EXAM. 3. Find the cube root of 8+45 or 4√5+8. B=8 Here A=4/5 · ‚ƈ — B2 = 80 −6416, which is not a cube number, and the least number which multiplied into it will produce a cube number is 4,(a) ..C=4, and (A-B') C=16×4=64; hence 64, and «=4. Now √(A+B2+2AB)C=✔✅ (80+64 +64 / 5)4=10+ƒ, * (4+B–24B)C=V(80+64–64 ..t=10+2=12, 5)4=2-f (a) In finding the least number by which a given number (a) must be multiplied so as to give a product which shall be a complete nth power, it may be observed, that if a be a prime number, it must always be multiplied by a"; thus, there is no other number by which 3 can be multiplied to make it a cube number, but 32 or 9, which gives the product 27; nor is there any other number by which 5 can be multiplied to make it a biquadrate number, but 53 or 125, which gives the product 625. But if the given number is resolvable into factors, one or more of which are square, cute, &c. numbers, then a less number than a will answer the purpose. Thus 12=3 × 43 × 2o; and if 3 × 2a be multiplied by 3a × 2, it gives 33 × 23, which is the cube of 3 × 2; i. e. if 12 be multiplied by 18 it gives 216 the cube of 6. Or in general, if the given number (a) be resolvable into factors Bry" &c. then if this number be multiplied u, ß, y, &c. such that a by a "-"-" &c. it gives a" 6" 7" &c. which is the nth power of ußy &c. Thus 360=8× 9 × 5=23 × 32 × 5; here m=3, p=2, q=1; and if it be required to find a multiplier which should make it a biquadrate number, then "=4, .'.n−m=1, n−p=2, n−q=3; hence the multiplier is 2 X 32× 532250, and we have 360 × 2250=810000, which is the fourth power of 2 × 3 × 5 or 30. If one or more of the indices m, p, q, &c. be greater than n, then, in finding the multiplier, such multiple of n must be taken as to make the indices of all the factors in the multiplier positive; thus if m be greater than n but less than 2n, then the multiplier to be taken is «2n-m ßr-p 2”-9, which gives for the product of it and «" 6" 7" the quantity a 6" 7", which is the nth power of aß y. XLVI. On the Comparison and Reversion of Series. Our observations upon the "comparison and reversion of series," will be confined to such as are of the form ax + bx2+cx3 +dx+ex+&c. referring the Reader to the 10th Chap. of the 2d Part of MACLAURIN's Algebra, and to other authors who have written upon the more advanced parts of the science, for the further illustration of the subject. 165. Previous to the reversion of this series, it will be necessary to shew the manner in which it may be raised to any power (n). This is done by separating the first term from the rest, and then applying the binomial theorem to the involution of the series so transformed; thus (ax+bx2+cx3+dx2+&c.)"=ax+(bx2+cx3+dx*+ &c.) n(n =a”x”+na”—'x”"-1 (bx2+cx3+dx*+&c.) + 2 (bx2+cx3+&c.)'+&c. n(n-1) =a”x”+na”—1x2¬' (bx2+cx3+dx1+&c.) + 2 (b2x*+2bcx3 + &c.) +&c. n We have here found only the four first terms of the involved series, but they are sufficient for our present purpose. 166. Let ax + bx2+cx3+dx1+&c. and ax+6x2+yx3+ dx+&c. be two series of this kind, in which a, b, c, d, &c. a, ß, y, d, &c. are invariable coefficients; if these two series be equal to each other, then will a=a; b=ß; c=v; d=d; &c, For, let the terms of each series be divided by x, then a+bx+ cx2+dx2+&c=a+ßx+yx2+dx+&c.; and since x may vary through all degrees of magnitude whilst the coefficients remain constant, let x=0, then a=a; but a and a are invariable quantities; if therefore they are once equal, they will always be equal. Again, since a=a, we have bx + cx2+dx3 +&c.=6x+yx2+dx3+ &c.; divide each side of the equation by x, then b+c+dx2+&c.=ß+yx+dx2+&c.; let x=0, then b=6; and so we might proceed till all the terms of the series were exhausted; we therefore infer, that when these two series are equal to each other, a=a; b=8; c=y; d=d; &c. &c. 167. By transposing the terms of the equation ax + bx2 + cx3 +dx*+ &c.= ax + b x2 + y x2 + dx+&c. we have (a-a) x + (b − ß) x2 + (c − y) x3 + (d−d) x1 + &c. = 0. Now whatever is true in the original equation, must also be true in the transposed equation; but it has already been proved, with respect to the former equation, that a=a; b=ß; c=y; d=d; &c.; hence a—a=0; b—ß=0; c—y=0; d-d=0; &c.; from which it follows that in an equation of the form (a−a)x+(b−ß)x2 + (c−y) x3 + (d−d)x1+&c. =0, whose coefficients consist of positive and negative quantities, these coefficients must all become equal to o at the same time. 168. Suppose now that x=ay+by+ cy3+ dy'+ &c. and that it be required to find the value of y in terms of x. Transpose x to the other side of the equation, then ay+by+ cy3+dy+&c.-x=0. Assume y=ax+ß x2+yx2+dx*+ &c.; and finding the value of the successive powers of y, by Art. 165, we have ad+2bay+b3*+3 ca2ß+da*=0, or d=—2 bay-b3-3 ca2ß-da a -5b3+5abc-a'd a7 Substitute these values for a, ß, %, d, &c. then y: C b x3 (2 b3 — a c) x3 _ (5 b3 — 5 a b c + a3d) x* + as a7 and if a=1, or x=y+by2+ cy3+dy*+ &c. then + &c.; y=x− b x2 + (2 b3 — c) x 3 — ( 5 b3 — 5 b c +d)x2+&c. (A). 179. In the following chapter it will be shewn, that if x be the logarithm of the number 1+n, x=n—}n2 + ƒn3 —‡n1 + &c.; suppose therefore it was required to find the number in terms of the logarithm, i. e. to find n in terms of x, then comparing the equation x=n—\n' + }n3 — \n' + &c. with the equation. x=y+by+cy3+dy'+&c. and substituting n for y in the equation (A), we should have n=x-bx2+(2b3 — c) x3 — (5 b3—5bc+d)x*+&c. where b=—', c=1, d= 1, − 1, &c. ; = 1 200 CHAP. X. ON LOGARITHMS, AND SUBJECTS CONNECTED WITH THEM. XLVII. Definition and Properties of Logarithms. 170. In the two following series of quantities, a, a*, a*", a*'"', &c. (A); x, x', x', x", &c. (B); where a is some given number greater or less than unity, and x, x', x", x"", &c. any variable quantities whatever, the several terms of the series (B) are called the logarithms of the several terms corresponding to them in the series (4). Thus if a=y, a"=y', a""=y", &c. then x=log. y; x'=log. y'; x"=log. y"; &c. 171. In adapting the series (A) to the numbers 1, 2, 3, 4, 5, 6, &c. the given number a must be greater than unity, the first index x must be equal to 0, and the several indices x', x", "", &c. must keep continually increasing. For in this case, since (by Art. 68.) a°=1, this series will increase from 1 to infinity; and by properly adjusting the values of x, x", x", &c. it is evident that the several quantities a*, a", a", &c. may be made to coincide with the numbers 2, 3, 4, 5, 6, &c. For instance, let a= 10; then (since 10°1 and 10'-=10), the indices of 10 which would give 10, 10*", 10", &c. equal to the numbers 2, 3, 4, 5, &c. must be fractions between 0 and 1. = Take |