The quotient of .397940 divided by .008363=47.5+ .6975 .8263 =47 nearly. The product of .017033 into 35= 35 .596155 The sum of log. of 7.000,000, and .596155=7.441253 Example 9. The logarithm of 15000= The logarithm of 9000= The diff. between 4.176091 and 3.954243= 4.176091 3.954243 .221848 Dividing .221848 by 12 the quotient is .018487, whose nat ural number is 1.044 very nearly. Therefore the rate of increase would be .4 per cent. .501780. The log. of 10,000,000 7.000000. 7.000000+ .501780 7.501780, whose natural number is 31,750,000. Example 11. 1+ 7 607 600 500 whose log. 2.783189 .005038. .005038×20=.100760. The log. of 10,000,000 =7.000000. 7.000000+.100760=7.100760, whose natural number 12,610,000. By Art. 34, the proportional part to be added to 12,610,000, is 13084, which added to the sum is 12,611,3084. Example 12. The logarithm of 4,500,000-6.653212, and the log. of 3,000,000=6.477121, and their difference is .176092. The ratio of births and deaths being and the ratio of immigration and emigration being 1 1 20 30 1 1 40 50 =.009309. The quotient of .176092, (as obtained above,) divided by .009309 18.9 with a remainder. In the next four examples, the period in which the population will double is given, and the several successive periods are of course in geometrical progression, where P=the first term, A=the last term, and n the number of periods. Hence, (Alg. 439.) A=P×2′′: Also, log. A=log. P+ nxlog. 2. Example 1. 1000÷25=40=the number of periods. The logarithm of 2, the 1st term=.301030, and this multiplied into 40 the number of periods, the product becomes 12.041200 Adding .301030, the log. of the 1st term, .301030 12.342230 The natural number belonging to 12.342230=2,199,000,000,000. By Art. 34, we obtain 102,278,129, which added to the above gives the answer nearly. Example 2. 300÷÷20=15=the number of periods. The logarithm of 2.301030 whose product into 15=4.515450, to which add .778151 the logarithm of 6, the first term, and the sum becomes 5,293601, whose natural number is 196,600. Example 3. The number of years from 1820 to 2020 would be 200, which divided by 25=8=the number of periods. The logarithm of 9,625,000 6.9834007. The log. arithm of 2.3010300, whose product into 8 is 2.4082400, which added to 6.98340079.3916407, the natural number of which is 2,464,000,000. Example 4. 1650÷50=33=the number of periods. The logarithm of 2 3010300, which multiplied into 33 = 9.9339900, and added to .3010300, the sum becomes 10.2350200, whose natural number is 17,170,951,759. EXPONENTIAL EQUATIONS. Art 63. Example 2. Reduce the equation 4x=100x3. By the given equation, 4x=100x3 Dividing each side of the equation by 4, x=25x3 Dividing each side by x3, x2-3=25 Taking the log. of each side, log. xxx-3=log. 25 Assume x=4.5, and 4.6: then the log. of 4.5=.6532125, which multiplied by 1.5 becomes .97981875, which subtracted from 1.3979400, the logarithm of 25, the remainder becomes .41812125, the 1st error. Again, the log. of 4.6= .6627578 which multiplied into 1.6, becomes 1.06041248: This subtracted from 1.3979400, the log. of 25, the remainder becomes, .33752752, the 2d error. Subtracting the 2d error from the 1st, we find their difference to be .08059373: Hence, by the rule; .08059373.01::.38752752.418 .418, the correction, added to the least assumed number, makes it 4.918, which equals the answer very nearly. Example 3. Reduce the equation x=9x. By the given equation, x=9x. Dividing by x, x2-1-9. Assume x=1.5 and x=1.6. The log. of 9=.9542425. The log. of 1.5=.1760913, which multiplied into 5= .08804565; this subtracted from .9542425, the log. of 9, leaves .8661969, the 1st error. The log. of 1.6=.2041200, which multiplied into .6=.12247200; this subtracted from .9542425, leaves .83177050, the 2d error. .86619695.83177050=.0342645, the difference of the errors. By the rule, .9342645 1.::.83177050.243. 1.5+.243=1.743 =1.8 nearly. In the second place, assume x=1.8, and 1.9. Proceeding as above, we find the first error to be .75002450, and the second .70336426, and their difference .04766084: Hence, by the rule, .04766084.1::.7036426.2 nearly. 1.8+.2=2. Example 5. Find the value of x, in the equation, =m. 1. Clearing of fractions, ba+d=cm 2. Transposing d, ba=cm-d ba+d C 3. Taking the logarithm of each side, log. ba*=log. (cm-d) 4. Dividing by the log. of b, a=log. (cm-d)÷log. b 5. By article 45, log. a ×x=log. (cm-d)-log. b 6. Dividing by log. a, x=. log. (cm-d)-log. b log. a Note. It will be observed that some of the answers found by the methods here pursued, differ somewhat from those given by Day. This, for the most part, is to be attributed to different tables which are used in the different operations. In some cases, however, a change in the order of the several steps of the solution may produce a small difference. In most instances, it would appear, that the results given in the text-book, are obtained from using logarithms carried to only 5 or 6 places of decimals. The object should be, to obtain the most correct answers, and in doing this, it is not always safe to make those which are appended to the questions a criterion, and that, for reasons already alluded to. From the complication that attends solutions of this nature, wholly arithmetical, the ordinary methods of proof cannot often be resorted to. DIFFERENT SYSTEMS OF LOGARITHMS, AND THE METHODS OF COMPUTING THEM. Art. 65. The first four or five pages of the section now commenced, furnish us with no examples or problems, but present some difficulties not readily understood. We shall give an explanation of the whole in one continued connection, after the manner of a solution of a problem, numbering the succes sive steps for convenience of reference in different parts of the operation. Explanation of the process of finding the base of Napier's system; comparison of his system with that of Briggs; and the computation of Napier's logarithms. If a N, then x is the logarithm of N, (Art. 2.) To find the value of x, in a series, let the quantities a and N be put into the form of a binomial, by making a=1+b, and N= 1+n: 1. Then, by article 2, (1+6)=1+n 2. Extracting the root y of both sides, (1+b); =(1+n}} 3. Expanding the equation, (Alg. 475,) 1+2 (6)+ y y x 1 4. Cancelling and within the parentheses, 1+b— 2 y y y y |