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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,
log 769 2.8859263

= 9:586839.

log 2 0:3010300 Ex. 2. Find x from the equation amx bu* = c. In this mx log a + nx log b = log c, or x =

log c

m log a + n log bo
= 541, to find x.
5

109
Here x log

log 54:5 = log or x =

= 17.9177. 2

log 1.25 Ex. 4. Given a = c to find x *. Put 6* = 2: then a = c, and hence z=

log c log à

Ex. 3. Given ()

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log (logo

a

log 6

log 6

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E.C. 5. Given at b3 = k, and x : 2:31:s, to find x and z.

These give x log a + z log b = log k, and 2 = *** Insert the value of 2 from the second of these results in the first : then

r

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S

{log a +

log 0}= log k, and resolving, we get

b

r

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r log k
; and hence z =

s log k
r log a + s log b

r log a + s log b
Ex. 6. Given at • bi k, and X: 2::o : s, to find x and z.
This gives, in the same way as before,
r log k

s log k
rlog a
s log

r log a

- s log bo 22 Ex. 7. Given m = a, and a : 2::-:s, to find x2 and zo.

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* 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 x 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 ?a and log 2c without the introduction of the imaginary symbol. The solution is, however, generally expressible symbolically by x =

log2 c— loga a

log b

Here, taking the log of the first equation, and reducing the second, we get, after substitution,

2r2 log a 2ro x2 log m + sa x2 log n = 2r2 log a, or x2 =

2r2 log m + sa log n s?r?

2s- log a and 22 =

r2 2r2 log m + sa log n

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Ex. 8. Given m® = a, and x : -::-;s, to find x and z.

Here we have xo log m + za log n = log a, and 2 =
By substitution r2 x? log m + so xa log n = yo log a, or

pa log a
x = +

;

and from this again, ga log m + sa log n

sa log a

pa log m + sa log n Ex. 9. Find the value of x in the equation xt = 100.

As an initial experiment, take x, 3, and x, = 4. Then 33 = 27, and 4' = 256, one of which being considerably too little, and the other considerably too great, we may take for x, 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 3.5. Also taking logs. we have x log. x = log. 100 = 2. First let X, = 3.5; then,

Second, let x, = 3:6; then, 3.5 log 3.5 = 1.9042380

3 6 log 3.6 = 2.002689 true no. = 2 0000000

true no. = 2.000000

too little by .095762

too great by .002689

2 =

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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 ones. 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

3.5973, we find it too great by '0000149, and for x = : 3.5972, 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 20* = 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 x + 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 x becomes known. The following example is given in illustration.

Ex. 10. Given 2* = 123456789 to find the value of x.

Here x log x = xy = log 123456789 = 8:0915148 = b; and y + log y = log b = .9080298.

Taking y, = 1, we have 1 + log 1 = = 1, too great by '0919702.
Taking y, = .9, we have .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.

Taking y = .93, we have y + log y = .8984829, too little by 0095469. Taking y, = .94, we have y + log y = .9131279, too great by '0050981.

a

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.

1 This method of solution becomes inapplicable when the equation is 2nt 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.

1 ,

a' ting log z = U, and we get z log a = log z = z; and taking the logs again, we find log 2 + logo a = log u, or log u — u= u loga a; an equation of the same nature as that of the last example, and soluble by a similar process *.

1

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Put <= : then we get ( )= . and thence a = z. Take the logs put

2

x =

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EXERCISES IN EXPONENTIAL EQUATIONS.

1. Find the value of x in the equation (?)* = 26.6%. Ans, a = 2-01374.

3

Ans. y

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(423)?

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3) ,

,2 =

(63)*= 1:75,

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x

318 2. Find y from 234 = 10, and x from (23)* . 9* = 47.

1.1073093, and x = '371606. 3. Resolve the equation 23* – : 24.

Ans. x = 1.26186. 3 4. Given-= and 3* =

Ans. x = '222885, y : 297181. Y

4 5. If v: 2: 2 : 4, and 422 = 930, what are the values of u and 2?

16/log. 2

8 log. 2 Ans. O=0,2 = 0; and v=

9 \log. 3 3 log. 3 6. In

show that 9:677291. . 7. Find x and y from all*+v1*= 66++), and a, y

+9 (+49) = 6,(sy). slog. a

Slog. a Ans. x =

, 1 /

and (log.blog. as

(log. 6 8. Find x from 2* = 5.

Ans, 2-129372. 9. Solve the equation z* = 2000.

Ans. 4.827822. 10. Solve the equation 2* = 123456789.

Ans. 8.640027. 11. Given n = 10%, and x

= 56

5", to find x and y. Ans.y = 0 and x indeterminate; and x = ·00117937, y = 2.92836.

= 1000; and 2-2 •5, to find z in both cases. Ans. 2=

: 2:384917 in the former; and z is imaginary in the latter.

2

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SIMPLE INTEREST. The interest of any sum for any time being proportional to that sum and the time, the interest of 11. 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 inay 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.

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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 il., 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

(1)

P a =p + prt

(3) pt

(2) 1 + rt

t =

(4)

тр 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 =

2pP_

as before. тр

a T=

a

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a - P

3

1

=

COMPOUND INTEREST.

a

IF r be the interest of ll. 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)? ; and this again put out for a third period equal to each of the former, becomes (1 + r)2 + (1 + r)?r, or (1 + r)3 ; and so on, till after t periods, the amount is (1 + r)'. As this is the amount of il. for t terms, the amount of p£ will be р

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, a=pR

(1)
R=

(3)

P (2)

log a – log p R

t =

(4)

log 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

log
log 2p – log p

log 2
t=

a

P=

a

log p

log R

log R

log R*

Thus, if the rate of interest be 5 per cent. per annum; then R=1:05; and hence,

log 2 •301030

= 14.2067 nearly ;

log 1.05 ·021189 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 il per annum in any number of Years.

Yrs.22 perCent. 3 per Cent. 3.1 perCent. 4 per Cent. 41 perCent. 5 per cent. 6 per Cent.

1 1.02500 1.03000 1.03500 1.04000 1.04500 1.05000 1.06000 2 1.05063 1.06090 1.07123 1.08 160 1:09203 1:10250 1.12360 3 1.07689 1.092731 1:10872 1.12486 1.14117 1:15763 1.19102 4 1:10381 1.12551 1.14752 1.16986 1:19252 1.21551 1.26248 5 1:13141 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 1:31593 1.36086 | 1:40710 1:50363 8 1.21840 1.26677| 1:31681 1:36857 1:42210 1.47746 | 1:59385 9 1.24886 1:30477 1:36290 1.42331 1.48610 1.55133 1.68948 10 1.28008 1•34392 1:41060| 1:48024 1:55297 1.62889 1.79085 11 1.31209 1.38423 1.45997 1:53945 1.62285 | 1071034 1.89830 12 1:34489 1.42576| 1:51107 1.60103 1.69588 1.79586 2:01220 13 1.37851 1:46853 1.56396 1.66507 1.77220 1.88565 2:13293 14 1.41297 | 1:51259 1.61869 1.73168 185194 1.97993 2.26090 15 1:44830 1.55797 1.67535 1.80094 1.93528 2.07893 2:39656 16 1.48451 1.60471 1.73399 1.87298 2.02237 2:18287 2:54035 17 1:52162 1.65285 1.79468 1.94790 2:11338 2.29202 2:69277 18 1:55966 1.70243 1.85749 2.02582 2.20848 2.40662 2:85434 19 | 1:59865) 1.75351 1.92250 2:10685 2:30786 2:52695 3 02560 20 1:63862 1.80611 1.98979 2:19112| 2:41171 2.65330 3:20714 21 1.67958 1.86029 2:05943 2:27877 2:52024 2-78596 3:39956

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 287601 3 22510 4 (4893 25 1.85394 2:09378 2:36324 2.66584 3:00543 3:38635 4.29187 26 1.90029 2:15659 2.44596 2.77247 3:14068 3.55567 4.54938 27 1.94780 2.22129! 2:53157 2.88337 3.28201 373346 4.82235 28 1:99650 2:28793 2:62017| 2:99870 342970 3:92013 5:11169 29 2.04641 2:35657 | 2:71188 3:11865| 3.58404 4:11614 5.41839 30 2 09757 2:42726 2.80679 3.24340 3:74532 4.32194 574349 31 2:15001 | 2:50008 2.90503 3.37313 3.91386 4:53804 6:08310 32 2:20376 2:57508 3.00671 3:50806 4:08998 4•76494 6.45339 33 2-25885 2.65234 3:11194 3.64838 4.27403 5:00319 6.84059 34 2-31532 2.73191 3 22086 3079432 4.46636 | 5.25335 7.25103 35 2:37321 | 2.81386 3•33359 3.94609 4.66735 5.51602 7.68609 36 2:43254 2.89828 3:45027 4:10393 4.87738 5:79182 8.14725 37 2-49335 2.98523 3.57103 4.26809) 5:09686 6:08141 8.63609 38 255568 3:074781 3 69601 4:43881 5:32622 6.38548 9.15425

22

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For example, let it be required to find, to how much 5231. 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 il, viz. 2.0789 ; and this multiplied liy the principal 523, gives the amount 1087.2647, or 10871 58 31d, and therefore the interest 5641 58 31d.

Note 1. When the rate of interest is to be determined to any other time than a year; as, suppose to į 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 ;

be determined in the following manner : 1st. For any time which is some aliquot part of a year. Find the amount of

it may

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