### Part I. Section I. Chapter 21. “Of Logarithms in general.”

220 Resuming the equation $$a^b=c$$, we shall begin by remarking that, in the doctrine of Logarithms, we assume for the root $$a$$, a certain number taken at pleasure, and suppose this root to preserve invariably its assumed value. This being laid down, we take the exponent $$b$$ such, that the power $$a^b$$ becomes equal to a given number $$c$$; in which case this exponent $$b$$ is said to be the logarithm of the number $$c$$. To express this, we shall use the letter L or the initial letters $$\log$$. Thus, by $$b = \textrm{L} c$$, or $$b =\log c$$, 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 $$\log \frac{1}{a^2} = -2$$; $$\log \frac{1}{a^3} = -3$$; $$\log \frac{1}{a^4}=-4$$, etc.

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 $$a^{1/2}$$, or of $$\surd a$$; consequently we have also $$\log \surd a= \frac{1}{2}$$; and we shall find, in the same manner, that $$\log \sqrt[3]{\vphantom{a}} a = \frac{1}{3}$$, $$\log \sqrt[4]{\vphantom{a}} a = \frac{1}{4}$$, etc.

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, $$a^{\textrm{L}c} = c$$.

225 Let us now consider another number d, whose logarithm has been represented in a similar manner by $$\log d$$; so that $$a^{\textrm{L}d} = d$$. Here if we multiply this expression by the preceding one $$a^{\textrm{L}c} = c$$, we shall have $$a^{\textrm{L}c+\textrm{L}d} =cd$$; 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, we shall obtain $$a^{\textrm{L}c-\textrm{L}d}=\frac{c}{d}$$; and, consequently,

$\log c - \log d = \log \frac{c}{d}.$

226 This leads us to the two principal properties of logarithms, which are contained in the equations

$\log c + \log d = \log cd,$

and

$\log c - \log d = \log \frac{c}{d}.$

The former of these equations teaches us, that the logarithm of a product, as $$cd$$, is found by adding together the logarithms of the factors; and the latter shows us this property, namely, that the logarithm of a fraction may be determined by subtracting the logarithm of the denominator from that of the numerator.

227 It also follows from this, that when it is required to multiply, or divide, two numbers by one another, we have only to add, or subtract, their logarithms; and this is what constitutes the singular utility of logarithms in calculation: for it is evidently much easier to add, or subtract, than to multiply, or divide, particularly when the question involves large numbers.

228 Logarithms are attended with still greater advantages, in the involution of powers, and in the extraction of roots; for if $$d=c$$, we have, by the first property, $$\log c + \log c = \log cc$$ or $$c^2$$; consequently, $$\log cc = 2 \log c$$; and in the same manner, we obtain $$\log c^3 = 3 \log c$$; $$\log c^4=4\log c$$; and, generally, $$\log c^n=n\log c$$. If we now substitute fractional numbers for $$n$$, we shall have, for exmaple, $$\log c^{\frac{1}{2}}$$, that is to say, $$\log \sqrt{\vphantom{c}} c$$, $$=\frac{1}{2}\log c$$; and lastly, if we suppose $$n$$ to represent negative numbers, we shall have $$\log c^{-1}$$, or $$\log \frac{1}{c}$$, $$=-\log c$$; $$\log c^{-2}$$, or $$\log \frac{1}{c^2}$$, $$=-2\log c$$, and so on; which follows not only from the equation $$\log c^n = n \log n$$, but also from $$\log 1=0$$, as we have already seen.

229 If therefore we had Tables, in which logarithms were calculated for all numbers, we might certainly derive from them very great assistance in performing the most prolix calculations; such, for instance, as require frequent multiplications, divisions, involutions, and extractions of roots: for, in such Tables, we should have not only the logarithms of all numbers, but also the numbers answering to all logarithms. If it were required, for example, to find the square root of the number $$c$$, we must first find the logarithm of $$c$$, that is, $$\log c$$, and next taking the half of that logarithm, or $$\frac{1}{2}\log c$$, we should have the logarithm of the square root required: we have therefore only to look in the Tables for the number answering to that logarithm, in order to obtain the root required.

230 We have already seen, that the numbers, 1, 2, 3, 4, 5, 6, etc. that is to say, all positive numbers, are logarithms of the root a, and of its positive powers; consequently, logarithms of numbers greater than unity: and, on the contrary, that the negative numbers, as -1, -2, etc. are logarithms of the fractions $$\frac{1}{a}$$, $$\frac{1}{a^2}$$, etc. which are less than unity, but yet greater than nothing.

Hence, it follows, that, if the logarithm be positive, the number is always greater than unity: but if the logarithm be negative, the number is always less than unity, and yet greater than 0; consequently, we cannot express the logarithms of negative numbers: we must therefore conclude, that the logarithms of negative numbers are impossible, and that they belong to the class of imaginary quantities.

231 In order to illustrate this more fully, it will be proper to fix on a determinate number for the root $$a$$. Let us make choice of that, on which the common Logarithmic Tables are formed, that is, the number 10, which has been preferred, because it is the foundation of our Arithmetic. But it is evident that any other number, provided it were greater than unity, would answer the same purpose: and the reason why we cannot suppose $$a=$$ unity, or 1, is manifest; because all the powers $$a^b$$ would then be constantly equal to unity, and could never become equal to another given number, $$c$$.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.