Elements of Algebra

CHAP. XXII. OF ALGEBRA. 67 233. We have already observed, that the logarithm of the number 1 is always 0; and we have also $10^0 = 1$; consequently, log. $1 = 0$; log. $10 = 1$; log. $100 = 2$; log. $1000 = 3$; log. $10000 = 4$; log. $100000 = 5$; log. $1000000 = 6$. Farther, log. $\frac{1}{10} = -1$; log. $\frac{1}{100} = -2$; log. $\frac{1}{1000} = -3$; log. $\frac{1}{10000} = -4$; log. $\frac{1}{100000} = -5$; log. $\frac{1}{1000000} = -6$. 234. The logarithms of the principal numbers, therefore, are easily determined; but it is much more difficult to find the logarithms of all the other intervening numbers; and yet they must be inserted in the Tables. This however is not the place to lay down all the rules that are necessary for such an inquiry; we shall therefore at present content ourselves with a general view only of the subject. 235. First, since log. $1 = 0$, and log. $10 = 1$, it is evident that the logarithms of all numbers between 1 and 10 must be included between 0 and unity; and, consequently, be greater than 0, and less than 1. It will therefore be sufficient to consider the single number 2; the logarithm of which is certainly greater than 0, but less than unity: and if we represent this logarithm by the letter $x$, so that log. $2 = x$, the value of that letter must be such as to give exactly $10^x = 2$. We easily perceive, also, that $x$ must be considerably less than $\frac{1}{2}$, or which amounts to the same thing, $10^{\frac{1}{2}}$ is greater than 2; for if we square both sides, the square of $10^{\frac{1}{2}} = 10$, and the square of $2 = 4$. Now, this latter is much less than the former; and, in the same manner, we see that $x$ is also less than $\frac{1}{3}$; that is to say, $10^{\frac{1}{3}}$ is greater than 2: for the cube of $10^{\frac{1}{3}}$ is 10, and that of 2 is only 8. But, on the contrary, by making $x = \frac{1}{4}$, we give it too small a value; because the fourth power of $10^{\frac{1}{4}}$ being 10, and that of 2 being 16, it is evident that $10^{\frac{1}{4}}$ is less than 2. Thus, we see that $x$, or the log. 2, is less than $\frac{1}{3}$, but greater than $\frac{1}{4}$: and, in the same manner, we may determine, with respect to every fraction contained between $\frac{1}{3}$ and $\frac{1}{4}$, whether it be too great or too small. In making trial, for example, with $\frac{2}{7}$, which is less than $\frac{1}{3}$, and greater than $\frac{1}{4}$, $10^x$, or $10^{\frac{2}{7}}$, ought to be $= 2$; or the seventh power of $10^{\frac{2}{7}}$, that is to say, $10^2$, or 100, ought to be equal to the seventh power of 2, or 128; which is consequently greater than 100. We see, therefore, that $\frac{2}{7}$ is less than log. 2, and that log. 2, which was found less than $\frac{1}{3}$, is however greater than $\frac{2}{7}$. F 2