### Part I. Section I. Chapter 22. “Of the Logarithmic Tables now in use.”

232 In those Tables, as we have already mentioned, we begin with the supposition, that the root $$a$$ is = 10; so that the logarithm of any number, $$c$$, is the exponent to which wc must raise the number 10, in order that the power resulting from it may be equal to the number $$c$$; or if we denote the logarithm of $$c$$ by $$\textrm{L}c$$, we shall always have $$10^{\textrm{L}c}=c$$.

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 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 ½, or which amounts to the same thing, $$10^{\frac{1}{2}}$$ is greater than 2; for if we square both sides, the square of $$10^{\frac{1}{2}}$$ equals 10, and the square of 2 equals 4. Now, this latter is much less than the former: and, in the same manner, we see that $$x$$ is also less than ⅓; 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 $$\log 2$$, is less than ⅓, but greater than ¼: and, in the same manner, we may determine, with respect to every fraction contained between ¼ and ⅓, whether it be too great or too small.

In making trial, for example, with ²⁄₇, which is less than ⅓, and greater than ¼, $$10^x$$, or $$10^{\frac{1}{2}}$$, 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 ²⁄₇ is less than $$\log 2$$, and that $$\log 2$$, which was found less than ⅓, is however greater than ²⁄₇.

Let us try another fraction, which, in consequence of what we have already found, ,ust be contained between ²⁄₇ and ⅓. Such a fraction between these limits is ³⁄₁₀; and it is therefore required to find whether $$10^{\frac{3}{10}} = 2$$; if this be the case, the tenth powers of those numbers are also equal: but the tenth power of $$10^{\frac{3}{10}}$$ is $$10^3 = 1000$$, and the tenth power of 2 is 1024; we conclude therefore, that $$10^{\frac{3}{10}}$$ is less than 2, and, consequently, that ³⁄₁₀ is too small a fraction, and therefore the $$\log 2$$, though less than ⅓, is yet greater than ³⁄₁₀.

236 This discussion serves to prove, that $$\log 2$$ has a determinate value, since we know that it is certainly greater than ³⁄₁₀, but less than ⅓; we shall not however proceed any farther in this investigation at present. Being therefore still ignorant of its true value, we shall represent it by $$x$$, so that $$\log 2=x$$; and endeavour to show how, if it were known, we could deduce from it the logarithms of an infinity of other numbers. For this purpose, we shall make use of the equation already mentioned, namely,

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

which comprehends the property, that the logarithm of a product is found by adding together the logarithms of the factors.

237 First, as $$\log 2=x$$, and $$\log 10=1$$, we shall have

 $$\log 20=x+1$$ $$\log 200=x+2$$ $$\log 2000=x+3$$ $$\log 20000=x+4$$ $$\log 200000=x+2$$ $$\log 2000000=x+6$$, etc.

238 Farther, as $$\log c^2 = 2 \log c$$, and $$\log c^3=3\log c$$, and $$\log c^4=4\log c$$, etc. we have

$\log 4=2x; \quad \log 8 =3x; \quad \log 16=4x; \quad \log 32=5x; \quad \log 64=6x, \quad \textrm{etc.}$

Hence we find also, that

 $$\log 40=2x+1$$ $$\log 400=2x+2$$ $$\log 4000=2x+3$$ $$\log 40000=2x+4$$, etc.
 $$\log 80=3x+1$$ $$\log 800=3x+2$$ $$\log 8000=3x+3$$ $$\log 80000=3x+4$$, etc.
 $$\log 160=4x+1$$ $$\log 1600=4x+2$$ $$\log 16000=4x+3$$ $$\log 160000=4x+4$$, etc.

239 Let us resume also the other fundamental equation,

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

and let us suppose $$c=10$$, and $$d=2$$; since $$\log 10=1$$, and $$\log 2=x$$, we shall have $$\log \frac{10}{2}$$, or $$\log 5=1-x$$, and shall deduce from hence the following equations:

 $$\log 50=2-x$$ $$\log 500=3-x$$ $$\log 5000=4-x$$ $$\log 50000=5-x$$, etc.
 $$\log 25=2-2x$$ $$\log 125=3-3x$$ $$\log 625=4-4x$$ $$\log 3125=5-5x$$, etc.
 $$\log 25=2-2x$$ $$\log 125=3-3x$$ $$\log 625=4-4x$$ $$\log 3125=5-5x$$, etc.
 $$\log 250=3-2x$$ $$\log 2500=4-2x$$ $$\log 25000=5-2x$$ $$\log 250000=6-2x$$, etc.
 $$\log 1250=4-3x$$ $$\log 12500=5-3x$$ $$\log 125000=6-3x$$ $$\log 1250000=7-3x$$, etc.
 $$\log 6250=5-4x$$ $$\log 62500=6-4x$$ $$\log 62500=7-4x$$ $$\log 625000=8-4x$$, etc.

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 $$\log 3$$ be represented by the letter $$y$$: then,

 $$\log 30=y+1$$ $$\log 300=y+2$$ $$\log 3000=y+3$$ $$\log 30000=y+4$$, etc.
$\log 9 =2y, \quad \log 27 = 3y, \quad \log 81= 4y, \quad \textrm{etc.}$

we shall have also,

$\log 6 = x+y, \quad \log 12 = 2x+y, \quad \log 18 = x+2y, \quad \log 15 = \log 3 + \log 5 = y + 1 - x.$

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 · 2 · 2 · 3 · 3 · 5 = $$2^3 \cdot 3^2 \cdot 5$$,

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.

#### 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.