Taking the logarithms of both sides of the equation, with respect f(x)= 2x. a4x-3 — 2x. ax-2 4. a3x-1+4 = 0. The left member of the equation may be factored thus: and, therefore, by logarithmic operation, in any system, 13. Having given 10log 2 =.3010300 and 10log 3.4771213, find 10log, 10log 60, 10log 4500, and 1olog /25. 14. Given 10log 3.4771213 and 10log 5 = .6989700, find 10log 3.75, 10log 1.28 and 10log (38 × 58/27). 15. Given 10log 5.6989700 and 10log 6.7781513, find 10log 324, 10log 1.458, and 10log.00432. = 16. Given 10log 5 .6989700 and 10log 7 = 10log 1.25, 10log 1.28, and 10log (23 × 74/55). .8450980, find 17. Given 10log 12 = 1.0791813 and 10log 18 = 10log 8 and 10log 9. 1.2552726, find 18. Given 10log 24 = 1.3802113 and 10log 36 = 1.5563026, find 10log 72 and 10log 432. 19. If a, b, c be the bases of any three systems of logarithms, prove that blog x = clog α. alog x. 20. Prove that, in a series of logarithmic systems, whose bases are respectively a, b, c, etc., 22. Given log 5 = 1.60944 and log 3 = 1.09861, find 5log 27, 3log 25, and log 15. 23. If 10log 30.6614: 1.48659 and 10log 3.1299.49553, what is the relation between 30.6614 and 3.1299 ? 24. If 10log 3.0501.4843141, 10log 9.458055 = .9758019 and 10log 3.1009.4914878, what is the relation between 3.0501, 3.1009, and 9.458055 ? 25. Given 10log 2 = .301030, find x from the equation 2x 10. 26. Given 10log 2 = .301030 and 1olog 5 = .698970, find x from the equation 2¤-1 = 10x. 28. Solve the equation 52. 5x+1 +60, obtaining two values of x in terms of the logarithms of 2, 3, and 5. 29. Solve the equation a3-1 — α2x−1 . 30. Solve the equation a+1 + b · ax 31. Solve the equation 2. 32x 3x 32. Solve the equation 3x2-2x+2 in terms of the logarithm of 3 to base 10. = a2 + 1 = 0. c = 0. 9.6x+9=0. 10, obtaining two values of x 33. Solve the simultaneous equations 2x+2y= 10, 53y-1 = 25x+y, obtaining x and y in terms of the logarithm of 2 to base 10. CHAPTER XXX. NATURAL LOGARITHMS. 301. The limit which the ratio ('logy'-'logy)/(y'- y) approaches when y' and y both approach the common limit 1 is called the modulus of the system of logarithms whose base is b. Let y' = ry; then this ratio becomes - and when y = 1 it is ('log r)/(r-1).* Hence the modulus may be defined as the limit which ('log r)/(r − 1) approaches when r approaches 1; or, symbolically, if M denote the modulus, When the modulus is 1 the corresponding system of logarithms is called the natural or Napierian system,† and the base of such a system is called natural or Napierian base,† * This modulus is otherwise described as the rate of change of 'log r at the instant when = 1. [See Stringham's Uniplanar Algebra, Art. 23.] †The logarithms of the natural system must not be confounded with the numbers of Napier's original tables, which were not calculated with respect to a base, and are therefore different from the so-called Napierian logarithms of modern tables. and is denoted by the letter e. We therefore have, by definition, 302. From this definition we are able to derive a formula by means of which an approximate value of the number e may be calculated, and this being done, the modulus corresponding to any base b will be given by the formula that is, since limit (log r)/(r−1)=1, M='loge=1/log b. In place of log we shall henceforth use the shorter symbol ln, made up of the initial letters of logarithm and of natural or Napierian. 303. We first derive a formula which determines e as a numerical limit. For this purpose let lnru, that is, r=e", and let u/(e" — 1)=h, or e"h=h+u. * This fraction is positive whether 1 be greater or less than 1, for the logarithm of any number less than 1 is negative [Art. 298]. Hencer may approach 1 through values either a little smaller or a little greater than 1. Then u=0 when r= 1, and h = 1 when u = 0, and we We have here supposed u to approach zero from the positive side, but the supposition was not a necessary If we assign to u a negative value, say -v, the formula for the value of e becomes one. 304. Now it may be shown that in whatever way u is made to approach zero, whether positively or negatively, the limit of (1 + u)/" is a finite number, and that when u passes continuously towards zero, (1+u)/" diminishes, and (1) increases, continuously towards a common limit; and that e is therefore a definite number. For a formal proof of these statements, which is not properly within the scope of this treatise, we refer the student to Chrystal's Algebra, Vol. II., Chapter XXV., § 13, and only present here a few numerical verifications. By ordinary arithmetical computation we easily find: (1 + 1)2 = 2.25. (1+1)=2.370. 2 (11)2 = 4. −3 (11) 3.5. = = (1-4) 3.1604903... (1 − 1)−5 (1)-10 = 3.0517578125. |