775 105 021189 021603022016 022428 022841 023252 023664024075 024486024896 106 025306025715026125 026533 026942 027350027757 028164 028571 028978! 107 029384 029789030195 030600 031004031408031812032216032619 033021 108 033424033826 034227 034628035029 035430035830036230 036629 037028 109 037426037825038223|038620 039017 039414/039811 040207|040602|040998| 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. 1n 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, and find their sum; the result will be the logarithm required. 1. Required the Observe that hence, EXAMPLES FOR PRACTICE. logarithm of 533.5. 533.5 106.7X5; log. 106.72.028164 log. 5 = .698970 2.727134, Ans. 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, 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 greater in the table, by the inferior places of the number, considered as a decimal; add this product to the former result, und the sum will be the logarithm required. EXAMPLES FOR PRACTICE. 1. Required the logarithm of 1.07632. This number is found between 1.076 and 1.077; hence log. 1.077—log. 1.076 = 404 = d1· In order to make use of table II, we proceed thus: 357935 102.25714+. 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. 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), 2. Given 5* = 4, to find the value of x. Raising both members of the given equation to the power denoted Taking the logarithms of both members, log. 25 x log. 3—x log. 7; whence, 3. Given ra2 = b3c, to find the value of x. Taking the logarithms of both members of the equation,we have, by (404, 3 and 5). |