Elements of Algebra

64 ELEMENTS SECT. I. we mean that $b$ is equal to the logarithm of the number $c$, or that the logarithm of $c$ is $b$. 221. We see then, that the value of the root $a$ being once established, the logarithm of any number, $c$, is nothing more than the exponent of that power of $a$, which is equal to $c$: so that $c$ being $= a^b$, $b$ is the logarithm of the power $a^b$. If, for the present, we suppose $b = 1$, we have $1$ for the logarithm of $a^1$, and consequently $log. a = 1$; but if we suppose $b = 2$, we have $2$ for the logarithm of $a^2$; that is to say, $log. a^2 = 2$, and we may, in the same manner, obtain $log. a^3 = 3$; $log. a^4 = 4$; $log. a^5 = 5$, and so on. 222. If we make $b = 0$, it is evident that $0$ will be the logarithm of $a^0$; but $a^0 = 1$; consequently $log. 1 = 0$, whatever be the value of the root $a$. Suppose $b = -1$, then $-1$ will be the logarithm of $a^{-1}$; but $a^{-1} = \frac{1}{a}$; so that we have $log. \frac{1}{a} = -1$, and in the same manner, we shall have $log. \frac{1}{a^2} = -2$; $log. \frac{1}{a^3} = -3$; $log. \frac{1}{a^4} = -4$, &c. 223. It is evident, then, how we may represent the logarithms of all the powers of $a$, and even those of fractions, which have unity for the numerator, and for the denominator a power of $a$. We see also, that in all those cases the logarithms are integers; but it must be observed, that if $b$ were a fraction, it would be the logarithm of an irrational number: if we suppose, for example, $b = \frac{1}{2}$, it follows, that $\frac{1}{2}$ is the logarithm of $a^{\frac{1}{2}}$, or of $\sqrt{a}$; consequently we have also $log. \sqrt{a} = \frac{1}{2}$; and we shall find, in the same manner, that $log. \sqrt[3]{a} = \frac{1}{3}$, $log. \sqrt[4]{a} = \frac{1}{4}$, &c. 224. But if it be required to find the logarithm of another number $c$, it will be readily perceived, that it can neither be an integer, nor a fraction; yet there must be such an exponent $b$, that the power $a^b$ may become equal to the number proposed; we have therefore $b = log. c$; and generally, $a^{log. c} = c$. 225. Let us now consider another number $d$, whose logarithm has been represented in a similar manner by $log. d$; so that $a^{log. d} = d$. Here if we multiply this expression by the preceding one $a^{log. c} = c$, we shall have $a^{log. c + log. d} = cd$; hence, the exponent is always the logarithm of the power; consequently, $log. c + log. d = log. cd$. But if, instead of multiplying, we divide the former expression by the latter,