Elements of Algebra

72 ELEMENTS SECT. I. consequently composed of two figures, are comprehended between $1$ and $2$, and therefore must be expressed by $1$ plus a decimal fraction, as $log. 50 = 1.6989700$; its value therefore is unity, plus $\frac{6}{10} + \frac{9}{100} + \frac{8}{1000} + \frac{9}{10000} + \frac{7}{100000}$: and it will be also easily perceived, that the logarithms of numbers, between $100$ and $1000$, are expressed by the integer $2$ with a decimal fraction: those of numbers between $1000$ and $10000$, by $3$ plus a decimal fraction: those of numbers between $10000$ and $100000$, by $4$ integers plus a decimal fraction, and so on. Thus, the $log. 800$, for example, is $2.9030900$; that of $2290$ is $3.3598355$, &c. 248. On the other hand, the logarithms of numbers which are less than $10$, or expressed by a single figure, do not contain an integer, and for this reason we find $0$ before the point: so that we have two parts to consider in a logarithm. First, that which precedes the point, or the integral part; and the other, the decimal fractions that are to be added to the former. The integral part of a logarithm, which is usually called the *characteristic*, is easily determined from what we have said in the preceding article. Thus, it is $0$, for all the numbers which have but *one* figure; it is $1$, for those which have *two*; it is $2$, for those which have *three*; and, in general, it is always one less than the number of figures. If therefore the logarithm of $1766$ be required, we already know that the first part, or that of the integers, is necessarily $3$. 249. So reciprocally, we know at the first sight of the integer part of a logarithm, how many figures compose the number answering to that logarithm; since the number of those figures always exceed the integer part of the logarithm by unity. Suppose, for example, the number answering to the logarithm $6.4771213$ were required, we know immediately that that number must have seven figures, and be greater than $1000000$. And in fact this number is $3000000$; for $log. 3000000 = log. 3 + log. 1000000$. Now $log. 3 = 0.4771213$, and $log. 1000000 = 6$, and the sum of those two logarithms is $6.4771213$. 250. The principal consideration therefore with respect to each logarithm is, the decimal fraction which follows the point, and even that, when once known, serves for several numbers. In order to prove this, let us consider the logarithm of the number $365$; its first part is undoubtedly $2$; with respect to the other, or the decimal fraction, let us at present represent it by the letter $x$; we shall have $log. 365 = 2 + x$; then multiplying continually by $10$, we shall