十分 +&c.)= ( +3.2+7+&c. 1 1 1 .. common log 2=.86858896 0.3010300 3 5.35 Log. 1 3 •86858896 + = 0.4771213 5.25 4 = 2 log 2 = 0·6020600 10 E5 = log =log 10-log 2 =1- log. 2 = = 0.6979700 2 :6 = log 3+log 2 = 0.7781513 1 1 1 7 6+ +log5=0.8450980 8 log. 23 = 3 log.2 = 0.9030900 9 log. 32 = 2 log 3 0.9542425 10 = log. = 1.0000000 1 - 86858896(+3.6+5.677.6 = 11 = -86858896(10 +3.00 +5.107&c.)+log 9=1-0413927 13 = -88858896(12+3-10 +5.19+) + logli=1-1139434 14 = 15 log. 42 19 +3.18 +5.18 )+ 18* log.3 12 = log. 3 + log. 4 = 1.0791812 十 = = 1.1461280 log. 5 + log.3 =1.1760913 16 log.4 1.2041200 1 1 1 17: = .86858896 (16 +3.16 +5.18+&c.)+log15=1.2304489 18 = log. 9 + log.2 1.2552725 1 1 1 •85858896 5.19 +-&c. +log17=1.2787536 20 log.10 + log 2 = 1.3010300 21 log. 7 + 1.3222193 22 = log. 11 log. 2 1.3424227 1 1 1 23= .86858890 22 = 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 l=0 and log 10=l; 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 num. ber 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. +3.22 +5.290+&c.) +log21=1-3617278 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 diminish. ed 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 1 period; the fractional part of the population which die in a year, (or ratio of mortality ;) the proportion of births in a year. Then, if A represents the state of the population at the end of n m m , ) years, log A =log P+nxlog(1+"m). P(1+ m-b = m mo mb The rate of increase of population in one year 1 1 ;.. 1:1+ ::P:P(1+ state of 6 mb mb the population at the end of the first year. But it is increased every year in the same proportion; therem-6 mfore 1:1+ :: 1 = state of the mb mb mb 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 Hence A=P1+. ; and log. A mb m m will be P(1+") P(1 = log. P +n X log(1+md). From which we deduce 'route nxlog(1+ ; log:(1+ log (1 (1+* log.P m-6 n mb mlog A – NXlog(1+ mb log A-log P log A - log P mb 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 oth 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 = P1 Log P=log A—nxlog (1+ m log (1+" mo 121 =log 10000000 x 100 x log = 7:3604200 = log.22931000. 120 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 been foth, and the number of births agth. What was the state of its population in the year 1700? Here A=27000000, n=92, m=30, b=26, = 196 mb =log 27000000—92 x log 1827.226985&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 Isth, and the annual proportion of births at 4th? Here A=16000000, P=5000000, m=45, b=24; m-6 log A - log P mb = 60.39.years. log 167 .0084636 4. A province in the year 1760 was estimated at 500000 per In the year 1800 it amounted to 720000. From the bills of mortality it appeared that, upon an average, both part of the population had died annually ; no register had been kept of the births. What was the annual proportion of them during this period ? Here A=720000, P=500000, n=50, n=40. m-o 50 1 mb 500 log 720000 --- log 500000 =.0039590=log 1.009. 9 50-6 9 500 1000 127. But in any country, under given circumstances of births and mortality, the fraction is always a given quantity. Let 1 it be represented by ; then the relation between the four quan. р sons. Log. (1+") or log (1+ n +1000 and 1450 m-b tities A, P, P, n, is expressed by A= If A=mP,we have mP=P(1+ ); =( Hence we deduce the six following =P(1+). 5) or m=(1+y); and taking the logarithm = log (1 + log A - log P A II. III. n= A log m 1 V. n= IV. Log (1 +5)=log A – log P n VI. Log (1 +)=log , for finding the rate, n for finding the period in which the population would be increase m times. m 1 at which 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 oth of the whole; what will be the state of its population at the end of 200 years ? 1 1 1 Here P=6, n= = 200, 1 = 1.0625 ; . 16 x = 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 been ths for fifty years previous. What was the 'state of its population in the year 1750 ? Ans. 1908930 persons. 1 7 1 7 367 Here A=500000, ; 1+=lto and n=50; 360 р 367 = 1+=+1 P A=log P+n log (2+2) = log 6+200 X log (1.0625) = log ... P=log 8–n Xlog (1 +-)=log 5000000—50xlog 360 р log, 5000000—50 X (log 367–log 360)6.2807900=log 1908930; hence A=1908930 persons. 7. An empire to be 40 millions, and the annual increase Tosth; how long will it be before it amounts to 50 millions ? Ans. 43.6 years. Here A=50000000, P=40000000; p=1$; 1+paltid=18$ log. A-log.P log 50000000—log 40000000 (log 196—log 185) 1+ log:(1+5) .096910 =43.6 years. .002221 4. What must be the rate of increase, that the population of a country may be changed from 1106400 persons to 5 millions in 100 years. Ans. About oth annually. Here A-5000000, P= 1106400, n = 100; .. log A-log Plog 50000000-1106400 -006550=log 1.015; 100 1 1 15 1 hence 1 + -1.015, and =.0153 nearly P р -1000 log m 5. By means of the formula n= verify the following Table. (1+) n 66' log(1+1) 277.4 years. 132.3 years. 52 Period of Period of Period of being P doubling. trebling. increased 10 times. Ito | 83.5 years. 36.3 120.8 years. 1 1 1 1 Here m=2; ; 1+ ; therefore р 120 logm 301030 831 years. log 121-log 120 ·003604 1 log 2 n = log (1+) 2 1321 years, 1 log 3 .477121 Here m=3; then log 121-log 120 .00364 log 10 1 nearly. And here m=10; then log 121-log 120 003604 1 1 1 53 277.3 years. Again, here m=2, ; i+=lt р 52 log 2 •301030 =36) years, nearly. log (1+) log 53—log 52 ·008273 log 3 .477121 52 •00827 1 52; n |