Therefore x = 7.124 nearly the root required. Note 5. The same rule also, among other more difficult forms of equations, succeeds very well in what are called exponential ones, or those which have an unknown quantity in the exponent of the power; as in the following example: Ex. 4. To find the value of x in the exponential equation x X = 100. For more easily resolving such kinds of equations, it is convenient to take the logarithms of them, and then compute the terms by means of a table of logarithms. Thus, the logarithms of the two sides of the present equation are xx log. of x 2 the log. of 100. Then, by a few trials, it is soon perceived that the value of x is somewhere between the two numbers 3 and 4, and indeed nearly in the middle between them, but rather nearer the latter than the former. Taking therefore first x = 3.5, and then 3.6, and working with the logarithms, the operation will be as follows: First Supp. x = 35. Log. of 3.50*544068 then 3.5 X log. 3:51.904238 the true number 2·000000 error, too little, -.095762 = Second Supp. x=3.6. Log. of 3.6 0.556303 then 3.6 × log. 3·6 = 2·002689 the true number 2·000000 error, too great, +002689 098451 sum of the errors. Then, As 098451: 1 :: 002689 : 0.00273 the correction On trial, this is found to be a very small matter too little. Take therefore again, x = 3.59727, and next = 3.59728, and repeat the operation as follows: First, Supp. x = 3.59727. Log. of 3.59727 is 0.555973 3.59727 X log. of 3.59727 1.9999854 the true number 2.0000000 error, too little, -0.0000146 -0.0000047 Second, Supp. x = 3.59728. Log. of 3.59728 is 0.555974 3.59728 x log. of 3.59728 = 1 9999953 the true number 2·0000000 error, too little, -0.0000047 0.0000099 diff. of the errors. Then,. As 0000099: 00001 :: 0000047: 00000474747* the cor. gives nearly the value of x = 3.59728474747 Ex. 5. To find the value of x in the equation x3 + 10x2 + 52 = 260. Ans. x 4.1179857. Ex. 6. To find the value of x in the equation 3 —2x = 50. Ans. 3.8648854. *The Author has here followed the general rule in finding as many additional figures as were known before: viz. The 6 figures 3-59728, but as the logarithms here used are to 6 places only, we cannot depend on more than 6 figures in the answer; we have no reason, therefore, to suppose any of the figures in 00000474747, to be correct. The log. of 3.59728 to 15 places is 0-55597 42431 34677 the log of 3.59727 to 15 places is 0.55597 30358 47267 which logarithms multiplied by their respective numbers give the following products: 1.99999 50253 435122 both true to the last figure. Therefore, the errors are 49746 56488 148773 37702 99026 81214. Now since only 6 additional figures are to be obtained, we may omit the last three figures in these errors; and state thus: as diff. of errors 9902681: diff. of sup 1: error 4974656: the correction 502354, which united to 3.59728 gives us the true value of x = 3.59728502354. Ex. 7. To find the value of x in the equation x2 + 2x2 -23x : 70. Ans. x = 5.13457. Ex. 8. To find the value of x in the equation x3 — 17x2 +54x= 350. Ans. x 14.95407. equation x 3.x 2 Ans. x = 10.2609. equation 2x4 — 16x3 Ans. x 1.284724. equation x5 + 2x4 Ans. x 8.414455. Ex. 9. To find the value of x in the -75x = 10000. Ex. 10. To find the value of x in the +40x2 - 30x = - 1. Ex. 11. To find the value of x in the + 3x3 + 4x2 + 5x = 54321. Ex. 12. To find the value of x in the equation x = Ans. x 123456789. 8.6400268. Ex. 13. Given 2x4 -7x8 + 11x2 3.r 11, to find x. Ex. 14. To find the value of x in the equation 5 (3x2 — 2 √ x + 1)3 — (x2 — 4x √√ x + 3 √ x)3 = 56. Ans. x === 18.360877. To resolve Cubic Equations by Cardan's Rule. THOUGH the foregoing general method, by the application of Double Position, be the readiest way, in real practice, of finding the roots in numbers of cubic equations, as well as of all the higher equations universally, we may here add the particular method commonly called Cardan's Rule, for resolving cubic equations, in case any person should choose occasionally to employ that method. The form that a cubic equation must necessarily have, to be resolved by this rule, is this, viz. z3 + az = b, that is, wanting the second term, or the term of the 2d power z2, Therefore, after any cubic equation has been reduced down to its final usual form, x3 + x2 + qx = r, freed from the co-efficient of its first term, it will then be necessary to take away the 2d term fix; which is to be done in this manner: Take, or of the co-efficient of the second term, and annex it, with the contrary sign, to another unknown letter z, thus z-; then substitute this for x, the unknown letter in the original equation x3 + phx2 + qx = r, and there will result this reduced equation z3 + az = b, of the form proper for applying the following, or Cardan's rule. Or take Ja, and db, by which the reduced equation takes this form, 23 + 3c z = 2. €= Then Then substitute the values of c and d in this form z=d+v (d2 + c3) + ŷd — √ (d2 + c3), C or z = 3d+✔ (d2 + c3) - d+ √(d2 + c3), 2 3 and the value of the root z, of the reduced equation z3+ az = b, will be obtained. Lastly, take x = x , which will give the value of x, the required root of the original equation x3+x2 + qx = r, first proposed. +qx One root of this equation being thus obtained, then depressing the original equation one degree lower, after the manner described p. 260 and 261, the other two roots of that equation will be obtained by means of the resulting quadratic equation. Note. When the co-efficient a, or c, is negative, and c3 is greater than d2, this is called the irreducible case, because then the solution cannot be generally obtained by this rule. Ex. To find the roots of the equation x3- 6x2 + 10x = 8. First, to take away the 2d term, its co-efficient being its 3d part is 2; put therefore x z +2; then x3z3+6z3 + 12z+8 --6, 6x2 = +10x = theref. the sum z3 Theref. 3d+v (d2 + c3) = 32 +√(427) = √2 + 10 √3 = 1.57735 then the sum of these two is the value of z = 2. + 24, one root of x in the Hence x 10x = 8. To find the two other roots, perform the division, &c, as in p. 261, thus: Hence x2-2x =— - 2, or x2 - 2x+1=— 1, and x=±√−1; x = 1 + √ 1 or = 1−√1, the two other sought. Ex. 2. To find the roots of x3 9x2+28x = 30. = Ans. x = 3, or = 3+1, or 3 — √ — I. Ex. 3. To find the roots of x3- 7x2 + 14x = 20. Ans. x = 5, or =1+ √3, or = 1 3. 1040 OF SIMPLE INTEREST. As the interest of any sum, for any time, is directly proportional to the principal sum, and to the time; therefore the interest of 1 pound, for 1 year, being multiplied by any given principal sum, and by the time of its forbearance, in years and parts, will give its interest for that time. That is, if there be put 7 = the rate of interest of 1 pound per annum, t = any principal sum lent, the time it is lent for, and a = the amount or sum of principal and interest; then is prt the interest of the sum f, for the time t, and conseq. +prt or p× (1 + rt) = a, the amount for that time. = From this expression, other theorems can easily be deduced, for finding any of the quantities above mentioned; which theorems, collected together, will be as below: Ist, a = p + prt, the amount. For Example. Let it be required to find, in what time any principal sum will double itself, at any rate of simple interest. In this case, we must use the first theorem, a = p + prt, in which the amount a must be made = 21, or double the principal, that is, p + prt = 2p, or firt =f, or rt=1; 1 and hence —, r |