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First Problem. Suppose there should be a hundred thousand inhabitants in a province, and that the popula tion augments there, every year a thirtieth part, it is asked, what will be the number of inhabitants in this province after a century?

To facilitate the calculation, let us call n the present population. At the end of the first year this population will be n + 35 • n = n ( ! ); at the end of the second year, it will be n (1 + 3'5) + 3'5 • n (1 +

); but let us remark, that in the two terms which compose this expression, we have n (1 + '%), which is a common factor; then we shall be able, therefore, to put the whole quantity under this form n (1+36) × (1 + 3) = n (1+); there will be found similarly, that at the end of the third year, the population will be n (13), so that at the end of the century, it will be n (1+)10°; putting in place of n its value, we shall have 100000 (1 + z ) 1o° = x_the number sought for.

100

But, if it was necessary to raise 1+, or, to the hundredth power by successive multiplications, it is very sensible, that the calculation would be of an extreme length; whereas, by making use of logarithms, we shall have immediately L x L 100000 +L() = L 100000 + 100 L ; now L.!

100

L31-L30 = 0,014241, and then 100 L }} = 1, 4241; afterwards as L 1000005, we shall have, therefore, Lr = 5 + 1, 4241 =6, 4241, and seeking the number to which this log. belongs, we shall find x = 2654800, which is the number of inhabitants that there would be in this province at the end of 100 years.

Second Problem. The earth having been repeopled after the deluge by the children of Noah and their three wives; we demand what must have been the increase each year, in order that the population of the earth might amount to one million at the end of two hundred years?

This increase is supposed to have been always a same part of the population of each year. Let us call the increase of each year.

At the end of the first year, the population shall have been 6+ 1 × 6 = 6 (1 +); by proceeding as in the preceding problem, it will be found, that at the end of the second year, it shall have been 6 (1 +1), so that at the end of 200 years, it will be found 6(1)2001000000, conformably to the enun ciation of the question.

2

From this equation we get 1+1=

and L (1+4) = 38ʊL (

1000

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20000); performing the -)

6

calculations expressed in the second member, we find L (1)=0,026109. Now, secking the fractional number, to which this log. answers, and pushing the operation to the sixth decimal figure, we shall have 1+1=1,061963, from which we get x+1= (1,061963)x, and x(0,061963) = 1, from that

1

0,061963

and finally, effecting the division, we

find a 16 nearly.

Therefore it was necessary, that the human race had increased every year '; which the robust health, and

the long days of our first parents, render sufficiently probable.

Third Problem! In what proportion should a people increase every year, to be twice as numerous at the end of every century?

Let n be the number of those that compose this people, letbe the quantity required; we shall have for every secular epoch, the equation (1)100= 2n, which gives L (1+) LT0, 003010; but the number to which this logarithm belongs, is 1,006955, which gives 1+1, 006955, from which we get x 144 nearly.

Thus, (according to the remark of the author from whom we have taken these examples) we ought to regard as very ridiculous the objections of those incredu lous persons, who deny that the earth could have been peopled in so short a time by a single man.

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PRACTICAL

ARITHMETIC.

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