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 59 
 
 UC-NRLF 
 
 B M EMT 2^D 
 
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LOGARITHMS. 
 
 ^-^.^^ 
 
 H. N. |WHEELER 
 
 CAMBRIDGE : 
 
 Charles AV. Sever, University Bookseller. 
 
 1882. 
 
LOGARITHMS. 
 
 BY 
 
 H. N. WHEELER. 
 
 3j®<0 
 
 CAMBKIDGE: 
 Charles W. Sever, University Bookseller. 
 
 1882. 
 
LOGAEITHMS 
 
 § 1, The logarithm of a number N is the exponent de- 
 noting the power to which a fixed number called the base 
 must be raised in order to produce N. 
 
 Thus, \i N=V, then is log^N^x; or, in words, when 
 the base is h the loo;aritlim of N is x. 
 
 What is logo 8? 
 
 8 = 2^ ; 
 
 .-. log28 = 3. 
 
 What is log. — ? 
 
 h^C^^ 
 
 .•.log,f^ = 4. 
 
 WhatisloggS? 
 
 3 = 9^; 
 
 •••log93 = ^. 
 
 What is log 10 1000? 
 
 1000=103; . 
 
 •. Iogiol000 = 3. 
 
 What is log3 27? 
 
 
 Ans. 3. 
 
 What is log 4 64? 
 
 
 Arts. 3. 
 
 What is the number whose loo-arithm is 5 when the base 
 
 is 2? 
 
 log2-?v^= 5 ; 
 
 N^'2^ 
 
 What is the number whose logarithm is 3 when the base 
 is 3? Ans. 27. 
 
 What is the number whose logarithm is 3 when the base 
 is 10? Ans. 1000. 
 
 What is the number whose logarithm is — 2 when the 
 baseisS? log3^=-2; .-. iY= 8"^ =JL = JL. 
 
2 LOGARITHMS. 
 
 What is the number whose logarithm is — 2 when the 
 base is 10? ^.^^ 1 , 
 
 ■ lOO' 
 2 
 What is the number whose logarithm is — - when the 
 
 base is 27? ^^^g 1. 
 
 ■ 9 
 
 § 2. The logarithms of all numbers referred to the same 
 base are said to belong to the same system of logarithms. 
 Thus, log]o6, log 10 7, logio84, logio768, all belong to the 
 denar}- or common system. 
 
 § 3. In any system of logarithms the logarithm of the 
 base itself is 1 . 
 
 For 2 = 2^ .-.logs 2 =1: 
 
 10=101; _._ iog,^iO=l: 
 
 and, in general, b = h^ \ .-. logj?> = 1 . 
 
 § 4. In every sj'stem of logarithms the logarithm of 1 is 0. 
 For 1=2"; .-. loggl =0: 
 
 1 = 10"; .-. logiol=0: 
 and, in general, \ — lP\ .-. log^l = 0. 
 
 § 5. In any system the logarithm of the reciprocal of a 
 number is the negative of the logarithm of the number. 
 
 Proof: — is the reciprocal of If ; now - = h~'' ; 
 If If^ 
 
 What is log. — ? 
 
 l = i, = 4-- 
 IG 4- 
 
 .*. logi — = log4~- = — 2 : or log. — = — log4lG = —2. 
 ° 16 ° 16 
 
LOGAEITHMS. 
 
 What is log 10 0.1? 0.1 =-i- = 10-1; ... iog,o0.1 = -1. 
 
 What is logs-;;? 
 
 Ans. 
 
 What is the number whose logarithm is — - when the 
 base is 16? ^ 
 
 16l 32 
 What is the number whose logarithm is — 3 when the 
 base is 10? Ans. 0.001. 
 
 § 6. In any s^'stem the logarithm of the product of two 
 or more numbers is equal to the sum of the logarithms of 
 the numbers. 
 
 Proof: If l = h% m = 6^ n = h% 
 
 then is logjZ = .v, log677r = i/, \ogj,n = z', 
 
 now I xm xn = h'' X b" X b^ = /j^ + ^ + ^ ; 
 
 .-. log(? X m X n) = x-\-y + z = log? + logm + logn. 
 
 It is evident that this principle ma}' be applied to any 
 number of factors. 
 
 What is log 2 (8 x4x 32)? 
 
 8 = 2^ .-. logo8 =3 
 4 = 22; _._ iog^4 ^2 
 32 = 2^ .-. log232 = 5 
 logo(8x4x 32) = log28+log24+log232 
 
 = 3 + 2+5 = 10. 
 1 
 
 Whatisloggf^ X 3 X 81 1? 
 
 Ans. -2 + 1+4 = 3. 
 
 If log 10 123 = 2.0899, what is log ,o 12300? 
 
 12300 = 100 X 123 = 10- X 10-"«^ ; 
 
 .-. Iogaol2300 = 2 + 2.0899 = 4.0899. 
 
 If log 10 2 = 0.3010, what is log lo 200? Ans. 2.3010. 
 
4 LOGARITHMS. 
 
 § 7. In any system the logarithm of a fraction is equal 
 to the logarithm of the numerator mimis the logarithm of 
 the denominator. 
 
 Proof: If / = h''. m = 6^, then is logj? = x, log^m = y ; 
 
 lb'' I 
 
 now - = - = b^ " ; .-. \ogi- = X - y = logt,l -logtm. 
 m ¥ m 
 
 What is logo — ? 
 
 logo — = log,8-lo2:.32 = 3-5= -2. 
 
 Given: logioS = 0.-4771 and log io2 = 0.3010 ; what is 
 log 107? Arts. 0.17G1. 
 
 Given: log^ 123 = 2.0899 ; what is logio0.123?_ 
 
 Ans. 2.0899-3 = 1.0899. 
 
 § 8. In any system the logarithm of any power of a 
 number is equal to the logarithm of the number multiplied 
 by the exponent of the poicer. 
 
 Proof: If I = b\ then is log^Z = .r. ?"■ = (6^)"' = b"^ : 
 
 .-. logZ" = mx — m X logZ. 
 
 Under this head may be brought a root, for a root may 
 be regarded as a fractional power. 
 
 7" = (6-)- = 5™; .-.10??^ = -=- xlog?. 
 
 ^ ^ m m ° 
 
 What is logo-i^? log24«= 6 X logoi = G x 2 = 12. 
 
 What is lo2,8*? Iogo8* = ^log,8 =^ X 3 = 2. 
 
 o- 3 °- 3 
 
 WhatislogioA'TOO? 
 
 A Too = vTo^ = lOf : . • . log 10 <'Tob = |- 
 
 What is log 2^/16"^? Ans. '^- 
 
LOGARITHMS, 
 
 COMMON LOGARITHMS. 
 
 § 9. The logarithms most convenient for practical use 
 are the so-called common logarithms in which the base is 
 10. The following table contains the common logarithms 
 of certain integral powers of 10, and also shows between 
 what limits the logarithms of certain other numbers must 
 He. 
 
 10000 = 10* ; 
 9648 = lO-' + a-iw™^; 
 1000 = l(f ; 
 
 7g8 = 10- ~ * decimal . 
 
 100 = 10- ; 
 
 88 = I0l^a.decmal. 
 
 10 = 10^ ; 
 
 J zzz l(jO — a decimal . 
 
 1 = W ; 
 
 0.66 = 10-1 -a decimal. 
 
 0.1= — =10-^ 
 
 lU 
 0.079 = 10-2-^ a decimal. 
 
 0.01 = — = —=10- 
 
 100 10^^ 
 0.00684 = 10-3 -"decimal. 
 
 0.001 = 
 
 1 
 
 =-L=io- 
 
 1000 10-^ 
 
 0.000795= 10-<H- a decimal; 
 
 0.0001 =- 
 
 1 
 
 10000 10* 
 
 =-l.=io- 
 
 log 10000 = 4. 
 
 log 9648 = .3 + a decimal. 
 
 log 1000 
 
 = .3. 
 
 log 768 
 
 = 2 + a decimal. 
 
 log 100 
 
 = 2. 
 
 log 88 
 
 = 1 -^ a decimal. 
 
 log 10 
 
 = 1. 
 
 log 7 
 
 = + a decimal. 
 
 logl 
 
 = 0. 
 
 log 0.66 
 
 = — 1 + a decimal. 
 
 log 0.1 
 
 = -!• 
 
 log 0.07 
 
 = — 2 + a decimaL 
 
 log 0.01 
 
 = — 2. 
 
 log 0.00684 
 
 = — 3 + a decimal. 
 
 log 0.001 
 
 = -.3. 
 
 log 0.000795= — 4 -f a decimaL 
 .log 0.0001= -4. 
 
6 LOGARITHMS. 
 
 § 10. From § 9 we see that the logarithm of a number 
 which is not tiu integral power of 10 is an integer plus a 
 decimal. Take 9G48 for example ; it is between 1000 [10'] 
 and 10000 [10^], therefoi'e its logarithm is between 3 and 
 4, i.e. its logarithm is 3 + a decimal. Take 0.000795 ; it 
 is between 0.0001 [lO"''] and 0.001 [10"=^], therefore its 
 logarithm, being between — 3 and — 4, can be written 
 either — 3 — a decimal, or — 4 + a decimal. The latter 
 method will be adopted. 
 
 Suppose the decimal part of the logarithm of 0.000795 
 to be 0.9004, tlien is log 0.000795 = - 4 + 0.9004. Since 
 we shall make the decimal part alwajs +, its sign need not 
 be written ; and in order to denote that the negative sign 
 applies to the integral part onl}', we write it above the 
 integer ; thus, log 0.000795 = 4.9004. 
 
 Let us take an exami)le to show the advantage to be 
 gained by making the decimal part of the logarithm always 
 positive. 
 
 Suppose log72G = 2.8009 ; find logO.0726. 
 
 0.072G = -I^ = I^; 
 10000 10* 
 
 .-.logo. 0726 = log 726 -log 10* = 2. 8009-4 = 2.8609. 
 
 Comparing tliis result with the given logarithm, we see 
 that the decimal parts are exactly the same ; this will not 
 be the case if log 0.0726 is expressed in the other way, for 
 tlien we shall get 
 
 -2 + 0. 8609 = -1.1391 or -1-0.1391. 
 
 Let us take another example. 
 Given : log 0.0726 = 2.8609 ; find log 7.26. 
 7.26 = 6.0726 X 100; 
 
 .-. log 7.26 = 2.8609 + 2 = 0.8609 ; 
 
LOGARITHMS. 7 
 
 and figain we see that the decimal part of the result is the 
 same as that of the given logarithm. If, then, we alwa3-s 
 make the decimal part of a logarithm +, the following 
 statement will be true : — 
 
 Numbers icliich consist of the same se^'ies of figures, i.e., 
 which differ only in resjject to the position of the decimal 
 2yoint, ivill have the same decimal ^xirts to their logarithms ; 
 and the integral parts of the logarithms ivill depend upon 
 the position of the decimal point; for a change in the posi- 
 tion of the decimal point is equivalent merely- to multiply- 
 ing or dividing by an integral power of 10, and this process 
 will only increase or diminish the logarithm (exponent) by 
 an integer. 
 
 § 11. The integral part of a logarithm is called its 
 characteristic, and the decimal part its mantissa. 
 
 The characteristic, when the number is not an integral 
 power of 10, Ynmy be found easil}^ by noting between what 
 adjacent integral powers of 10 the number lies. 
 
 Thus, 9648 (see § 9) lies between 10^ [1000] and 10* 
 [10000] ; therefore the characteristic of log9G48 is 3 : 
 
 0.0007G5 lies between 0.0001 [10-*] and 0.001 [10-^] ; 
 therefore the characteristic of log 0.000765 is —4. 
 
 Thus we see that the characteristic of the logarithm of a 
 number depends wholly on the })osition of the decimal 
 point, and not at all on the figures which the number 
 contains. 
 
 From what precedes we can readily deduce the following 
 rule for finding the characteristic of the logarithm of any 
 number : — 
 
 The characteristic of the logarithm of a number is equal 
 to the number of places by which the first significant figure 
 of the number is removed from the units' place,, and is posir 
 
8 LOGARITHMS. 
 
 live for numbers greater than unity (i.e., where the first 
 significant figure is to the left of the decimal 2'>oint) , and 
 negative for mimhers less than iinity (i.e., tchere the first 
 significant figure is to the right of the decimal 2'>oint) ; it is 
 whe7i the first significant figure is in the units' place. 
 
 Examjyles: Find by the above rule the characteristics 
 of the logarithms of the following numbers : — 
 
 3 2 10 2 10 
 
 (1)1689; (2)168.9; (3)1.689; 
 
 (4)0.'l689; (o)0"o~i689; (6) o7o~o'o'l689. 
 
 Answers. (1)3; (2) 2; (3)0; 
 (4)-l; (5)-2; (6)-4. 
 
 § 12. The mantissas., or decimal parts of the logarithms 
 of numbers, have been computed to different degrees of 
 accurac}', the results for some numbers having been carried 
 out to more than 25 places of decimals. A collection of 
 these mantissas is called a logarithm-tahle. Tables are 
 characterized by putting before the word logarithm the 
 expression Z-place, 4-place, 5-place, 1-place., d-jylace, etc., 
 according to the number of places of decimals to which the 
 results are given. Thus, a 4-place logarithm-table is one 
 in which the results are given to four places of decimals. 
 
 § 13. Examples of finding logarithms from the tables. 
 
 Find log 495000. 
 
 Referring to p. 2 of Peirce's Tables, we seek 49 at the 
 left of the table in the column headed n, and run along 
 the horizontal line in which this lies until we reach the ver- 
 tical column which has 5 at the top, where we find 6946, 
 and this is the mantissa or decimal part (accurate to 4 
 
LOGARITHMS. 
 
 9 
 
 places) of the logarithms of all numbers whose significant 
 figures form the series 495 ; 
 
 .-. mantissa of log 495000 = .G946. 
 By the rule of § 11 the characteristic is 5 ; 
 
 - .-. log 495000 = 5.6946. 
 Find the logarithms of 
 (1) 9.84; (2) 0.729; (3) 6690; 
 
 (4) 0.066 ; (5) 0.00000543 ; (6) 0.007. 
 
 Answers. (1)0.9930; (2)1.8627; (3)3.8254; 
 (4) 2.8195; (5) 6.7348; (6) 3.8451. 
 
 What is log 241 50? 
 
 "We do not find this number in the table ; we do find, 
 however, log 24100 = 4.3820, 
 
 and log 24200 = 4.3838. 
 
 The given number 24150 is half-way between 24100 and 
 24200 ; i.e., it is equal to the first of these numbers plus 
 half the distance between the first and second. Assuming* 
 that the required logarithm is half-wa}' between 4.3820 and 
 
 * The assumption that numbers vary proportionally with their 
 logarithms — that in the above example, for instance, if a num- 
 ber is half-way between two numbers, its logarithm is half-way 
 between the logarithms of these numbers — is only approxi- 
 mately correct ; but, when the difference between the numbers 
 is small, the error to which this assumption leads us is smaU; 
 and in every properly-constructed table the differences or inter- 
 vals between the successive numbers of the table are so small 
 that the error in a logarithm fouud as above will not affect the 
 last place of decimals. 
 
 This method of finding the logarithm of a number which lies 
 between two successive numbers of a table is called the method 
 of interpolation. 
 
 If we refer to a 7-place table, we find from the table, without 
 interpolation, log 24150 = 4.3829171, and this carried to only 4 
 places of decimals is 4.3829, which is the result obtained from 
 the 4-place table by interpolation. 
 
to 
 
 4.383&, i.e. Hiatiki&e^paLto thfrSrsfr of these losarrthms 
 ploji half the cS&can£& fisam. th& first 1 og,irtt.hin to thfr see-^ 
 Offli, we gpc 
 
 We csfflc if we wish*. S»i .5 of O.OOl!^ tbont the pattr i )i" 
 tijtj loi^attthm oi" itlOO uWiLwe riauih. tha ^t-r tun 
 
 'Wba^ is- lo^:5t>7T 
 
 T^ 5680 
 
 *ttct the dttfertjiKti ,s '-^ 5. 'tjor. 
 
V.,,. ,. 
 
12 LOGARITHMS. 
 
 Find the logarithms of 
 
 (1)10.09; (2)190.1G; (3)1000.4; (4)0.12168. 
 Answers. (1) 1.0039 ; (2) 2.2791 ; 
 (3) 3.0002 ; (4) 1.0852. 
 
 Examples : 
 
 Find the logai 
 
 ithms of the following numbers : 
 
 1.834 
 
 19.98 
 
 0.01592 
 
 1.2899 
 
 54.97 
 
 1587.1 
 
 10.041 
 
 1000 
 
 9990 
 
 9999 
 
 0.09998 
 
 0.00000000010007 
 
 79930000 0.0001001 
 
 50090 
 
 0.6394. 
 
 Ansivers : 
 
 
 
 
 0.2G34 
 
 1.3006 
 
 2.2019 
 
 0.1106 
 
 1.7402 
 
 3.2006 
 
 1.0018 
 
 3.0000 
 
 3.999G 
 
 4.0000 
 
 2.9999 
 
 l0.0003_ 
 
 7.9027 
 
 4.0004 
 
 4.6997 
 
 f.8058. ' 
 
 § 14. To find the number ichich corresjyonds to a given 
 logarithm. 
 
 Of what number is 1.8156 the logarithm? 
 
 We find the mantissa .8156 on p. 3, in the same hori- 
 zontal line witli 65 (on the left) and in the vertical column 
 headed 4, therefore the significant figures of our number 
 are 654 ; now the characteristic of our logarithm being 1, 
 the first significant figure 6 of our number must be the first 
 to the left of the units' place, hence 5 is in the units' place, 
 and the required number is 65.4. 
 
 Find the munbers whose logarithms are 
 
 (1) 2.7789; (2) 4.3160; (3) 0.5478. 
 
 Ansivers. (1) 0.0601; (2) 20700; (3) 3.530. 
 
 Find the number (iV) Avhose logaritlim is 0.5506. 
 
 We do not find tliis logarithm in the table ; we do find, 
 however, two successive logarithms between which this 
 lies. Thus, 
 
LOGARITHMS. 13 
 
 log 3.550 = 0.5502, 
 logN =0.5506, 
 log 3.560 = 0.5514. 
 The difference between 0.5502 and 0.5506 is 4 (in the 4th 
 place), and the difference between 0.5502 and 0.5514 is 12 ; 
 .-. logiV^=log3.550+y42.of (log 3.560 -log 3.550); 
 hence, by the method of interpolation, 
 
 iVr= 3.550 + y% (3.560 - 3.550) 
 
 = 3.550 + -j^ (0.010) = 3.550 + 0.003^ 
 = 3.553= 
 ^ of 10 may be found in the table of proportional parts ; 
 thus, in the same horizontal line with our logarithm we 
 find 4, and vertically above this at the top we find 3, which 
 is the fourth significant figure of the number whose log- 
 arithm exceeds 5502 by 4. 
 
 Given: log.V= 3.5449 ; find N. 
 
 Seeking in the table the logarithm Clearest the given 
 logarithm, we find log 0.003510 = 3.5453, which exceeds 
 the given logarithm l;)y 4 ; and aliove 4 in the table of pro- 
 portional parts we find 3, which is in tliis case to be sub- 
 tracted from the fourth significant figure of the number 
 whose logarithm is 3.5453 ; 
 
 .-. iVr= 0.003510 — 0.000003 = 0.003507. 
 
 Examples : Find the numbers which correspond to the 
 following logarithms : — 
 
 1.8359 
 
 2.4089 
 
 1.8058 
 
 3.4429 
 
 3.4631 
 
 4.0003 
 
 4. 099 7 
 
 9.9016 
 
 1.0020 
 
 1.1109 
 
 0.0001 
 
 3.7402. 
 
 Ansivers : 
 
 
 
 
 68.53 
 
 0.025G4 
 
 0.6394 
 
 2773 
 
 0.002905 
 
 0.00010008 
 
 50090 
 
 0.000000007972 
 
 10.046 
 
 12.91 
 
 1.0002 
 
 5498. 
 
14 
 
 LOGARITHMS. 
 
 § 15. Examples of Products ill ast rating § G. 
 1. Find the product of 78.05, 0.6178, 341U0, 10.009, 
 and 0.0009. 
 
 Solution : We find by the tables, 
 
 log 78.05 
 logO.G178 
 log 34 100 
 log 10.009 
 los 0.0009 
 
 1.8924 
 
 9.7909 — 10* 
 4.5328 
 1.0004 
 0.1)542 - 10 
 
 log of prod. = 24.1707 - 20 = 4.1707 ; 
 
 .-. the requh'ed product is 14815. 
 
 *In order to avoid the use of negative characteristics, com- 
 puters generally add 10 to each negative characteristic, and then 
 write —10 after the logarithm. 
 
 2. Find the product of 
 
 (1) 4.31, 0.39, G4, and 1.02; 
 
 (2) 3G1, 0.043, and 0.00621 ; 
 
 (3) 3.81, 97.6, 3120, and 0.00081 ; 
 
 (4) 2.358, 4.321, 0.8765, and 1.12. 
 
 Aiisicers. (1) 109.75; (2) 0.09G4 ; 
 (3) 939.8; (4) 10. 
 
 § 16. The Arithmetical Complement of a logarithm^ 
 and examples of reciprocals illxstrating § 5. 
 
 The arithmetical complement of a logarithm — written 
 colog — is tlio remainder which we get after subtracting the 
 logarilhui IVoui 10. 
 
 Thus, if log78.05 = 1.8924, 
 
 then is colog 78.05 = 10-1 .8924 
 
 = 8.1076. 
 Find colog 6.912. 
 
 log6.912 = 0.8396; 
 
 .-. colo<r6.912 = 9.1604. 
 
LOGARITHMS. 15 
 
 The arithmetical complement of a given logarithm may 
 be obtained with great ease if we note that subtracting a 
 logarithm from 10 is equivalent to subtracting the last sig- 
 nificant figure from 10, and each of the other figures from 
 9. In getting colog6.912, for instance, 
 
 10.0000 
 loa:6.912 = 0.8396 
 
 colog6.912 = 9.1004 
 we can begin on the left, and saj, from 9 is 9, 8 from 9 
 is 1, 3 from 9 is 6, 9 from 9 is 0, and 6 from 10 is 4. 
 
 Findcologsof (1) 7.3G ; (2) 0.0134 ; (3) 81.2 ; (4) 3041. 
 Ansivers. (1) 9.1331; (2) 11.8729; 
 (3) 8.0904; (4) 6.5170. 
 We have, in general, 
 
 colog^= 10 — logiV; 
 subtracting 10 from both sides of this equation we get 
 
 colog^— 10 = — log^, 
 and from tliis we see that cokxj N diminislied by 10 is the 
 same as the negative of log N. 
 From § 5 we have 
 
 log — = — log^= (by the above) cologiV— 10. 
 
 Hence, the logarithm of the reciprocal of a number is equal 
 to the arithmetical complement of the logarithm diminished 
 by 10. 
 
 Find losjc — •• 
 "273 
 
 log— = colog273 - 10 = 7.5638 - 10 = 3.5638. 
 *273 
 
 Using the colog here has enabled us to obtain very easily 
 
 a positive mantissa for loa; 
 
 ^ ^273 
 
16 LOGARITHMS. 
 
 Find the values of 
 
 (1)^' (2) Jt^; (3):^; (4) 
 
 319' ^ ^ 0.0173' ^ ' 36.18' ^ ' 0.U00123 
 
 Answers. (1) 0.003135; (2) 57.81 ; 
 (3) 0.02764; (4) 8130. 
 
 § 17. Examples o/ Fractions illustrating § 7. 
 1. Divide 0.01478 by 0.9243. 
 Solution: We have, by the tables, 
 
 log0.01478 = 8.1697 -10 
 logo. 9243 =9.9658-10 
 
 subtracting, we get log of quotient = 2.2039 
 
 .-. the required quotient is 0.01599. 
 
 Remembering that cologiV"— 10 = —log ^^, we can do 
 the above example as follows : 
 
 logo. 01478 =8.1697-10 
 
 colos 0.9243 - 10 = 0.0342 
 
 and adding, we get log of quotient = 8.2039 — 10 
 
 = 2.2039 
 
 „ T- 1 ^-1 1 P 894x0.17.S6x41 
 2. Innd the value of 
 
 0.86x0.0007952x9001 
 
 log894 = 2.9513 logO.86 = 9.9345-10 
 
 logo. 1786= 9.2519-10 logO.0007952 = 6.9005-10 
 log41 = 1.6128 los9001 = 3.9542 
 
 log numer. = 13.8160 - 10 logdeuom. = 20.7892 - 20 
 = 3.8160 = 0.7892 
 
 0.7892 
 
 .logquot. = 3.0268 and tlie required quotient is 1063.8. 
 
LOGARITHMS. 17 
 
 We will now do the above by the aid of cologs. 
 
 log894 = 2.9513 
 
 logo. 1786 = 9.2519-10 
 
 log41 = 1.6128 
 
 cologO.86-10 = 0.0655 
 
 colog 0.0007952 - 10 = 3.0995 
 
 » coloo-9001 - 10 = 6.0458 - 10 
 
 23.0268 - 20 
 = 3.0268 
 A comparison of these two methods of solution shows 
 the advantage to be gained by the use of cologs. 
 
 Find the values of 
 
 .J. 3176 X 4.31 X .023 . .,-,. 731x9.16 . 
 ^ ^ 6.4 X. 231 ' ^"^ 2113x27 ' 
 
 .g. 6. 19 X 37000 X. 002 . ,^. 49880 X .03754 X 68.1 . 
 ^ ^ 31.96x40.1 ' ^ ^ 7.816x578.9x28.43' 
 A7iswers. (1)212.9; (2)0.1174; 
 (3) 0.3575 ; (4) 0.9910. 
 
 § 18. Examples o/ Powers and Roots illustrating § 8. 
 
 1. Find the fourth power of 0.9857. 
 Solution : We have b}' the tables, 
 
 log 0.9857 = 9.9937-10 
 multiply by 4 
 
 log (0.9857)* = 39.9748 - 40 = 1.9748 ; 
 
 .-. (0.9857)^ = 0.9436. 
 
 2. Find the values of 
 
 (1) (7.013)^ (2) (.3009)^ 
 
 (3) (1.04)1^ (4) (2.314)«. 
 
 Answers. (1)344.9; (2) .002469 ; 
 (3) 1.663; (4) 153.4. 
 
18 LOGARITHMS. 
 
 3. Find the cube root of G3. 
 Solution : We have bj- the tables, 
 
 log 63 =1.7903, 
 dividing this b^- 3 (multiplying l)y i) we get 
 
 logv'63 = 0.5998; 
 
 .-. ^v/63=: 3.979. 
 
 4. Find the fifth root of 0.02814. 
 
 Solution : We have by the tables, 
 logO.02814 = 2.4493 ; 
 
 if we divide the logarithm by 5 we get —^ for a character- 
 istic ; we will therefore increase tlie characteristic — 2 by 
 50* and write —50 after tlie logarithm ; thus, 
 
 logo. 02814 = 48.4493 - 50 ; 
 now, di\ading by 5, we get 
 
 log VO. 02814 = 9.6899 - 10 
 = 1.6899; 
 
 .-. V0.02814 = 0.4897. 
 
 * A number of tens equal to the exponent of the root ; i.e., in 
 this example, 5 teiK, or 50. 
 
 5. Find the cube root of 0.27. 
 
 Iog0.27 = 1.4314; 
 increasing the characteristic —1 b}' 3 tens, or 30, we get 
 logo. 27 =29.4314-30; 
 .-.log ^0.27= 9.8105-10 
 = 1.8105; 
 
 .•.-v/o:27 = 0.6464. 
 
 6. Find log (0.1 684)1. 
 
 First find log (0.1684)-, and then divide this logarithm 
 by 3. Ans. 0.3049. 
 
LOGARITHMS. 
 
 19 
 
 Find the values of 
 
 7.1; (2)V:0041; (3)V3l430; 
 {A)A/lAm; (5)^v^37J:9; (6) -^0.0000492. 
 Answers. (1) 1.922; (2) 0.2530; (3) 7.933; 
 (4) 1.072; (5) 1.321; (G) 0.3321. 
 
 § 19. Miscellaneous Problems and Examples. 
 
 1 . Given the exponential equation 20 = 2'' ; find x. 
 By § 8 we have 
 
 log 20 = a; X log 2 ; 
 ,^ log20 ^ 1.3010 
 * ' ^~ log 2 ~ 0.3010' 
 
 . • . log X = log (log 20 ) — log (log 2 ) 
 = log 1.3010 -logo. 3010. 
 logl.3010 = 0.1143 
 logo. 3010 = 1.4786 
 
 logo; =0.6357 
 
 .-. a; = 4.322. 
 
 We see from the definition of a logarithm (§ 1) that the 
 above value of x is the logarithm of 20 to the base 2 ; 
 
 .-. Iog2 20 = 4.322. 
 
 We can in like manner work out, by the aid of common 
 logarithms, the logarithms of all numbers to any base. 
 
 2. Find loggO.G. 
 
 Put 0.6 = 2^; .-. is logO.G =a; X log2; 
 
 .^ logo. 6 ^ 1.7782 
 ' ' '^~ iog2 ~ 0.3010' 
 
 In order to obtain the value of x from this equation, we 
 must observe that the form 1.7782, although sometimes 
 convenient for purposes of addition and subtraction, is not 
 
20 LOGARITHMS. 
 
 suitable for purposes of inultipliea,tion or division ; we will 
 therefo]"c reduce it to its equivalent, 
 
 -1 +0.7782= -0.2218; 
 
 ^0.2218' 
 
 ._ -0.2218 
 ^~ 0.3010 
 
 o.;30io 
 
 The value of the part in brackets, being positive, can be 
 found in the usual manner. 
 
 Thus, logo. 2218 = 1.3400 
 
 los0.3010 = T.478G 
 
 1.SG74 
 
 2-?18 
 ..-. ^-^ =0.7368; 
 0.3010 
 
 .-. x= — 0.73G8. 
 
 * Negative iiumljcrs have uo real logarithms ; for we can give 
 to 10 no exponent whicli will produce a negative number, since 
 10 with any exponent (positive or negative) gives always a pos- 
 itive quantity, greater than uuitj' if the exponent is positive, and 
 between zero and unity if the exponent is negative. 
 
 3. The number 2.7183 is the base of the so-called natu- 
 ral system of logarithms, a sj'stem much used in the higher 
 mathematics. 
 
 Find, Avith this number for a base, the logarithms of 
 0.7, 0.8, and 0.9. 
 
 Answers. — 0.35G6, —0.2231, and —0.1055. 
 
 4. How many digits are there in the integral part of 
 
 (1.04)«»°? 
 
 Put (1.04 )*■'"«' =.V, 
 
 then 6000 log 1 . 04 = log X, 
 
 GOOO X 0.0170 = 102.00= log.V. 
 
 The first significant figure of X is removed 102 steps 
 from the imits' place, and there are thei'eforc 103 digits in 
 the integral part of JSf. 
 
LOGARITHMS. 
 
 21 
 
 5. Show that there is onl}' one digit in the integral part 
 of (1.02y\ 
 
 6. Show that there are 20 digits in 2^\ 
 
 7. If $1 be put at d(/o interest, in how many years will 
 it amount to $2, if the interest is compounded annually? 
 
 The amount at the end of one year is $1.03 ; this amoimt 
 now forms a new principal, so tliat, at the end of the sec- 
 ond year, ever>/ dollar of this $1.0;l amounts in turn to 
 $1.08, and the tvJtoIe amount is $(1.03) (1.03) = $(1.03)^. 
 
 At the end of the third 3'ear the amotuit is evidently 
 $(1.03)% and at the end of m years, $(1.03)'". 
 
 In this example we are required to find a value m such 
 that 
 
 (1.03)- 
 mlosrl.OS 
 
 2, 
 log 2; 
 Tog 2 
 
 0.3010 
 
 = 23.5. 
 
 log 1.03 0.0128 
 In general, if $P be put at interest at a I'ate r, com- 
 pounded annuall}', the amount at the end of m years will 
 
 ^® $^=:$P(l+r)"'. 
 
 8. In how many years Avill $1000 amount to $1800, at 
 5 per cent compound interest? 
 
 1800 = 1000(1.05)"'; 
 
 log 1-8 10 1 
 
 •'• "^^^^ i 1 n' = 1^' "early. 
 logl.Oo 
 
 9. Show that a sura of mone}' at compound interest will 
 double itself in al)out 7:}- years if the rate be 10 per cent, 
 and in about 1 7t} years if the rate be 4 per cent. 
 
 10. Show that, if the population of a given State increases 
 by o'oth of itself in every year, tlie number of inhabitants 
 at the end of 14.2 years will l)e twice as great as it was at 
 the beofinnins: of the interval. 
 
 ■"«'^*'*4<*niiv«wv*ivMii>iimi 
 
 R'Wf.WPflKD.'iaiJ/^^P';*' 
 
22 LOGARITHiVIS. 
 
 [11.* logO. 
 
 — = 10-^; .-. loii —=-X. 
 
 Let a; = CO (ineauiug by this that x is a vai"ial)le quantity which 
 is supposed to iucrease iucleflnitely, tluit is, to increase without 
 limit), theu lO-' will become oo, and we shall have loi;;— = — oo. 
 
 — , or — when x = oo, is not ; it is a quantity which may be 
 
 made to approach as nearly as we please, but it can never be 
 made to reach 0; such a quantity is called an infinitesimal, and 
 is denoted by some authors by the symbol o. (I believe that 
 this notation was first introduced by Professor C. H. Judson of 
 South Carolina.) 
 
 Zero has no logarithm; many authors say, however, that the 
 logarithm of zero is — co, meaning tliereb}^, perhaps, that — oo is 
 the logarithm of an infinitesimal, that is, of a variable which has 
 for its limit. 
 
 I very much prefer to use some symbol — a liorlzontal zero, 
 for instance — for an infinitesimal; I can then say with strict 
 accuracy that log o = — co. I also express this by ])utting 
 logO = — ob, using the dot over the sign of infinity to denote 
 that log is impossible, and retaining the symbol — » because 
 it expresses the logarithm of a quantity whose limit is 0. I be- 
 lieve that this mode of expression can be used advantageously 
 in other cases, and that its adoption, or the adoption of some 
 other method whereby we are not compelled to regard a vari- 
 able as reaching its limit, will make pcrfectlj^ clear to the stu- 
 dent of mathematics many things which he now accepts on 
 faith. 
 
 The following are examples of cases to which I would apply 
 what has been stated above : — 
 
 (1) i = 00 [not i = ccl ; i = ob. 
 
 (2) 00 X o = rt, not 00 X — «, for od x = 0. 
 
 (3) tan 90° = ob, not tan 90° = oo. 
 
 (4) Two parallel lines do not meet at infinity ; thev may be, 
 however, the limits approached by two lines whose point of inter- 
 section recedes indefinitely.] 
 
 * This is intemled lor llie toaclu'r r;itlK'r lliaii lor tlie stiuioiit. 
 
LOGARITHMS. 
 
 9?. 
 
 12. Find the value of 
 
 a/0.0432 X (3.122)f X (0.0001)^' 
 (18G2)2x syo,0048 
 
 logo. 0432 =58.6355-60 logVO.0432 = 9.7726-10 
 
 [log3.122 = 0.4945 
 
 ilog(3.122)« = 3.9560 log(3. 122)7 = 0.5651 
 
 logo. 0001 = 4.0000 log(0. 0001)1= 1.0000 
 
 nogl862 = 3.2700 
 
 !log(1862)- = 6.5400 colog(1862)--10= 3.4600-10 
 
 I log 0.0048 = 3.6812 
 
 (log ^0.0048 = 1.2271 cologSyo.0048 - 10= 10.7729 - 10 
 
 23.5706—30 
 or 7.5706 
 .-. the reqiured value is 0.0000003721. 
 
 Find the values of 
 
 13. ( {O-^O'T^m Ans. 0.5549. 
 
 V0.4x67 
 
 14. W (8.763)-^ XIQO ;^^ ^^^^_ g_ 
 
 >'9X VO.1109 X (4.9)i 
 
 15. ^/(0.01-2)-x0.27. ^4,^^._ ,_o,3i_ 
 
 \(64)2x 0.00651 
 
 16. Jv-27.4 + (0.9)^^ ^^^^ jg4_97_ 
 
 ^^ 0.0001021 
 
 17 1^ (134.9)- X a/Tg Y Ans. 0.0003164. 
 
 V 10000 X 46.49 / 
 
 18. 
 
 0.001019004_x^0.j999 ^^^_ 0.000004442. 
 
 760 X a/0.02751 
 
24 LOGARITHMS. 
 
 § 20. The Trigonometric Functions. 
 
 The trigonometric functions of angles between 0° and 90°, 
 at intervals of 10', are given in the tables on pages 22-27. 
 Let us find the functions of 15° 12' for example. Under 
 <f>, in the left-hand column of the tJmxl division of page 22, 
 we find 15° immediately followed b}- 10' ; in the same 
 horizontal line with the latter, and in the column headed 
 sin </), we find .2616, which is sin 15° 10'. 
 
 .-. sin 15° 10' = 0.2616, and similarly 
 sin 15° 20' =0.2644 
 
 difference = 28. 
 Now, by the method of interpolation (see § 13), the 
 amount to be added to 0.2616 in order to produce 
 sin 15° 12' is .2 of 28, or 5.6, which we call. 6; 
 
 . . sm lo 12' = -^ ,,,, , 
 -f .0006 
 
 = 0.2622. 
 In like manner, from page 22 and from pages 24 and 26, 
 cos 15° 12'= 0.9650* 
 
 tanl5° 12' = 0.2717 sec 15° 12' = 1.036 
 
 ctn 15° 12' = 3.681 esc 15° 12' = 3.814. 
 
 By the above method we find : — 
 
 sin 13° 28' = 0.2328 sin 28° 13' = 0.4728 
 
 cos 13° 28' = 0.9725 cos 28° 13' = 0.8812 
 
 * Cos 15° 20' being less than cos 15° 10', .2 of the difference 
 was subtracted from cos 15° 10'. The student should carefully 
 consider whether the amount obtained by the method of inter- 
 polation is to be added or subtracted. 
 
 We learu, either from Trigonometry or from an examination 
 of the tables, that the sine, tangent, aud secant iiicrease, and 
 that the cosine, cotangent, and cosecant decrease, as the angle 
 increases from 0° to 90°. 
 
LOGARITHMS. 25 
 
 tan 13° 28' = 0.2395 tan 28° 13' = 0.5365 
 
 ctn 13° 28' = 4.176 ctn 28° 13' = 1.864 
 
 sec 13° 28' = 1.028 see 28° 13' = 1.135 
 
 CSC 13° 28' = 4.294 esc 28° 13' = 2.115, 
 
 and in a similar way are to be found the functions of all 
 angles heticeen 0° ayid 45°. 
 
 § 21. We will now show how to find the trigonometric 
 functions of angles l)etween 45° and 90°. 
 
 Let us take 58° 24' for example. At the bottom of the 
 right-hand column of the middle division of page 23 we 
 find ; we run up this column luitil we find 58°, and im- 
 mediately above this 10', 20', etc. ; in the same horizontal 
 line with 20', and in the column which has sin^ at the hot- 
 torn^ we find .8511, which is the sine of 58° 20', Now, hy 
 the method of interpolation, we get sin 58° 24' = 0.8517. 
 
 Similarl}^ : — 
 
 cos 58° 24' = 0.524 sec 58° 24' = 1.909 
 
 tan 58° 24' = 1.625 ctn 58° 24' = 0.6152 
 
 CSC 58° 24'= 1.174 
 
 sin 84° 13' = 0.9949 sin 46° 28' = 0.7250 
 
 cos 84° 13' = 0.10077 cos 46° 2S' = 0.6888 
 
 tan 84° 13' = 9.88 tan 46° 28' = 1.053 
 
 ctn 84° 13' = 0.10128 ctn 46° 28' = 0.9501 
 
 sec 84° 13' = 9.93 sec 46° 28' = 1.452 
 
 CSC 84° 13' = 1.0051 CSC 46° 28' = 1.380. 
 
 From the above we see that when the angle is between 
 45° and 90° it is to be sought in the right-hand column, 
 that the name of the required function is to be found at 
 the bottom of the page, and that the columns are to be 
 read upwards. 
 
26 LOGARITHMS. 
 
 § 22. When a function of an angle is given, the angle 
 can be fonnd from the tables l^y a method simihir to that 
 alread}' explained for finding a nnmber from its logarithm. 
 Thus, if sinvl = 0.G792, we find on page 23, 
 
 sin 42° 40'=0.G777 
 
 sin 42° 50' =0.6799 
 
 difference = 0.0022 
 
 and sin 42° 50'- sin^l = 0.0007 ; 
 
 .-. ^ = 42° 50'-— X 10' = 42°47'. 
 22 
 
 Examples : Find A from the tables in each of the fol- 
 lowing cases : — 
 
 (1) sin ^=0.9621; (4) sec ^1 = 1.111 ; 
 
 (2) cos^l = 0.12G8; (5) ctnyl = 0.0079 ; 
 
 (3) tanJ. = 2.172; (6) esc J. = 9.872. 
 Answers: (1) 74° 10'; (3) 65° 17'; (5) 89° 27' ; 
 
 (2) 82° 43'; (4) 25° 50' ; (G) 5° 49'. 
 
 § 23. Referring to the tables, pages 22-27, we notice 
 that the angles in the right-hand columns are the comple- 
 ments of the angles in the left-hand columns. Thus, oppo- 
 site 41° 10' on the left we find 48° 50' on the right, and we 
 
 also find 
 
 sin 41° 10' = 0.G583 = cos 48° 50' 
 
 cos 41° 10' = 0.7528 = sin 48° 50'. 
 
 These results are in accordance with the principles of 
 
 Trigonometry, for Trigonometry tells us that the functions 
 
 of any angle (<^) are the complementary functions of its 
 
 conn)lemont (90°—^). 
 
 § 24. Tlie lunctions of angles greater than 90° can be 
 expressed in terms of the functions of angles less than 90°, 
 b}- the aid of the formulas of §§ 31 and 32, Plane Trigo- 
 nometry. 
 
LOGARITHMS. 27 
 
 Examples : 
 sin 259° 25'= sin (180°+ 79° 25') = -sin 79° 25'= - 0.9<S30 
 cos259°25'=cos(180°+79°25') = -cos79°25'=-0.1836 
 sin 123° 32'= 0.8336 ctu258°19'= 0.20G8 
 
 tan 123° 32' = - 1.509 sin 258° 19' = - 0.9792 
 
 sec 300° 18'= 1.982 cos 174° 12' = - 0.9949 
 
 cos 300° 18'= 0.5045 ctn 174° 12' = - 9.85. 
 
 Find A m each of the following cases : — 
 
 (1) sin.l= 0.5G21 ; (7) esc ^1 = 2.842; 
 
 (2) sin A = - 0.9021 ; (8) esc A = - 1 .4G9 ; 
 
 (3) tan.l= 1.451; (9) ctn.l= 0.4592; 
 
 (4) tan.l = -0.G921 ; (10) ctn .1 = - 0.9221 ; 
 
 (5) sec^l= 1.0084; (11) cos yl = 0.4G82 ; 
 
 (6) sec^=-2.9G4; (12) cos .1 = - 0.8196. 
 Answers : 
 
 (1) 34° 12' or 145° 48'; (7) 20° 36' or 159° 24' ; 
 
 (2) 244° 26' or 295° 34' ; (8) 222° 54' or 317° 06' ; 
 
 (3) 55° 26' or 235° 26'; (9) 24° 40' or 204° 40' ; 
 
 (4) 145° 19' or 325° 19'; (10) 132° 39' or 312° 39' ; 
 
 (5) 7° 25' or 352° 35'; (11) 62° 05' or 297° 55' ; 
 
 (6) 109° 43' or 250° 17' ; (12) 145° 02' or 214° 58'. 
 
 § 25. "When logarithms were invented the}' were called 
 artificial numbers, and the originals, for which logarithms 
 were computed, were accordingly called natural numbers." 
 Dk MouctAN. 
 
 The tables just explained, §§ 20-24, are therefore called 
 tables of the natural functions of an angle. 
 
 § 26. Logarithms of the Trigonometric Functions. 
 
 Having found the natural functions of an angle by the 
 methods just explained, we can find the logarithms of 
 
28 LOGARITHMS. 
 
 these functions from the table of the logarithms of num- 
 bers (pages 2-0 of the Tables) : it more often happens, 
 however, that the natural functions themselves are of no 
 service to the computer ; he generally wishes only the 
 logarithms of these functions. If, for example, 
 
 a = 16 sin 28° esc 32°, 
 we have log a = log 1 G + log sin 28° + log esc 32° ; 
 and hence, in order to fuid the value of a we do not need 
 sin 28° and esc 32° but their logarithms. For the above 
 reason tables have l)een constructed (see pages 8-15 of the 
 Tables) , liy the aid of which the logarithms of the trigono- 
 metric functions can be found directbj from the angles. 
 
 § 27. Angles between G° 00' and 8i° 00' inclusive. 
 (Tables: pages 10-1.k) 
 
 To the student who is familiar with the tables already 
 discussed [Tables : pages 2-5 and pages 22-27] only two 
 things need be explained in regard to pages 10-15. 
 
 First. In order to avoid the use of negative cliaracter- 
 istics, each logarithm whose characteristic is negative has 
 been increased by 10. Thus, from page 10 we find 
 log sin 7° 00' = 9.0859, and this, the tabular logarithm of 
 sin 7° 00', is 10 too large. 
 
 .". true log sin 7° 00' = tabular log sin 7° 00' — 10 
 = 9.0859-10 = 1.0859. 
 
 Now, the characteristic of the logarithm of any trigono- 
 metric function will be negative when the function itself 
 is less than 1, and positive when the function is greater 
 than 1 ; and we know either from Trigonometry or froui 
 the tables oH natural functions, pages 22-27 of the tables, 
 that the sines and cosines of all angles, the tangents of 
 angles between 0° and 15°, and the cotangents of angles 
 
LOGAEITIIMS. 29 
 
 between 45° and 00°, arc less than 1, while the secants and 
 cosecants of all augk's, the tangents of angles between 
 45° and 90°, and the cotangents of angles between 0° and 
 45°, are greater tlian I. Consequeutl}', the tabular loga- 
 rithms of all sines and cosines, of the tangents of angles 
 between 0° and 45°, and of the cotangents of angles be- 
 tween 45° and 1)0°, are 10 too large. 
 
 Examx)les : true log sin 81° 20' = 9.9950 - 10 
 true logtan81° 20' = 0.81G9 
 true log tan 42° 10' = 9.9570 - 10 
 true logctn42° 10' = 0.0930. 
 
 Second. The numbers printed in smaller type are dif- 
 ferences to be used in interpolating ; we find, for example, 
 between log sin 7° 00' and log esc 7° 00', the number 103 in 
 smaller type, and this is the difterence to be used in flnd- 
 nig, l)y the method of interpolation, the logarithms of the 
 sines and cosecants of angles between 6° 55' and 7° 05', 
 i.e., of angles that are nearer to 7° 00' than they are to the 
 angle which precedes or follows it. 
 
 Let us lind log sin 7° 02' and log esc 7° 02'. 
 
 log sin 7° 00' = 9.0859 log esc 7° 00' = 0.9141 
 
 .2 of 103= 21 .2 of 103= 21 
 
 .-. log sin 7° 02' = 9.0880 ; .-. log esc 7° 02' = 0.9120. 
 
 Find log sua 7° 08'. 
 
 7° 08' is nearer to 7° 10' than it is to 7° 00', therefore we 
 use for our difference the number 100 (oi)i)osite 7° 10'). 
 
 Iogsin7°10' = 9.09(;i 
 .2 of 100 = 20 
 
 •. Io2sin7°08' = 9.0941. 
 
30 LOGARITHMS. 
 
 Find logctn 7° 08' and log cos 7° 07'. 
 
 log ctn 7° 10' = 0.9005 log cos 7° 10' = 9.9966 
 
 .2 of 102 =20 .3 of 2 = 1 
 
 .-. log ctn 7° 08' = 0.9025 ; .-. log cos 7° 07' = 9.99G7. 
 
 Suppose we wish to find log sin 7° 05' ; this angle hciiig 
 half way hetween 7° 00' and 7° 10', we will take for onr 
 difference lOH, which is half way between 103 (opposite 
 7° 00') and 100 (opposite 7° 10'). 
 
 log sin 7° 00' = 9.0859 
 .5oflOU= 51 
 
 .-. log sin 7° 05' = 9.0910. 
 
 The student has donT)tless noticed tliat the number 103 
 (opposite sine and cosecant of 7° 00') is not the actual 
 difference between two successive logarithms of our table, 
 
 for log sin 7° 00' - log sin 6° 50' = 104, 
 
 and log sin 7° 10' - log sin 7° 00' = 102 ; 
 
 if Ave refer to a 7-place table, however, we find 
 
 log sin 7° 05' = 9.0910 082 
 log sin G° 55' = 9.0807 189 
 
 difference =0.0102 893 
 
 and this difference carried out to four })laces we call 103. 
 
 103, then, is a more accurate difference; for us to use \\\ 
 finding the log sine and the log cosecant of angles whicli 
 fall between 6° 55' and 7° 05' than tlie actual difference 
 between successiA'^e logarithms of our table would be. We 
 see from the aboA-e hoAv tlie lunnbers in smaller type iiavc; 
 been obtained ; tliese numbers neA'cr differ much, and in 
 most cases not at all from the differences l)etween succes- 
 sive logarithms of our table. 
 
LOGARITHMS. 
 
 
 '61 
 
 = 9.1959 log cos 
 
 14° OG' 
 
 = 9.9867 
 
 = 9.9400 log cos 
 
 59° 04' 
 
 = 9.7110 
 
 = 9.9855 log cos 
 
 70° 15' 
 
 = 9.5288 
 
 § 28. Examples: 
 
 log sin 9° 02' 
 log sin 62° 01' 
 log sin 75° 18' 
 
 logsinl28°06' =9.8959 log[-cos210°26']*= 9.9357 
 
 log[-sin200° 16']*= 9.5395 log[-cos 99° 18']*= 9.2085 
 
 logtan 24° 02' =9.6493 logctn 18° 04' =0.4865 
 
 logtan 60° 12' =0.2421 logctn 48° 12' =9.9514 
 
 log[-tanl20° 04'] = 0.2374 log [ - ctn 99° 02'] = 9.2013 
 
 logtan260° 18' =0.7672 log ctn210° 16' =0.2339 
 
 log sec 16° 02' =0.0173 log esc 22° 01' =0.4261 
 
 logsec 70° 16' =0.4715 log esc 48° 02' =0.1287 
 
 log [-seclOO° 18'] = 0.7476 log esc 102° 58' =0.0112 
 
 log [- sec 200° 02'] = 0.0271 log [- ese 216° 46'] = 0.2229 
 
 Find A in each of the following cases : — 
 
 (1) log sin ^ =9.6021 ; (8) log[-tan^]= 9.9966 ; 
 
 (2) log sin yl = 9.8588 ; (9) logctn^ = 0.2174 ; 
 
 (3) log [-sin^] = 9.8120; (10) log [-ctnyl]= 9.7218 ; 
 
 (4) logcos^ =9.6821; (11) logsec^l =0.1261; 
 
 (5) log cos ^ =9.8266; (12) log [-sec ^1] = 0.6879 ; 
 
 (6) log [-cos^] = 9.7692; (13) log esc ^1 =0.6428; 
 
 (7) log tan .1 =0.6868; (14) log [-csc.4] = 0.6999. 
 
 Answers : 
 
 or 156° 25'; 
 or 133° 45'; 
 or 319° 33'; 
 
 or 298° 45'; 
 or 312° 08'; 
 
 * See Example 2, pp. 19, 20, and note, p. 20. 
 
 (1) 
 
 23° 35' 
 
 (2) 
 
 46° 15' 
 
 (3) 
 
 220° 27' 
 
 (4) 
 
 61° 15' 
 
 (■5) 
 
 47° 52' 
 
 (6) 
 
 126° 00' 
 
 (7) 
 
 78° 23' 
 
 {^) 
 
 135° 14' 
 
 or 
 
 224° 46' 
 
 ('^>) 
 
 31° 07' 
 
 or 
 
 211° 07' 
 
 (10) 
 
 117° 47' 
 
 or 
 
 297° 47' 
 
 (11) 
 
 41° 35' 
 
 or 
 
 318° 25' 
 
 (12) 
 
 101° 50' 
 
 or 
 
 258° 10' 
 
 (13) 
 
 13° 09' 
 
 or 
 
 166° 51' 
 
 (14) 
 
 191° 31' 
 
 or 
 
 348° 29'. 
 
32 LOGARITHMS, 
 
 Angles between 0° and 6° and between ,84° and 90°. 
 (Tables: pages 8, 9.) 
 
 § 29. When au angle is near 0° or 90°, the difference 
 between successive logarithms of certain of its functions is 
 comparatively' large, and when the interval in angle is as 
 great as 10', this difference is so large that we cannot 
 obtain accurate results b}" the method of interpolation for 
 angles which lie within the interval. (See note on page 9, 
 and also last paragraph of page 11.) 
 
 We lind, therefore, on pages 8 and 9 of the tal)les, a sep- 
 arate table for angles between 0° and G° and between 84° 
 and 90°, where the interval between the successive angles 
 of tlie table is only 1'. 
 
 § 30. log sin, log tan, and log sec of angles between 0° 
 and G°. 
 
 ((() log sin . Let us find log sin 1° 35'. At the top of the 
 middle division of page 8 of the tables we find 1°, 1 sin, 
 1 tan, and 1 sec ; running down the column headed 1 sin 
 until we reach the horizontal line which has 35' on the left, 
 we find 8.4414, which is log sin 1° 35'. 
 
 Find log sin 2° 18. C : in the third division of j)age 8 of 
 the tables we find 
 
 log sin 2° 19' =8.6006, 
 and on the right, in smaller type, the difference SI : 
 .4 of 31 = 12; 
 
 .-. logsin 2° 18'.6 = 8.6054. 
 
 Find log sin 3° 30'. 2. Ans. 8.7861 . 
 
 (b) log tan. The log tan of an angle between 0° and 
 6° diflfers little from its log sin ; they are, in general, the 
 same, except in the last two figures of the mantissa. 
 
LOGARITHMS. 33 
 
 Therefore, in order to save space, onl}^ the last two figures 
 of the log tan are given in the table. 
 
 Find log tan 1° 35' : on the right of log sin 1° 35', in the 
 eohinni headed 1 tn, we find 16, the last two figures of the 
 required mantissa ; therefore, remembering that the char- 
 acteristic and first two figures of the mantissa are the same 
 as in log sin 1° 35', we get log tan 1° 35' = 8.4416. 
 
 Find log tan 2° 18'. Ans. 8.6038. 
 
 Find log tan 5° 13'. We find in the third division of 
 page 9 of the tables, in the column headed Itn, the figures 
 05, and just to the left of these figures a star ; this star 
 indicates that the characteristic and first two figures of the 
 mantissa are to l)e taken 7iot from the logarithm to the left 
 of Ob, but from the following logarithm; thus, 
 
 log tan 5° 13' = 8.9605. 
 
 In this case log tan 5° 13' is a little above 8.9600, 
 and has therefore 8.96 for its first three figures, while 
 log sin 5° 13', being a little below 8.9600, has 8.95 for its 
 first three figures. 
 
 Find log tan 3° 42'. Ajis. 8.8107. 
 
 Where there are two stars in succession the first three 
 figilres are to be taken from the log sin which follows the 
 second star ; thus, log tan 5° 20' = 8.9701 , 
 logtan5°35' = 8.9901. 
 Find, by interpolating, 
 
 logtan0°54'.4. Ans. 8.1994. 
 
 'logtanr02'.8. Ans. 8.2617. 
 
 (c) log sec. The characteristic, and the first two figures 
 of the mantissa, of log sec of any angle between 0° and 6° 
 are zeros, and are not given in the table. 
 
34 
 
 LOGARITHMS. 
 
 Find log sec 1° 35' ; in the secant (sc) column for 1°, and 
 opposite 35 '"on the left, we find 02, the last two figures of 
 log sec ; prefixing zeros for the other figures we get 
 log sec r 35' = 0.0002. 
 Find (1) log secO° 35' ; (3) log sec 2° 18'. 
 (2) log sec 4° 29'; 
 Answers: (1) 0.0000*; (2) 0.0013; (3) 0.0004. 
 
 § 31. log cos, log ctn, and log esc of angles betiveen 84° 
 and 90°. 
 
 We proceed as in § 30, except that the angles and names 
 of the functions are to be found at the bottom of pages 8 
 and 9 of the tables, while the minute column is to be found 
 on the right, and all the columns are to be read upwards. 
 Thus : 
 
 logcos84°lG' =8.9996, logctn84°16' =9.0017, 
 logcsc84°16' =0.0022, logcos 89° 05'.6 = 8.1993, 
 logctn89° 05'.6 = 8.1994, log esc 89° 05'.6 = 0.0001. 
 
 (4) log tan .1 = 9.0103; 
 
 (5) log tan .4 = 8.9903. 
 
 § 32. To find an angle (betiveen 0° and C)°) from its 
 log sin, log tan, or log sec. 
 Given : 
 
 (1) log sin .4 = 8.8140; 
 
 (2) log sin .4 = 8.9116; 
 
 (3) logtan^l = 8.6424; 
 Find .4 in each case. 
 
 Answers: (1)3°44'.2; (2) 4° 40'.8 ; (3) 2° 30'.8 ; 
 (4) 5° 50'.8 ; (5) 5° 35'.2. 
 
 * If an ancle is near 0° its secant is near 1. and tliercfore tlie 
 lo2:sec is near 0; the table gives log sec 0° 35'= 0.0000, which 
 means that log sec 0° 35' ditters from by au amount too small 
 to affect the fourth place of decimals. 
 
LOGARITHMS. 
 
 35 
 
 Given: log sec ^d = 0.0016 ; find ^. 
 
 Referring to page 8 of the tables, we find that 0.0016 is 
 the log sec of all angles from 4° 51' to 4° 59' inclusive ; we 
 cannot, therefore, find an angle with definiteness from its 
 log sec when tlie angle is near 0°. 
 
 § 33. To find an angle (between 84° and 90°) from its 
 log cos, log ctn, or log esc. 
 Given : 
 
 (1) log cos ^1 = 9.0076 
 
 (2) log cos ^ = 8.5360 
 
 (3) log ctn ^1 = 8.4190 
 Find A in each case. 
 
 Answers: (1) 84° 09 '.5; 
 (2) 88° 01'.9 ; 
 
 (4) log ctn yl = 8.6870; 
 
 (5) log esc yl = 0.0005; 
 
 (6) log CSC ^1 = 0.0012. 
 
 (3) 88° 29 '.8 ; 
 
 (4) 87° 12'.9 ; 
 
 (5) Any angle* from 87° O8'to 87° 23', inclnsive. 
 
 (6) Any angle * from 85° 40' to 8o° 49 ', inclusive. 
 
 §34. 
 
 log CSC, log ctn, and log cos of angles between 0° 
 and 6° ; and log sec, log tan, and log sin of angles between 
 84° and 90°. 
 
 We cannot find these logarithms directly from the table ; 
 we know from Trigonometry, however, tliat 
 
 1 . . 1 ,1 
 
 csc<^ 
 sec^ 
 
 sin ^ 
 1 
 
 ctn ci = • 
 
 tan ^ = 
 
 tan </) 
 1 
 
 cos <^ 
 sin </) = 
 
 1 
 
 cos (j> ctn ' esc cji 
 
 From § 16 we Ivnow that the logarithm of the reciprocal of 
 a quantity' is the colog of the quantit}^ minus 10, therefore : 
 log CSC cji = colog sin ^—10 log sec (f> = colog cos </> — 10 
 log ctn ^ = colog tan ^—10 log tau c^ = colog ctn (/> — 1 
 log cos cf) — colog sec ^ — 10 log sin <^ = colog esc (/> — 1 
 
 * An angle near 90° cannot be found with defluiteuess from 
 its lo" CSC. 
 
36 
 
 LOGARITHMS. 
 
 Let US find log esc 2° IG' ; from tlie taV)le (p. S) we get: 
 tabular* log sin 2° IC = 8.5'J72 ; 
 .-. true log sin 2° 16' = 2.5972 ; 
 .-. log CSC 2° 1G' = 11.4028- 10 =1.4028. 
 What was done above is eqni\'alent to subtracting tabular 
 log sin from 10, for 
 
 log CSC 2° 16' = colog sin 2° 16' - 10 
 = (10 - 2.r>972)-10 
 = 10 -(2.5972 + 10) 
 = 10 - 8.5972 (tab. log) = 1.4028 ; 
 >•. log esc 2° 16' = 10 - tab. log sin 2° 16' ; 
 and in the same way it ma}' be shown that when ^ is an^^ 
 angle between 0° and 6°, 
 
 ( 1 ) log CSC <j!) = 10 — tab. log sin <^, 
 
 ( 2 ) log ctn <;()=: 1 — tab. log tan <^, 
 and when is between 84° and 90° 
 
 (;3) log sec <;^ = 10 — tal). log cos <^, 
 
 (4) log tan (ji — 10 — tab. log ctn cjy. 
 
 If we also put 
 
 (5) log cos (^ = 10 — tab. log sec ^, 
 
 (6) log sin 9^ = 1 — tab. log esc <^, 
 
 we shall get results which, although 10 too large, will accord 
 with other tabular logarithms. (See § 27, first part.) 
 
 After a little practice the student can tell at a glance 
 whether a logarithm is 10 too large ; he will then be able 
 to use the following rule : — 
 
 To get the logarithm of the reciprocal of a function sub- 
 tract the tabular logarithm of the function from 10, and the 
 result ivill be either the required logarithm or one which 
 differs from it by 10. 
 
 * See § 27, first part 
 
LOGAEITIIMS. 
 
 37 
 
 Examples : 
 log CSC 0° 30' = 2.0592 
 logctn 0° 30' = 2.0591 
 log cos 0° 30' = 0.0000 
 ]ogsec«()° 18'= 1.1902 
 logtaii8G° 18'= 1.1893 
 lossm8(r 18' = 9.9991 
 
 log CSC 2° 53' =1.2984 
 logctn 2° 53' = 1.2979 
 log cos 2° 53' =9.9994 
 log sec 88° 02'.4= 1.4660 
 logtan88°02'.4 = 1.4657 
 log sill 88° 02'. 4 = 9.9997. 
 
 § 35. To find cm angle {between 0° and 6°) from its 
 log CSC, logctn, or log cos; and an angle (between 84° and 
 90°) from its log cos, log tan, or log esc. 
 
 (1) Given: log esc ^= 1.2458 ; find ^. 
 From § 34 (1) log esc ^= 10 — tab. log sin ^ ; 
 
 .-. tab.logsinyl = 10-logcsc^= 10-1.2458 
 = 8.7542. 
 
 .-. from the table, .4 = 3° 15 '.3. 
 
 (2) Given : log etn A = 1.444 ; find ^. 
 
 A71S. (from § 34 (2) and then from table) ^ = 2° 03'.5. 
 
 Find A in each of the following cases : — 
 
 (3) log cos .1 = 9.9988; (G) log cos ^1 = 9.9976 ; 
 
 (4) logcsc^l= 1.7921 ; (7) log esc J. = 1.1111 ; 
 
 (5) log etn .4 = 1.3221 ; (8) logctn^ = 1.1086. 
 
 A7iswers : 
 
 (3) Any angle from 4° 11' to 4° 20', inclusive ; * 
 
 (4) 0°55'.2; (5) 2° 43'.9 ; 
 
 (6) Any angle between 5° 58' and 6° 04', inclusive ;* 
 
 (7) 4°26'.4; (8) 4° 27'.2. 
 
 * An angle near 0° can not be found with deflniteuess from 
 its log COS. 
 
38 LOGARITHMS. 
 
 § 36. Explanation of the formulas by the aid of lohich 
 the small table, at the toj) of jxige 10 of the tables, may be 
 used. 
 
 We know from Trigonometry tliat, in a circle with radius 
 unity, when ^ is small, sin cf) and tan <fi are very nearh' 
 equal to arc^, or, where <^ has l)een reduced to minutes, 
 that sill (fy = arc </> = ^ arc 1 ', approximately', 
 
 and tan ^ = arc 4> — 4> ^i"<^ 1 ' ^ approximately ; 
 
 .'. (1) log sin ^ = log<^ + log arc 1'. approximately. 
 (2) logtan(/) = log^ + log arc 1', approximately. 
 
 We know also that sin ^ < arc ^ and that tan^ > arc^ ; 
 therefore, log sin ^ obtained by (1) will be a little too 
 small, and log tan (^ obtained by (2) will be a little too 
 large ; let us denote the amount b}' which log sin ^ is too 
 small by e and the amount b}- which log tan <^ is too large 
 by e', then we get 
 
 log sin (^ = log cf) + log arc 1 '— e, 
 log tan (f) = log (f) + log arc 1 ' + «'• 
 
 Now, denoting logarcl'— e by S and log arc 1'+ c' by 
 T, we get 
 
 I. log sin (j!) = log ^ + S . 
 II. log tan (j> = log c^ + T. 
 
 These are the formulas by the aid of which the small 
 table is to be used. 
 
 Let us compute S and T wlien <^ = 0° 44'. 
 From T. , S = log sin ^ — log cf). 
 
 and from II . , T = log tan ^ — log <^ ; 
 
 logsinO° 44' = 8.1072 logtan 0°44' = 8.1072 
 log 44 =1.6435 log 44 =1.6435 
 
 S = 6.4637 T = 6.4637. 
 
LOGARITHMS. 
 
 39 
 
 Let us now compute log arc 1'. 
 Arc 360° = 2 7r>- = 2 TT : . • . arc 1 ' 
 
 ISO 
 
 arcl' 
 
 log 3.1416 
 lo2 10800 
 
 60 X 180 
 0.4971 
 4.0334 
 
 3.1416 
 10800 ■ 
 
 log arc 1' =6.4637 
 and this is the same as the vakies of S and T found above. 
 Hence we see that when ^ = 0° 44' (or less) , S and T 
 differ from log arc 1' hy quantities too small to affect the 
 fourth place of decimals. In like manner have been com- 
 puted the different values of 8 and T given in the small 
 table. In this table the first three figures of S and T are 
 printed in full-faced type at the top of the column, while 
 the last two figui'es are to be found l)elow. Referring to 
 this table we see that, as a small angle ^ increases, S and 
 T vary very slowly and differ little from log arc 1' [6.4637]. 
 
 § 37. The logarithms of the fimcfio)ts of angles near 0° 
 or 90° by the aid of the small table at the top of page 10 of 
 the tables. 
 
 I. Angles between 0° and 6°. 
 
 log sec (f> is readily found from the third division of the 
 small table, where the first three figures of the logarithm 
 are printed in full-faced type at the top of the column 
 and the last two figures are given below. Thus, to get 
 log sec 0° 18' we refer to the column headed cf) in the third 
 division of the table, and find that 0° 18', falling between 
 0°00' and 0° 52', has a log sec equal to 0.0000. In like 
 manner we find 
 
 log sec 1° 02' = 0.0001 log sec 4° 12' = 0.001 2 
 los sec 0° 50' = 0.0023 los sec 2° 50' = 0.0005. 
 
40 LOGARITHMS. 
 
 log cos (ji cannot be found directh' from the table ; we 
 
 know, however, that cos (f> = •> therefore by § 34 we 
 
 sec (/) 
 
 can get log cos by first finding log sec (/> and tlien sub- 
 tracting the latter from 10. Thus, 
 
 log cos 0° 18' = 10.0000 log cos 4° 12' = 9.9988 
 log cos 1° 02' = 9.9999 log cos 5° 50' = 9.9977. 
 
 log sin <^ can be found b}' the aid of the left-hand division 
 of the table and the formula 
 
 log sin (f> = log ^ + S. 
 
 Thus, to get log sin 0° 00'.;"), we find from the table of 
 the logarithms of numbers log 0.,') = 1.0990 ; we find from 
 the small table that for all angles between 0° 00' and 0° 51' 
 S = 6.4G37 ; . Y ^. y^ 
 
 lo£rsin0°00'.5 = j ^ '•,;'': 
 
 = 6.1627 
 
 and, in like manner, 
 
 log sin 2° 18' = log sin 138' = 
 
 2.1399 
 + 6.4636 
 
 8.6035 
 
 log sin 1°52'.6 = 8.5151 
 log sin 0°58'.8 = 8.2331. 
 
 log CSC (j> is not given in the small table ; it must there- 
 fore be obtained bv first finding log sin </> and then sub- 
 tracting the latter from 10. (See § 34.) 
 
 Examples : 
 
 log CSC 0° 15' = 2.3602 log esc 2° 18'= 1.3965. 
 
 I 
 
 I 
 
LOGARITHMS. 41 
 
 Jog tan (ft can be found by tlie aid of the second division 
 of the small table and the formula 
 
 log tan (ji = log </) + T. 
 
 .ogta„2M8' = {^^ 
 
 = 8.G039 
 logtanl°52'.G = 8.5154 
 log tan 0° 58'. <S = 8.2332. 
 log ctn <f> is not given in the small table ; it must be 
 ol)taiuod, therefore, by first finding log tan ^ and then sub- 
 tracting the latter from 10. (See § 3-i.) 
 Examples : 
 
 log ctn 0° 15'= 2.3602 log ctn 2° 08'= 1.4289. 
 
 II. To find (ji ivhen the logarithm of any one of its 
 functions is given. 
 
 Examples. Find ^ in each of the following cases : 
 
 (1) log sin (^ = 8.9824. 
 Using the formula log sin <^ = log ^ -}- S, we get 
 log ^ = log sin (/) — S. 
 
 Now, from the small table we see that when log sin <f> is 
 between 8.9845 and 8.9498, S = 6.4031, 
 , , f 8.9824 
 
 . • . log <h — J. 
 
 = 2.5193 
 
 .-. <^ = 330'.6=5°30'.6. 
 
 (2) log sin <^= 7.2982; c^ = 0° 0G'.8. 
 
 (3) logtan<^ = 7.()947 ; in this case 
 7.6947 
 
 log<^ = logtan,^-T , _ (. ^gg- 
 
 1.2310; .-. ^ = 0°17'.02. 
 
42 LOGARITHMS. 
 
 (4) logtiui</) = 8.1624; <^ = 0°49'.95. 
 
 (5) logsec</> = 0.0012; 
 
 (}> may be an}' angle between 4° 10' and 4° 20', inclusive.* 
 When logcscc/), logetn<^, and log cos ^ are given, we 
 must sul)traet them from 10 in order to get log sin ^, 
 log tan c/), and logcos</), respectively, and then from these 
 latter logarithms <^ can be found, as in the preceding- 
 examples. 
 
 (6) log CSC <^ = 2.2352; <^==0°20'. 
 
 (7) log ctnc^ =1.2250; </) = 3°24'.3. 
 
 (8) log cos </) = 9.9997; 
 
 (f> may be any angle between 1° 57' and 2° 17', inclusive.* 
 
 III. Angles between 84° and 90°. 
 
 When 4, is between 84° and 90° we must subtract it from 
 90°, and then get by the methods just explained the com- 
 plementary functions of the remainder. 
 
 Thus : log sin 8G° 18' = log cos 3° 42' = 9.9991 
 log tan 86° 18' = logctn3° 42' = 1.1893 
 log see 86° 18' = log esc 3° 42' =1.1902 
 log ctu 86° 18' = log tan 3° 42' = 8.8107 
 log CSC 86° 18' = log sec 3° 42' = 0.0009 
 log cos 86° 18' = log sin 3° 42' = 8.8098. 
 
 § 38. TJie advantage to be <jained by using the small 
 table at the top of page 10 0/ the tables for angles near 0° 
 or 90°. 
 
 AYlicn an angle is given only to the nearest minute, 
 nothing can be gained by using the small table, for in such 
 cases the logarithms can be taken directly from pages 8 
 and 9. Wlien. however, an angle which differs from 0° or 
 
 * An aiiirle near 0° cannot be found with defiuiteness from its 
 log sec, or from its log cos. 
 
LOGARITHMS. 
 
 43 
 
 90° b3^1ess than 1° is given to decimals of a minute, greater 
 accuracy can be obtained b}' using tlie small table. Let us 
 take the angle 0° 04'. 8 for example. From the small ta])le 
 log sin 0° 04'. 8 = 7.1433, and this result is accurate to the 
 fourth place of decimals ; whereas from page 8, by the 
 method of interpolation, log sin 0° 04'. 8 = 7.1449. 
 
 If we take an angle which differs from 0° or 90° b}' less 
 than 1', 0° 00'. G for example, we are compelled either to use 
 the small table or to express our angle only to the nearest 
 minute. By the small table we get log sin 0° 00'. fi = G.2419, 
 whereas, using 0° 01' for 0° 00 '.G, we get from page 8 
 log sin 0° 01' = G.4G37, thereby making an error of 0.2218. 
 Angles near 0° or 90° are of frequent occurrence, especially 
 in Astronomical work, and it is often necessaiy, even when 
 using 4-place tables, to express such angles to decimals of 
 a minute. 
 
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