86858896 (+3.20 = 2 log 2 10 log 2 1 0.3010300 0.4771213 =0.6020600 log 10-log 2 = 1 — log. 2 - =0·6979700 = log 3+log 2 86858896 = log. 23 9 = 10 = log. 1 6 +3.63 +5.6+7.67) +log5=0.8450980 =3 log.2 1 1 5 =0.9030900 0.9542425 = 1.0000000 11 = ·-86858896 10 +3.103 + 5.10+&c.+log 9—1·0413927 17= 86858896 3 (18 +3.16 +5.168+ &c. ) + log 15—1.2304489 16 18= log. 9+ log.2 = 1.2552725 19 = ·85858896(1+3.15 +5.15+&c.) +log17=1-2787536 +5.22+&c.) +log21=1·3617278 The logarithms of 24, 25, 26, 27, 28, can also be found by addition; but the log of 29 must be calculated by the series, as must the log of every prime number. See Index. The logarithm of 10 and log 10=1; hence the logarithm of any number between 1 and 10 is less than 1. Also the log of 100=2; therefore the log of any number between 10 and 100 will be 1, with some decimal part annexed The log of any number between 100 and 1000 will be 2, with some decimal part, These whole numbers are called the indices of the numerals, and can be readily found from the rule just given; they are not given in logarithmic tables, but are left to be supplied by the student. The great advantages attending the common system of logarithms arise from the facility with which we can find the index of the logarithm of any number; and that if any number be multiplied or divided by 10, 100, &c., its logarithm is increased or diminished by 1, 2, &c.; the decimal part remaining the same. On the method of finding the Increase of Population in any country under given circumstances of Births and Mortality. 126. Let P represent the population of any country at a given period; 1 m the fractional part of the population which die in a year, (or ratio of mortality ;) 1 the proportion of births in a year. b Then, if A represents the state of the population at the end of n years, log A=log P+nXlog(1+b). mb The rate of increase of population in one year 1 1 mb Ъ m mb m-b mb the population at the end of the first year. state of But it is increased every year in the same proportion; there m-b P(1 fore 1: 1+ :: P(1+: mb m population at the end of the second year. In the same manner we may prove that the state of the population at the end of n years will be P(1+ P(1+) Hence A=P(1+); and log.A log(1+). From which we deduce = log. P+n × log(1+ log.P = log A — n×log(1+ m ; log.(1+ mb m mb n mb Of the quantities A, P, m, b, n, any four being given, the fifth may therefore be found. 1. Suppose the population of Great Britain in the year 1800 to have been ten millions; that th part die annually; that the births are to the deaths as 40 to 30; and that no emigration takes place during the present century. What will be the state of its population in the year 1900? Here P=10000000, n=100, m-40, b-30, and =log 10000000X100Xlog 121 7.3604200 = log.22931000. Hence A-22931000. Ans. 2. Suppose the population of France in the year 1792 to have been 27000000; the ratio of mortality during the eighteenth century to have beenth, and the number of births th. What was the state of its population in the year 1700? Here A=27000000, n=92, m=30, b=26, m-b 196 ..1+ mb 195 Log P=log A―nXlog (1+ m mb =log 27000000-92xlog 18-7.2269858-log 16864980, nearly; therefore P=16864980. 3. Suppose the population of North America to have been five millions in the year 1800; in how many years will it amount to sixteen millions, taking the ratio of mortality at th, and the annual proportion of births at th? Here A 16000000, P-5000000, m45, b-24; sons. log 16000000—log 5000000 = 60.39.years. at 500000 perFrom the bills th part of the been kept of the 4. A province in the year 1760 was estimated In the year 1800 it amounted to 720000. of mortality it appeared that, upon an average, population had died annually; no register had births. What was the annual proportion of them during this pe riod? Here A=720000, P-500000, n=50, n=40. Hence 1 + 506 n -.0039590-log 1.009. 9 506 50-6 9 1.009=1+: and 506 1000 ... 50000-1000b-450b, or b=50000=34.4. The annual proportion of births therefore was about 127. But in any country, under given circumstances of births m-b and mortality, the fraction is always a given quantity. Let it be represented by then the relation between the four quan tities A, P, p, n, is expressed by A = P(1 = P(1+2). If A=mP,we have mP—P(1+1); or m=(1+1); and taking the logarithm log m=n × log (1+1). Hence we deduce the six following formulæ ; I. Log. A=log P+n log (1 +2). · II. Log P=log A—n log (1+1). IV. Log (1+1) log A-log P III. n= log (1 + 2) for finding the period in which the population would be increase m times. the population would be increased m times in n years. The following questions are intended to illustrate the use of these formulæ, in the order in which they stand. 5. Suppose the population of a country to begin with six persons, and to increase annually by th of the whole; what will be the state of its population at the end of 200 years? A=log P+n log (1+1)=log 6 + 200 × log (1.0625) = log 6.0439510 log 1106400. Hence A-1106400 persons. 6. If, as stated in the third Example, the population of North America was five millions in the year 1800, and the rate of increase has beengths for fifty years previous. What was the state of its population in the Ans. 1908930 persons. 7 367 1 7 1750? year 1 Here A=500000, :; 1+. P 360 360 360 and n=50; 367 .. P=log 1—n X log (1 +2)= log 5000000—50×log 360 p log 5000000-50X (log 367-log 360) 6.2807900=log 1908930; hence A=1908930 persons. 7. An empire to be 40 millions, and the annual increase th; how long will it be before it amounts to 50 millions? Ans. 43.6 years. log. A-log.P Here A-50000000, P-40000000; p=5; 1+2=1+85=188 log 50000000--log 40000000 (log 196-log 185) ...n=n= .096910 .002221 log. (1 + 1/2)) = 43.6 years. Ρ 4. What must be the rate of increase, that the population of a country may be changed from 1106400 persons to 5 millions in Ans. About th annually. 100 years. Here A-5000000, P. 1106400, n = 100; .. 006550=log 1.015; 15 1 -1000 66' log(1+1) verify the fol Period of being increased 10 times. 277.4 years. 120.8 years. Ρ Period of Period of 120 83.5 years. 132.3 years. 57.6 years. |