When, however, the unknown appears as an index, and in any other character, for instance, as a base, or as an addend; or if two exponentials be connected by other signs than those of multiplication or division: then the solution can only be attained, in general, by means of trial and error, or some mode of approximation tantamount to it, having previously simplified the expressions by means of the logarithmic formulæ. The following examples will illustrate the processes to be employed. Ex. 1. Given 2* = 769, to find x. Here we have, as in formula above, Ex. 5. Given a* b* = k, and x: z::r: s, to find x and z. Insert the value of z from the second of these results in the first: then S {log a + log b}x = log k, and resolving, we get r * In reading expressions of this form, it will be very important to keep in mind the signification of the notation. In the present case, it signifies a raised to the power of b*; and however many, n, successive exponents there be, the valuation is supposed by raising the (n − 1)th exponent to the power denoted by the nth; then raising the (n-2) exponent to the power denoted by the last-mentioned; and so on, till we come to the base, which is raised to a power denoted by the result of all the previous involutions. In numerical solutions of such quantities, it will be convenient to calculate, in all cases, the value of z before proceeding to the final equation which gives the value of x: but in certain cases it will be essential to do so, as it may happen that c and a, being numbers less than unity, the real values of their logs. being then negative, we cannot express log 2a and log 2 without the introduction of the imaginary symbol. The solution is, however, generally expressible log2 c-log2 a log b symbolically by x= Here, taking the log of the first equation, and reducing the second, we get, after substitution, 2r2 x2 log m + s2 x2 log n = 2r2 log a, or x2 = 2r2 log a 2r2 log m+s2 log n and z2 = = x2 Ex. 8. Given m n = a, and x: z::rs, to find ≈ and z. ᏚᏆ Here we have a2 log m + z2 log n = log a, and z = r Ex. 9. Find the value of x in the equation * = 100. As an initial experiment, take x, = 3, and x = 4. Then 3327, and 41 = 256, one of which being considerably too little, and the other considerably too great, we may take for æ, or x, a number midway between 3 and 4, with a prospect of a near approximation; and from the result, judge of the other value whether greater or less than 35. Also taking logs. we have a log. x = = log. 100 2. First let 1 = 3.5; then, ones. Hence, pp. 203, 204, x = 3·59727 nearly; the extent of the approximation, however, being less clear in equations of this class, than in purely algebraical It has been often the case, that approximations have been trusted to too far; as in the example above given, for instance, though a rather favourable one for rapidity of approximation. By forming the value of the expression for x = 3′5973, we find it too great by '0000149, and for a 35972, too little by *0000841. From these values we may obtain a further correction by a repetition of the use of the formula. = This method may, for practical purposes, be a little improved. For since x+ = a, we have, as before, x log x = log a. Put now log x = y, and log a =b: then we have xy = b; and taking the logs we get log + log y log b, or since log x = y, this is y + log y log b. This may be solved for y by Trial and Error, and the value thus found will be log x, and a becomes known. The following example is given in illustration. = Ex. 10. Given * 123456789 to find the value of x. log b log 123456789 = 8.0915148b; and y + log y = Here a log x = xy = 9, we have +log 1 = 1, too great by 0919702. 9+ log 9 ='8542425, too little by 0537873. Hence by the rule we have y = 93 nearly; and we may proceed to a second approximation. 8984829, too little by 0095469. 9131279, too great by 0050981. Hence again by the rule we have y = ·93652 nearly; and we may again proceed to approximate still more closely. The next step gives y = 93651503; and from this we have x = 8.640026, which is true in the last figure. This method of solution becomes inapplicable when the equation is a* = a' and a not less than unity; but it is easy to transform it so as to find the reciprocal of x, and thence x itself. Put <==: then we get (+) = 1, and thence a = z. Take the logs put x= 2 α ting log zu, and we get z log a = log zu; and taking the logs again, we find log z + log2 a = log u, or log u — u = u log2 a; an equation of the same nature as that of the last example, and soluble by a similar process *. EXERCISES IN EXPONENTIAL EQUATIONS. 2x 1. Find the value of x in the equation (~7) = 24. 6. Ans. ≈ = 2·01374. 2. Find y from 23 = 10, and æ from (23)* . 9* = 4°%• and x = 8. Find x from x = 5. 9. Solve the equation ** = 2000. 10. Solve the equation * = 123456789. 11. Given = 10", Ans. 2 129372. Ans. 4.827822. Ans. 8-640027. 13 5", to find x and y. 12. Given = 1000; and 2-2 ̄* Ans. z= = 2.384917 in the former; and z is imaginary in the latter. ⚫5, to find z in both cases. SIMPLE INTEREST. THE interest of any sum for any time being proportional to that sum and the time, the interest of 17. for 1 year, being multiplied by the principal and time, will give the interest for that time at the specified rate per cent. * An elegant method of solving such equations may also be seen in Mr. Charles Bonnycastle's Appendix to his Father's Algebra, published in 1823. A very elegant, though practically more laborious method of solution, may also be seen in foreign elementary works on algebra, by means of continued fractions. For the sake of expressing this algebraically, put p for the principal lent at interest, t the time of its continuance, r the rate of interest, or periodical interest upon 17., and a the amount of the principal and interest at the end of the given time, and lastly, i the interest itself. Then, obviously, we have the fundamental theorems, prt = i, and p + prt = p(1 + rt) = a. From these equations we may find any one of the values in terms of the other three and taking all together, we have Ex. Find in what time any principal will double itself at any given rate of simple interest. Here by equation (1) 2p = a=p+prt, or p = prt, and t = or, by equation (4), t = 2p-P_ = as before. COMPOUND INTEREST. Irr be the interest of 11. for a given period, the amount at the end of that period will be 1 + r; and this put out for an equal period at the same rate, will amount at the end of this period to (1 + r) + (1 + r)r, or (1 + r)2; and this again put out for a third period equal to each of the former, becomes (1+r)2 + (1 + r)2r, or (1 + r)3; and so on, till after t periods, the amount is (1 + r)'. As this is the amount of 17. for t terms, the amount of p£ will be p times as much, viz. p(1 + r)'; since each of the p pounds produces the same final amount. Hence, generally, adopting the notations of simple interest so far as they are common to both, and putting R = 1 + r, The equations (2, 3, 4) being obtained from (1) by mere common processes. It is usual to call R the ratio, meaning the ratio of increase at compound interest. Example. Suppose it be required to find in how many years any principal sum will double itself, at any proposed rate of compound interest. In this case the 4th theorem must be employed, making a = 2p ; and then it is Thus, if the rate of interest be 5 per cent. per annum; then R=1.05; and hence, t= log 2 •301030 = log 1.05 ⚫021189 14.2067 nearly; that is, any sum doubles itself in less than 14 years, at the rate of 5 per cent. per annum compound interest. The following Table will very much facilitate calculations of compound interest on any sum, for any number of years, not exceeding 38, at various rates of interest: it being the value of (1 + r)' for various values of r and t. The Amount of 17 per annum in any number of Years. Yrs. 2 perCent. 3 per Cent. 33 perCent. 4 per Cent. 4 perCent. 5 per Cent. 6 per Cent. 1 2 3 4 5 1.04500 1.02500 1.03000 1.03500 1.04000 1.07689 1.09273 1.10872 1.12486 1.14117 115763 1.19102 1.10381 1.12551 114752 1.16986 1.19252 1.21551 1.26248 113141 1.15927 1.18769 1.21665 1.24618 1.27628 1.33823 6 1.15969 1.19405 1.22926 1.26532 1.30226 1.34010 1.41852 7 1.18869 1.22987 1.27228 131593 1.36086 1.40710 1.50363 8 9 20 1.77220 1.88565 2.13293 1 97993 2.26090 2.18287 2.54035 1.21840 1.26677 1.31681 1.36857 142210 1.47746 1.59385 1.24886 1.30477 1.36290 142331 1.48610 1.55133 1.68948 10 1.28008 1.34392 141060 148024 1.55297 1.62889 1.79085 11 131209 1.38423 145997 1.53945 1.62285 1.71034 1.89830 12 134489 142576 1.51107 1.60103 1.69588 1.79586 201220 13 1.37851 1.46853 156396 1.66507 14 141297 151259 161869 1.73168 1 85194 15 144830 1.55797 1.67535 1.80094 1.93528 16 1.48451 1.60471 1.73399 1.87298 2.02237 17 152162 1.65285 1.79468 1.94790 2.11338 18 1.55966 1-70243 1.85749 2.02582 2.20848 19 159865 175351 1.92250 2.10685 2.30786 1.63862 1.80611 1.98979 2.19112 2:41171 21 1-67958 1.86029 2:05943 2.27877 2.52024 2.78596 3.39956 22 1.72157 1.91610 2 13151 2.36992 2·63365 | 2·92526 3.60354 23 1.76461 1.97359 2.20611 2.46472 2.75217 3.07152 3.81975 24 1-80873 2.03279 2.28333 2-56330 2.87601 3 22510 4 04893 25 185394 2.09378 2.36324 2.66584 3.00543 26 190029 2.15659 2.44596 2.77247 314068 27 1.94780 2.22129 2.53157 2.88337 3.28201 3.73346 482235 1:99650 2.28793 2.62017 2.99870 3:42970 3:92013 5:11169 29 2.04641 2.35657 2.71188 3.11865 3.58404 4:11614 541839 30 2:09757 2.42726 2.80679 3.24340 3 74532 4.32194 5.74349 31 2.15001 2.50008 2.90503 3:37313 3.91386 4.53804 6.08810 32 2-20376 2.57508 3.00671 3.50806 408998 4.76494 6.45339 33 2.25885 2.65234 3.11194 3.64838 4.27403 500319 6.84059 28 34 2-31532 2.73191 3 22086 3.79432 4:46636 35 2.37321 2.81386 3.33359 3.94609 4 66735 36 2.43254 2.89828 3.45027 4.10393 4.87738 37 2.49335 2.98523 3.57103 4.26809 509686 38 2-55568 3.07478 3 69601 4:43881 5·32622 3.38635 4.29187 For example, let it be required to find, to how much 5237. will amount in 15 years, at the rate of 5 per cent. per annum compound interest. In the table, on the line 15, and in the column 5 per cent. is the amount of 17, viz. 2·0789; and this multiplied by the principal 523, gives the amount 1087.2647, or 10877 5s 34d, and therefore the interest 5647 5s 3ld. Note 1. When the rate of interest is to be determined to any other time than a year; as, suppose to a year, or of a year; the rules are still the same: but then t will express that time, and R must be taken the amount for that time also. Note 2. When the compound interest, or amount, of any sum, is required for the parts of a year; it may be determined in the following manner: 1st. For any time which is some aliquot part of a year. Find the amount of |