Therefore x = 7·124 nearly the root required. Note 5. The same rule also among other more difficult forms of equation, 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 exponental 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 = 35, and then = 3.6, and working with the logarithms, the operation will be as follows: First supp. x = 3·5. Log. of 3.5 = 0·544068 then 35 x log. 3 5 1 904238 the true number 2.000000 error, too little, —095762 ⚫002689 Second Supp. x = 3.6. Log. of 3.6 = 0.556303 then 3'6 × log 36 = 2·002689 the true number 2·000000 error, too great + 002689 098451 sum of the errors. Then, As 0984511 :: 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 359728, 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 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, -0.0000146 error, too little, -0.0000047 -0.0000047 As 0000099 : 00001 :: 0000047: 0000 474747* the cor. 3-59728000000 added to gives nearly the value of x = 3.59728474747 Ex. 5. To find the value of x in the equation x3 + 10x2 50 = 260. Ans. 4.1179857. Ex. 6. To find the value of x in the equation x3 —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 3.59:28, 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 4241 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 : 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 mited to 3 59728 gives us the true value of x = 3*59728502354. Ex. 7. To find 'the value of x in the equation2 + 2x2 -23 x = 70. Ans. 5 13457. Ex. 8. To find the value of x in the equation ≈3 +54x350. Ex. 9. To find the value of x in the 17 x 2 Ans. 14.95407. equation x• — 3x2 Ans x 10.2609. equation 2x* — 16x Ans. x 1.284724. equation 3 + 2x♣ Ans. x 8.414455. Ex 12. To find the value of x in the equation r 123456789. I = Ans. x 8 6400268. 3x 11, to find x. Ex. 13. Given 2x-7x3 + 11x3 (3x2 — 2 √ x + 1)} − (x3 — 4x √ x + 3 √x)# = 56. Ans. 18-360877. To resolve Cubic Equations by Carden'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 Carden'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 + px2 + qx r, freed from the co-efficient of its first term, it will then be necessary to take away the 2d term p2; 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 the unknown letter in the original equation x3 + px2 + qx = r, and there will result this reduced equation z3 + az b, of the form proper for applying the following, or Carden's rule. Or. take ja, and d 6, by which the reduced equation takes this form z3 + 3c z = 2d. = Then Then substitute the values of c and d in this az = form z=d+ √ ¿d2 + c3) + 3d −√ (d2 + c3), and the value of the root z, of the reduced equation z3 + , which = b, will be obrained. Lastly, take x = z will give the value of x, the required root of the original equation x3 + px2 + qx =r, first proposed. 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 First, to take away the 2d term, its co-efficient being its 3d part is 2; put therefore x = z + 2; then x3 = z3 + 6z2 + 12z +8 24 622 24Z Here then a = 2, b -- 4, c= ૐ, d - 2. theref. the sum z3 = 8. 6, Theref. 3+ (d2 + c3) = 32 +√(4−387)=32+ √2+= 3/2 + √3 = 157735 and d-(d2+c3)=32—√ (4 — 24 ) = $2. 32 then the sum of these two is the value of z = Hence x=x+2= 4, one root of x in the eq, x3 10x = 8. To find the two other roots, perform the division, &c. as in p. 261, thus: -4)x3 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 the rate of interest of 1 pound per annum, f any principal sum lent," = t = the time it is lent for, and a = the amount or sum of principal and interest; then is prt the interest of the sum 2, for the time t, and conseq. +prt or x (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: 1st, a = p + prt, the amount, pr For Example. Let it be required to find, in what time. any principal sum will double itself, at any rate of simple in terest. In this case, we must use the first theorem, a = = p + prt, in which the amount a must be made 2, or double the principal, that iɛ, p + firt 2p, or prt = p, or rɛ. = 1; and 1 hence, Here, |