logarithm is reduced to 0.0392346; consequently, if we represent the number corresponding to this logarithm by N, the required number will be expressed by 10000 10 x 10 × N 10000 1000 x N. = Because 0.03923460 = 0.09034105='N, we get, from the exponential theorem, N=1 +0.09034105 + 0.00408075 + 0.00012289 +0.00000277 + 0.00000005 + etc. = 1.09454751. Since 10000 1000 56234.1266 nearly, we get 61550.92 for the sought number, correct to two decimals. 0.243. 2. To find x and y from 7 = 342 and 23o Taking the logarithms of the members of the equations, we get the more simple forms al7=1342 and y123 = 10.243. 2.5340261 Hence, from the tables of logarithms, we get x = 0.8450980 = 2.9984996 and y = 0.4511868. 1.3856063 1.3617278 0.6143937 = 1.3617278 3. Let P represent any principal at compound interest for t years, a the amount of a unit of P in one year; then, if A is the amount of P, it is proposed to find the equation which connects a, P, A, and t. Because the amounts of different principals in the same time are clearly as the principals, it is manifest that we shall have 1 a: a ; a2 = the amount of a unit of P in two years, and in the same way we have 1: a :: a2: a3 = the amount of a unit of P in three years, and so on; consequently, a' = the amount of a unit of P in t years. We also have the proportion 1: a' :: P: A, which gives A the equivalent equation a2 = which, by taking the logaP' rithms, gives the required equation tlalA-IP; consequently, if any three of the four quantities a, P, A, and t are known, this equation will enable us to find the fourth. = Thus, if a $1.06, P = $235.62, and A = $1925.32, the equation gives 1.06 1925.32 1235.62, or t = - 71925.32 1235.62 71.06 0.9122907 = 0.0253059 4. Given, ; consequently, from the tables, we have t 36.0505 years. 32, to find x. By taking the logarithms of the members of the equation, we get xlx = 132, which clearly gives a a very little greater than 3.08. Putting a (a + y) = al(a 3.08+ y = a + y, we have xlx = (a + y)l + y) + yl(a+y)= 132 = b; and since l(a + y) + etc., we get (by omitting = la + ml (1 + 2) = = la + my m + la a = = Since 132 = 1.5051500, ala 3.0813.08 1.5047361, b = and m (the modulus) = 0.4342944, we get y = 0.000448; consequently, we shall have x = a + y = ; = 3.080448 very nearly; and it is manifest, if we put 3.080448 for a, and repeat the process, and so on, that we shall get x to any required degree of exactness. By taking the logarithms of the members (since is the exponent of x) of the equation, we have ale 117, and taking the logarithms (of the members of this) we have alx +l(lx) = l(117) or xlx = l(l17) — l(lx); and taking the logarithms (of the members of this) we finally get lx + l(x) l[l(l17) — l(lx)]. = Since it is easy to perceive that a 2 very nearly, we shall put x = 2 + y = a+y; consequently, we shall have lx=l(a + y) = la + + etc., and l(lx) = l(la + Hence, lx + (x) = l[l(117) — Ux)] is easily reduced to etc.], or (by putting (717) — l(la) = b, and rejecting y3, y3, m m2 0.2171472, 1+ + = 1 + 1.4426947 + 1.0246976 = la 3.4673923, and m(1 that y= 0.0067233 0.75293453 bla m m2 1 + + = 0.75293453, it is clear la bla = 0.00893 nearly, and of course x = 2 + y = 2.00893 nearly, which (on trial) will be found to satisfy the proposed equation to a great degree of exactness. 6. Supposing equal principals P and P to be put at interest at the same time, the one at 6 per cent. compound interest, and the other at 7 per cent. simple interest; then it is required to find the time, t, when the amounts will be equal. Since P(1.06) is the amount at compound interest, and P+0.07Pt that at simple interest, we get (1.06)' = 1 + 0.07t or tl(1.06) = (1 + 0.07). If a = a near value of t, we have (by putting t= a + y) al(1.06) + yl(1.06) = (1 + 0.07a + 0.07y); and by putting 1+0.07a=b, and retaining only the simple power of y, we 0.07my. have al(1.06) + yl(1.06) = lb + b Hence, we easily get y [11.06) — 0.07m] = (1.06)); = After a few trials, we find that it will answer our purpose to put a 6.12, which gives b=1.4284, c = 0.0038129, b (1.06) = = - 0.0000223. Hence, we easily get y=0.00585 very nearly, and, thence a+ y = 6.12 0.005856.11415 years, very nearly. 7. Find a from the equation "100100, or from = (10)0 (" Since 0.02 equals the common logarithm of x, if we divide this by the modulus 0.43429448, we get 0.04605170 for the hyperbolic logarithm of x; consequently, from the exponential theorem, we get = 1.0471285, which is correct to seven decimal places. 8. To find the 53d root of 5843 1.1686 x 5000. Since 3 and 0.6989700043 are the common logarithms of 1000 and 5, we shall have 3.6989700043 = 0.0697918868 for 53 the common logarithm of (5000). Dividing this logarithm by 0.4342944819, we get 0.1607017581 for the corresponding hyperbolic logarithm; consequently, from the exponential theorem, we have (5000) = 1.174334677. If we calculate the hyperbolic logarithm of 1.1686, we may in like manner find the value of (1.1686); but it will be more simple to use the binomial theorem, which gives (1.1686)51.002944063. Hence, we shall have (5843)35 1.174334677 × 1.002944063 = 1.177791992, which is correct to eight decimal places. = Remark. We have given this and the preceding example for the purpose of showing the use of the exponential and binomial theorems in the extractions of the higher roots of numbers. SECTION XXI. RESOLUTION OF WHOLE NUMBERS INTO THEIR FACTORS. (1.) IF N stands for any proposed integer, then if N is even, it will be exactly divisible by 2; consequently, if 2m. is the greatest integral power of 2 which exactly divides N, N it is clear that must be an odd integer. Hence, the reso2m lution of any proposed integer into factors may be considered as being reduced to the resolution of an odd integer into factors. (2.) If N stands for an odd integer, and we divide it by 4, the remainder must clearly be either 1 or 3, since 1 and 3 are the only odd integers which are less than 4. Consequently, if n stands for the quotient of the division of N by 4, it follows from the nature of division that N will equal 4n+ 1 or 4n+ 3; and N is said to be of the form 4n + 1 or 4n + 3. (3.) If we multiply two integers of the forms 4n + 1, 4n' + 1, or 4n + 3 and 4n' + 3, by each other, the product will be of the form 4n" + 1. For (4n+1). (4n' + 1) 1). (4n' + 1) = 16nn' + 4(n + n) + 1, and (4n + 3). (±n' + 3) = 16nn' + 12(n + n') + 9, products, which are manifestly of the form 4n" + 1. (4.) The product of an integer of the form 4n + 1, and of an integer of the form 4n' + 3, will be an integer of the form 4n" +3. For (4n + 1). (4n' + 3) = 16nn' + 4(3n + n') + 3 is evidently of the form 4n" + 3. (5.) If the odd integer N is of the form 4n+ 1, and has factors, then n must be the product of two factors which are either both of the form 4n' + 1, 4n" + 1, or of the form 4n' + 3 and 4n" +3. |