Elements of Algebra

CHAP. XXIII. OF ALGEBRA. 69 $log. 50 = 2 - x,$ $log. 500 = 3 - x$ $log. 5000 = 4 - x,$ $log. 50000 = 5 - x, \&c.$ $log. 25 = 2 - 2x,$ $log. 125 = 3 - 3x$ $log. 625 = 4 - 4x,$ $log. 3125 = 5 - 5x, \&c.$ $log. 250 = 3 - 2x,$ $log. 2500 = 4 - 2x$ $log. 25000 = 5 - 2x,$ $log. 250000 = 6 - 2x, \&c.$ $log. 1250 = 4 - 3x,$ $log. 12500 = 5 - 3x$ $log. 125000 = 6 - 3x,$ $log. 1250000 = 7 - 3x, \&c.$ $log. 6250 = 5 - 4x,$ $log. 62500 = 6 - 4x$ $log. 625000 = 7 - 4x,$ $log. 6250000 = 8 - 4x, \&c.$ and so on. 240. If we knew the logarithm of $3$, this would be the means also of determining a number of other logarithms; as appears from the following examples. Let the $log. 3$ be represented by the letter $y$: then, $log. 30 = y + 1,$ $log. 300 = y + 2$ $log. 3000 = y + 3,$ $log. 30000 = y + 4, \&c.$ $log. 9 = 2y, log. 27 = 3y, log. 81 = 4y, \&c.$ we shall have also, $log. 6 = x + y, log. 12 = 2x + y, log. 18 = x + 2y,$ $log. 15 = log. 3 + log. 5 = y + 1 - x.$ 241. We have already seen that all numbers arise from the multiplication of prime numbers. If therefore we only knew the logarithms of all the prime numbers, we could find the logarithms of all the other numbers by simple additions. The number $210$, for example, being formed by the factors $2, 3, 5, 7$, its logarithm will be $log. 2 + log. 3 + log. 5 + log. 7$. In the same manner, since $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5$, we have $log. 360 = 3 log. 2 + 2 log. 3 + log. 5$. It is evident, therefore, that by means of the logarithms of the prime numbers, we may determine those of all others; and that we must first apply to the determination of the former, if we would construct Tables of Logarithms. CHAP. XXIII. Of the Method of expressing Logarithms. 242. We have seen that the logarithm of $2$ is greater than $\frac{3}{10}$, and less than $\frac{1}{3}$, and that, consequently, the exponent of $10$ must fall between those two fractions, in order that the power may become $2$. Now, although we know this, yet