The Construction of the Wonderful Canon of Logarithms/Remarks on Appendix

by Henry Briggs

SOME REMARKS

BY THE LEARNED

HENRY BRIGGS

On the foregoing Appendix.

The relations of Logarithms and their natural
numbers to each other, when the Logarithm
of unity ts made 0.

⁠[A]Wo numbers with their Logarithms being given; of both Logarithms be divided by some common divisor, and if each of the given numbers be multiplied by itself continuously, until the number of multiplications ts exceeded, by unity only, by the quotient of the Logarithm of the other number, two equal numbers will be produced. And the Logarithm of the number produced will be the continued product of the quotients of the Logarithms and their common divisor.

				Logarithms.
Let the given numbers be			25118865	4
			39810718	6

Let the common divisor be							1
The first multiplied	by	itself	5	times		makes	251188649
The secondultiplied”		itself”	3	times”
							1000000

Continued Proportionals.			Logarithms.
1	(0)		0thims‍
25118865	(1)	First power	4thims‍
63095737	(2)	Second power	8thims‍
158489331	(3)	Third power	12thims‍
39810718⁠	(4)	Fourth power	16thims‍
100000000⁠	(5)	Fifth power	20thims‍
251188649⁠	(6)	Sixth power	24thims‍
Continued Proportionals.			Logarithms.
1	(0)		0thims‍
39810718	(1)		6thims‍
158489331	(2)		12thims‍
630957379	(3)		18thims‍
251188649⁠	(4)		24thims‍

Another Example.

			Logarithms.
Let the given numbers be	$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$	316227766	5
		50118724	7
Let the common divisor be			1

The first multiplied	by	itself	6	times	$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$	makes 316227766
The secondultiplied”		itself”	4	times”

		Logarith.				Logarith.
⁠1	(0)	0rith.‍		⁠1	(0)	0rith.‍
⁠316227766	(1)	5rith.‍		⁠50118724	(1)	7rith.‍
⁠1000000000	(2)	10rith.‍		⁠251188649	(2)	14rith.‍
⁠100	(4)	20rith.‍		⁠630957376	(4)	28rith.‍
1000	(6)	30rith.‍		316227766	(5)	35rith.‍
316227766	(7)	35rith.‍

⁠It should be observed that if the common divisor be unity, as in both the preceding examples, the product of the given Logarithms is the Logarithm of the number produced, because multiplication by unity does not increase the thing multiplied.

Third Example.

			Logarithms.	Quotients.
Let the given numbers be	$\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$	343	2.53529412	3
		823543	5.91568628	7
Let the common divisor be			84509804

Number of Places
⁠	1	(0)	⁠0
⁠3	343	(1)	⁠2.53529412
⁠6	117649	(2)	⁠5.07058824
⁠8	40353607	(3)	⁠7.60588236
11	3841287201	(4)	10.14117648
18	558545864083284007	(7)	17.74705884
⁠6	823543	(1)	⁠5.91568628
12	678223072849	(2)	11.83137256
18	558545864083284007	(3)	17.74705884

⁠As the quotients of the given Logarithms are 3 and 7, their product is 21, which, multiplied by 84509804 the common divisor, makes 17.74705884 the Logarithm of the number produced.

⁠It should be observed that the cube of the second number, and its equal the seventh power of the first (which some call secundus solidus), contain eighteen figures, wherefore its Logarithm has 17. in front, besides the figures following. The latter represent the Logarithm of the number denoted by the same digits, but of which 5, the first digit to the left, is alone integral, the remaining digits expressing a fraction added to the integer, thus $\scriptstyle {\frac {5854586408}{10000000000}}$ &c.has for its Logarithm 74705884. Again, of four places remain integral, 3. must be placed in front of the Logarithm, thus $\scriptstyle 5585{\frac {4586408}{10000000}}$ &c. has for its Logarithm 3.74705884.

Hence from two given Logarithms and the sine of the first we shall be able to find the sine of the second.

⁠Take some common divisor of the Logarithms, (the larger the better); divide each by it. Then let the first sine multiply itself and its products continuously until the number of these products is exceeded, by unity only, by the quotient of the second Logarithm; or until the power is produced of like name with the quotient of the second Logarithm. The same number would be produced if the second sine, which ts sought, were to multiply itself until it became the power of like name with the quotient of the first Logarithm, as is evident from the preceding proposition. Therefore take the above power and seek for the root of it which corresponds to the quotient of the first Logarithm; thereby you will find the required second sine. Also the Logarithm of the power itself will be the continued product of the quotients and the common divisor.

⁠Thus let the given Logarithms be 8 and 14, and the sine corresponding to the first Logarithm be 3. A common divisor of the Logarithms is 2; this gives the quotients 4 and 7. If 3 multiply itself six times, you will have 2187 for the power which, in a series of continued proportionals from unity, will occupy the seventh place, and hence it may, without inconvenience, be called the seventh power. The same number, 2187, is the fourth power from unity in another series of continued proportionals, in which the first power, $\scriptstyle 6{\frac {838521}{1000000}}$, is the required second sine. The product of the quotients 4 and 7 is 28, which, multiplied by the common divisor 2, makes 56, the Logarithm of the power 2187.

Continued Proportionals.		Logarithms.	Continued Proportionals.		Logarithms.
218‍1	(0)	⁠0	218‍1	(0)	⁠0
218‍3	(1)	⁠8	218‍6838521	(1)	14
218‍9	(2)	16	21‍46765372	(2)	28
21‍27	(3)	24	2‍31980598	(3)	42
21‍81	(4)	32	2187	(4)	42
2‍243	(5)	40
2‍729	(6)	48
2187	(7)	56

⁠It will be observed that these Logarithms differ from those employed in illustration of the previous Proposition: but they agree in this, that tn both, the Logarithm of unity is 0; and consequently the Logarithms of the same numbers are either equal or at least proportional to each other.

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[B] If a first sine divide a third, )

⁠The first must divide the third, and the quotient of the third, and each quotient of a quotient successively as many times as possible, until the last quotient becomes less than the divisor. Then let the number of these divisions be noted, but not the value of any quotient, unless perhaps the least, to which we shall refer presently. In the same manner let the second divide the same third. And so also let the fourth be divided by each.

Thus let the	$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \ \end{matrix}}\right.}$	first	sine	be	2
		second	sine”	be”	4
		third	sine”	be”	16
		fourth	sine”	be”	64

The first, 2. divides the third, 16. four times; and the quotients are 8, 4, 2, 1. The second, 4. divides the same third, 16. two times; and the quotients are 4, 1. Therefore A will be 4, and B will be 2.

⁠In the same manner the first, 2. divides the fourth, 64. six times; and the quotients are 32, 16, 8, 4, 2, 1. The second, 4. divides the fourth, 64. three times; and the quotients are 16, 4, 1. Therefore C will be 6, and D will be 3.

⁠Hence I say that, as A, 4. is to B, 2. so is C, 6. to D, 3. and so ts the Logarithm of the second to the Logarithm of the first.

⁠If in these divisions the last and smallest quotient be everywhere unity, as in these four cases, the numbers of the quotients and the Logarithms of the divisors will be reciprocally proportional.

⁠Otherwise the ratio will not be exactly the same on both sides; nevertheless, if the divisors be very small, and the dividends sufficiently large, so that the quotients are very many, the defect from proportionality will scarcely, or not even scarcely, be perceived.

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Hence it follows that the logarithm )⁠[C]

⁠Let two numbers be taken, 10 and 2, or any others you please. Let the Logarithm of the first, namely 100, be given; it ts required to find the Logarithm of the second. In the first place, let the second, 2. multiply itself continuously until the number of the products ts exceeded, by unity only, by the given Logarithm of the first. Then let the last product be divided as often as possible by the first number, 10. and again in like manner by the second number, 2. The number of quotients in the latter case will be 100, (for the product ts its hundredth power; and if a number be multiplied by itself a given number of times forming a certain product, then it will divide the product as many times and once more; for example, of 3 be multiplied by itself four times it makes 243, and the same 3 divides 243 five times, the quotients being 81, 27, 9, 3, 1.) In the former case, where the product is continually divided by 10, it is manifest that the number of quotients falls short of the number of places in the dividend by one only. Therefore (by the preceding proposition) since the same product is divided by two given numbers as often as possible, the numbers of the quotients and the Logarithms of the divisors will be reciprocally proportional, But, the number of quotients by the second being equal to the Logarithm of the first, the number of quotients by the first, that is the number of places in the product less one, will be equal to the Logarithm of the second.

Number of Places
	1	0
1	2	1
1	4	2
2	16	4
3	256	8
4	1024	10
7	1048576	20
13	1099511627776	40
25	1208925819614	80
31	1267650600228	100
61	16069379676⁠	200
121	25822496318⁠	400
241	66680131608⁠	800
302	107150835165⁠	1000
603	114813014767⁠	2000
1205	131820283599⁠	4000
2409	17316587168⁠	8000
3011	19950583591⁠	10000

⁠Here we see that if we assume the Logarithm of 10 to be 10, the number of places in the tenth power is 4, wherefore the logarithm of 2 will be 3 and something over. The number of places in the hundredth power is 31; in the thousandth, 302; in the ten thousandth, 3011; and generally the more products we take the more nearly do we approach the true Logarithm sought for. For when the products are few, the fraction adhering to the last quotient disturbs the ratio a little; but if we assume the Logarithm of 10 to be 10,000,000,000, and if 2 be multiplied by itself continuously until the number of products is exceeded, by one only, by the given Logarithm; then the number of places, less one, in the last product, will give the Logarithm of 2 with sufficient accuracy, because in large numbers the small fraction adhering to the last quotient will have no effect in disturbing the proportion.

THE END

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