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c2 + y2 = 25. The solution of this system and of many somewhat similar systems can be more easily carried out, however, by the methods which follow.
x2 + y2 (1) 1. Solve ху 18. (2) HINT. Here the second equation could be solved for x in terms of y and the solution completed by substitution. A simpler method is to combine them in such a way as to obtain numeric values for x + y y as follows: 2 xy = 36. (3) (1) +(3), 2c2 + 2 xy + y2 81. (4) From (4), 2 + y (5) (1)-(3), 32 – 2 xy + y2 = 9. (6) From (6), X – Y (7) (5) and (7) combined give the four linear systems : 2 + y = 9, (8) c. S x + y = 9, A. (12) 3. (13) D. x + y 9, (14) y = - 3. (11) - 3. (15) These four systems are equivalent to the original one. The solution of these systems is left to the student. The pairs of roots found for the four systems A, B, C, D will check in the original system.
2. x2 + y2 65, ху 3. 4 x2 + y2 104, ху 10. 4. x2 + xy + ya 19, = 6.
6. x2 + xy + y2 = 5, 20. 22 + xy + y2 = 244. Page 3
13. 24 + x2y? + y4 = 21, 14. x2 ya 22 + xy + y2 = 7. x + y = a. 15. 23 + 43 a3 – 8, x + y = a 2.
Solve by any method and pair results. If any system cannot be solved by methods previously given, solve it graphically. 1.2 + y = 4, 11. 23 - y3 = 152, 2y = 3. x2 + xy + y2 = 19. 2. x2 Xy 15, 12. h2 - hk = 160, 2 x - y = 14. hk - K2 = 96. 3. x2 + xy + y2 7, 13. Y – xy = 72, , 2x + 3y = 0. 3 x + xy = 4. x2 + y = 38, 14. $2 + t2 = 100, s2 t2 28. 5. 82 - t2 = 24, 15. x12 + x1x2 = 200, 8 + 2 t = 3. X1X2 + x22 100. 6. x2 + y2 100, 16.2 + y = 1.7, 48. ху .52. 7. 2+ + yt 5, 17. .1s +.1 t = .015, ข .2 s2 + .2 t2 = .0025. 8. u3 + 03 = 35, 18. + 2 y = .94, u +0 = 5. .2 x2 Xy = - .742. 9. ♡c - Vy = 1, 19.2 + y = 2 a, cy = a2 - 62. 10. yz + 2 = 20, 20. 22 + xy + y2 = 3 a?, yz + y = 18. 32 + xy = 2 a. Page 4
6. The sides of a triangle are 25, 33, and 52. Find the altitude on the side 33.
7. The paral ly lel sides of a trapezoid are 25 and 36 respec -11+X tively. The nonparallel sides are respectively 13 and 20. Find the altitude of the trapezoid. HINTS. x2 + y2 169, (11 + 2)2 + y2
8. The two parallel sides of a trapezoid are 30 and 51 respectively. The other two sides are 10 and 17 respectively. Find the altitude of the trapezoid and its area. 9. The sides of a trapezoid are 25, 30, 27, and 81 respectively. The sides 30 and 81 are parallel. Find the altitude of the trapezoid and its area. 10. The sides of a trapezoid are 14, 18, 40, and 48. The sides 18 and 48 are the bases. Find the altitude and the area of the trapezoid. 11. The sides of a trapezoid are 45, 53, 90, and x. The latter is perpendicular to the sides 45 and 90. Find x and the area of the trapezoid. 12. The hypotenuse of a right triangle is 125 and its area is 2574. Find the other two sides. 13. The area of a right triangle is 96. The difference between the sum of the smaller sides and the hypotenuse is 8. Find the three sides.
14. A rectangular tank is 8 feet 6 inches long and 6 feet 8 inches wide. A board 10 inches wide is laid diagonally on the floor. What DOR two equations must be solved to determine the length of the longest board that can be thus laid?
HINT. Let DR = x and DK y. The triangle DKR is similar to the triangle AKL. If the student desires A to find the length of LB the board, he must solve the system obtained graphically. This is because an attempt to use the other methods presented in this chapter leads to an equation of the fourth degree which cannot be solved by factoring or any simple method. If the student studies advanced algebra, he will learn to solve such equations readily by Horner's method.
114. Powers of binomials. The following identities are easily obtained by actual multiplication: (a + b)2 a2 + 2 ab + 62 (1) (a + b)3 a3 + 3 a2b + 3 ab2 + 63 (2) (a + b)4 = ax + 4 a3b + 6 a2b2 + 4 abs +_64 (3) (a + b)5 = 25 + 5 a4b + 10 a3b2 + 10 a2b3 + 5 ab4 + 65 (4)
If a + b is replaced by a – b, the even-numbered terms in each of the preceding expressions will then be negative and the odd-numbered terms will be positive.
115. The expansion of (a + b)^. The form of the expansion for the case where n is an integer is as follows: The first term is a" and the last is bo. The exponents of a decrease by 1 in each term after the first. The exponents of b increase by 1 in each term after the second. The product of the coefficient in any term and the exponent of a in that term, divided by the exponent of b increased by 1, gives the coefficient of the next term. The sign of each term is + if a and b are positive; the sign of each even-numbered term is if b alone is negative.
According to the above statement we have
(a + b)= = q* +1 00-16+ an-b3 +
This expresses in symbols the law known as the binomial theorem. The theorem holds for all positive values of n and also, with certain limitations, for negative values.
NOTE. The coefficients of the various terms in the binomial expansion are displayed in a most elegant form as follows:
1 11 1 2 1 1 3 3 1 1 4 6 4 1
In this arrangement each row may be derived from the one above it by observing that each number is equal to the sum of the two numbers, one to the right and the other to the left of it, in the line above. Thus 4 = 1 + 3, 6 3 + 3, etc. The next line is 1 5 10 10 5 1. The successive lines of this table give the coefficients for the expansions of (a + b)n for the various values of n. The numbers in the lowest line of the triangle are seen to be the coefficients when n = 4; the next line would give those for n 5. This array is known as Pascal's triangle and was published in 1665. It was probably known to Tartaglia nearly a hundred years before its discovery by Pascal.
Expand by the rule: c)? 5. (a + 2)6. 3. (a + 1). 6. (a + 3)5.
7. (3 + a).
Obtain the first four terms of:
14. (a - 2)18 15. (a – 5) 30.
Expand : HINTS. To avoid errors in the exponents write (a2)5 + (a ) (3 c)+ (a)(3 c)2 + (a ) (3 c) + (a)(3 c)* + (3 c)5 Then in the spaces left for them put in the coefficients of the terms according to the binomial theorem. Finally simplify each term.
Obtain in simplest form the first four terms of:
31. Write the first six terms of the expansion of (a + b)", and evaluate it for n = 1, n = 2, n = 3, n 4. How does the number of terms compare with n? What is the value of each coefficient after the (n + 1)th? Why does not the expansion extend to more than five terms when n = 4? Page 5
6. Check the result of Exercise 5 by extracting the square root by the method of $ 64. 7. Show that (65) 4.0207.
Find to three decimals by the binomial theorem: 8. (26). 10. (38) 9. (27) 11. (83)*. 12. (62) HINT. (62)} = (64 – 2). Here (64 - 2); yields more accurate results with fewer terms than (49 + 13). 13. (28) 14. (61) 15. (29) 16. (520) 17. (17) 18. Check the result of Exercise 17 by finding the value 1 of by the ordinary method of square root. Here V17 1 V17 etc. V17 17
Expand to five terms: 19. (1 + x) 20. (2 - x) 21. (3 + x)? 22. (2 + x). 24. (1 – x) 23. (8 - x) -5. 25. (1 + x) =3.
117. The factorial notation. The notation 5! or 15 signifies 1.2.3.4.5, or 120. Also 4! = 1.2.3.4 = 24. In general, n! = 1.2.3.4 ... (n.- 2)(n − 1)n. The symbol n! orn is read “ factorial n." Page 6
5. log10 100 = ? log10 1000 = ? log10 10 = ? log10 1 ? 6. 4° = ? log, 1 = ? loge 1 ? log, 1 = ? NUMBER BASE LOGARITHM NUMBER BASE LOGARITHM 7. 25 5 ? ? 10 2 8. 81 3 ? 23. ? 10 1 9. 81 9 ? ? 10 0 10. 64 2 ? 25. ? 10 3 11. 125 5 ? 26. ? 10 1 12. 216 6 ? 27. ? 10 - 2 13. 343 7 ? 28. ? 10 3 14. 27 ? 3 4 ? - 를 15. 8 ? 3 30. 8 ? 16. 128 ? 7 31. 27 ? 17. 625 ? 3 32. 5 25 ? 18. 1024 ? 5 6 216 ? 19. ? 4 4 34. 16 ? 1 ? 3 5 35. 81 ? 21. ? 5 3 36. 25 ? Read in the notation of logarithms : 37. 73 = 101.863. 44. 165 = 102.2175. 38. 50 45. 16.5 101.2175. 39. 1 10o. 46. 1.65 10.2175 40..1 10-1 47. .165 10-1+.2175 41..2 = 10-.699. 48. .0165 = 10-2+.2175. 42. .089 10-1.05. 49. 200 102.301 43. 1650 = 103.2175 50. 20 = 101.301 Page 7
10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 05690607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 35 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 30 15 1761 1790 1818 1847 1875 | 1903 1931 1959 1987 2014 28 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 26 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 25 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 24 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 22 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 20 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 19 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 18 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 18 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 17 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 16 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 16 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 15 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 15 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 14 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 14 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 13 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 13 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 13 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 11 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 11 406021 6031 6042 6053 6064 | 6075 6085 6096 6107 6117 11 41 6128 6138 6149 6160 6170 | 6180 6191 6201 6212 6222 10 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 10 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 10 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 652210 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 | 10 46 6628 6637 6646 6656 6665 | 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 9 50 6990 6998 7007 7016 7024 | 7033 7042 7050 7059 7067 9 51 7076 7084 7093 7101 7110 | 7118 7126 7135 7143 7152 8 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 | 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 | 7364 7372 7380 7388 7396
55 | 7404 7412 7419 7427 7435 | 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 | 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 | 7745 7752 7760 7767 7774 60 | 7782 7789 7796 7803 7810 | 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 1 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 | 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 | 8129 8136 8142 8149 8156 | 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 840 8407 8414 8420 8426 8432 8439 8445 70 | 8451 8457 8463 8470 8476 | 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 | 8663 8669 8675 8681 8686 74 | 8692 8698 8704 8710 8716 | 8722 8727 8733 8739 8745 75 | 8751 8756 8762 8768 8774 | 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 | 8837 8842 8848 8854 8859 77 8865 887 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 | 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 | 9031 9036 9042 9047 9053 | 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 892 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 91919196 920 9206 9212 9217 9222 9227 9232 9238 84 | 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 | 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 | 9518 9523 9528 9533 9538 90 | 9542 9547 9552 9557 9562 | 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 | 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 | 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 | 9708 9713 9717 9722 9727 94 | 9731 9736 9741 9745 9750 | 9754 9759 9763 9768 9773 95 | 9777 9782 9786 9791 9795 | 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 | 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 | 9978 9983 9987 999 9996
ORAL EXERCISES Find the logarithm of: 1. 74. 5. 63. 2. 381. 6. 21. 3. 485. 7. 5. 4. 739. 8. 50.
Solution. The characteristic of .487 is 1 and the mantissa is .6875. Hence log .487 = -1 + .6875. This is often written in the abbreviated form, 1.6875. The mantissa is always kept positive in order to avoid the addition and subtraction of both positive and negative decimals, which in ordinary practice contain from three to five figures. Negative characteristics, being integers, are comparatively easy to take care of. (The student should note that log .487 is really negative, being - 1 +.6875, or - .3125.)
13. .638. 14. .0782. 15. .00964. 16. .0307.
17. 63800. 18. 27000. 19. 3.75. 20..784.
125. Interpolation. If a number contains four or more significant digits, its mantissa is not given in the table on page 244. It is possible, however, to calculate the mantissas of four- and five-digit numbers from mantissas which are given in the table. This process or the reverse of it is called interpolation. If we desire the logarithm of a number not in the table, say 3575, we know that it lies between the logarithms of 3570 and 3580, which are given in the table. Since 3575 is halfway between 3570 and 3580, we assume, though it is only approximately true, that the required logarithm is
halfway between their logarithms, 3.5527 and 3.5539. In order to find log 3575 we first look up log 3570 and log 3580 and then take half (or .5) their difference (this difference may usually be taken from the column headed D) and add it to log 3570. This gives log 3575 = 3.5527 +.5 x .0012 = 3.5533
Were we finding log 3578 we should take .8 of the difference between log 3570 and log 3580 and add it to log 3570 as follows: log 3578 = 3.5527 +.8 x .0012 = 3.5527 +.00096 = 3.5527 +.0010 3.5537
Observe that in using four-place tables one should not carry results to five figures. If the fifth figure of the logarithm is 5, 6, 7, 8, or 9, omit it and increase the fourth figure by 1; that is, obtain results to the nearest figure in the fourth place. For finding the logarithm of a number not given in the table we have the
RULE. Prefix the proper characteristic to the mantissa of the first three significant figures. Then multiply the difference between this mantissa and the next greater mantissa in the table (called the tabular difference) by the remaining figures of the number preceded by a decimal point. Add the product to the logarithm of the first three figures, taking the nearest decimal in the fourth place.
In this method of interpolation we have assumed that the increase in the logarithm is directly proportional to the
increase in the number. As has been said, this is not strictly true, yet the results thus obtained are nearly always correct to the fourth decimal place.
126. Antilogarithms. An antilogarithm is the number corresponding to a given logarithm. Thus antilog 3 equals 1000. If we desire the antilogarithm of a given logarithm, say 4.5132, we proceed as follows: The mantissa .5132 is found in the row which has 32 in column N and in the column which has 6 at the top. Hence the first three significant figures of the antilogarithm are 326. Since the characteristic is 4, the number must have five digits to the left of the decimal point. Thus antilog 4.5132 = 32600. Therefore, if the mantissa of a given logarithm is found in the table, its antilogarithm is obtained by the RULE. Find the row and the column in which the given mantissa lies. In the row found take the two figures which are in column “N” for the first two significant figures of the antilogarithm and for the third figure the number at the top of the column in which the mantissa stands. Place the decimal point as indicated by the characteristic. Page 8
If mantissa of a given logarithm, as in 1.4571, is not given in the table, the antilogarithm is obtained by interpolation as follows:
The mantissa 4571 lies between .4564, the mantissa of the sequence 286, and .4579, the mantissa of the sequence 287.
Therefore the antilogarithm of 1.4571 lies between 28.6 and 28.7. Since the tabular difference is 15 and the difference between .4564 and .4571 is 7, the mantissa .4571 lies of the way from .4564 to .4579. Therefore the required antilogarithm lies is of the way from 28.6 to 28.7. Then antilog 1.4571 28.6 + ?(.1) and 28.6 +.046 = 28.65.
Therefore to find the number corresponding to a given mantissa when the mantissa is not found in the table we have the
RULE. Write the number of three figures corresponding to the lesser of two mantissas between which the given mantissa lies. Subtract the less mantissa from the given one and divide the remainder by the tabular difference to two decimal places. If the second digit is 5 or more, increase the first digit by 1; if less than 5, omit it. Annex the resulting digit to the three already found and place the decimal point where indicated by the characteristic.
Find the antilogarithms of: 1. 1.4860. 5. 1.3626. 2. 2.4796. 6. 9.8448 3. 4.9481. 7. 6.0748 4. 0.3727. 8. 5.8153
9. .6187. 10. 7.5257 - 10. 11. 8.4230. 12. 8.6510 – 10.
127. Multiplication. Multiplication by logarithms depends on the THEOREM. The logarithm of the product of two numbers is the sum of the logarithms of the numbers. That is, for the numbers x and z log, (2 • 2) = logo x + logo 2.
It follows from this theorem that the logarithm of the product of more than two numbers is the sum of the logarithms of the factors. Thus log (x • Yoz) = log x + log y + log 2.
Perform the indicated operation by logarithms : 1. 16 x 35.
Solution. log (3470 X .0485) = 3.5403 + 8.6857 – 10 - 12.2260 – 10 = 2.2260 Antilog 2.2260 = 168.3. Since the mantissa is always positive, any number carried over from the tenths' column to the units' column is positive. This occurs in the preceding solution, where .6 +.5 = 1.*, giving + 1 to be added to the sum of the characteristics + 3 and – 2, in the units' column. Mistakes in such cases will be less frequent if the logarithms with negative characteristics be written as in the 8–10 notation.
14. 438 X.725. 15. 637 x .0354.
16. 91.7 x .00637.
18. (4316)(- 3.81) HINT. Determine by inspection the sign of the product. Then calculate as if all signs were positive and give to the result the proper sign.
19. 23(- 284).
21. 4.83 X 2.36 x .971. 22. 57(-28)(-3.65).
128. Division. Division by logarithms depends on the THEOREM. The logarithm of the quotient of two numbers is the logarithm of the dividend minus the logarithm of the divisor. That is, for the numbers x and 2
logo x - logo 2. It follows from the multiplication and the division theorems that if the product of two or more numbers is divided by the product of two or more others, the logarithm of the resulting quotient is the sum of the logarithms of the factors of the dividend minus the sum of the logarithms of the factors of the divisor. Thus abc log log a + log b + log c – (log x + log y + log 2). Xyz
Using logarithms, perform the indicated operations: 1. 765 ; 34.
129. Raising to a power by means of logarithms. Raising to a power by means of logarithms depends on the THEOREM. The logarithm of the rth power of a number is r times the logarithm of the number. Page 9
That is, for the numbers r and x, logo ac= r logo x. logo x = 1. Then b?. Raising both members of (2) to the rth power, 207 - brl. Therefore logo x = rl. From (1) and (4) logom", = r logo 2.
Compute, using logarithms : 1. (3.42)4 Solution. log 3.42 = .5340. log (3.42)* = 40.5340) = 2.1360. Therefore .(3.42)4 = antilog 2.1360 = 136.8.
2. (4.59) 4. (3.812) 3. (47.91). 5. (.0738)3. Solution. log .0738 2.8681 = 8.8681 10 log (.0738)3 = 3(2.8681) = 3(8.8681 – 10) = 4.6043 26.6043 – 30. Therefore (.0738)3 = antilog 7.6043 .0004021.
6. (.0874)4 7. (.007495) 8. (.9284)3. 9. 6984(842) (43)2(8381) 11. 10. 482(31.54) 0.097)2(435) 12. .967 · 0.0723).
130. Extracting a root by means of logarithms. Extracting a root by means of logarithms depends on the
THEOREM. The logarithm of the real rth root of a number is the logarithm of the number divided by r. That is, for the real numbers r and n, logo Vn = 1 logo n. Proof. Let logo n = l. (1) Then 6. (2) Extracting the rth root of both members of (2), (n) = (boy} = B? (3) 1 Therefore logo (n)
Compute, using logarithms : 1. 0495. Solution. log 495 2.6946. log 04953(2.6946) = .8982. Therefore 495 = antilog .8982 = 7.91. 2. 876. 4. 1732. 3. 1925. 5. 9.0785. Solution. log .0785 = 2.8949. If one divided 2.8949 as it stands by 3, he would be likely to confuse the negative characteristic and the positive mantissa. This and other difficulties may be easily avoided by adding to the characteristic and subtracting from the resulting logarithm any integral multiple of the index of the root which will make the characteristic positive. Thus log .0785 7.8949 – 9. Dividing by 3, log v.0785 2.6316 - 3. Therefore Ď.0785 = antilog 1.6316 = .4282.
6. V.0853. 8. V.005937. 10. (5.37) 7. Ņ.0007625. 9. $.7648. 11. (5.368) 12. (- 7.953). 15. V361. 13. V49.5(7.38) 16. V97742. 14. V.0395) 518 217
18. Determine the logarithms of 4267, 426.7, 42.67, and 4.267 to the base 10 and to the base 100. Compare the results. What fact about logarithms do these results emphasize ? NOTE. The preceding four-place table will usually give results correct to one half of one per cent. Five-place tables give the mantissa to five decimal places of the numbers from 1 to 9999 and, by interpolation, the mantissa of numbers from 1 to 99,999. Such tables give results correct to one twentieth of one per cent, a degree of accuracy which is sufficient for most engineering work. Six-place tables give the mantissa to six decimals for the same range of numbers as a five-place table, but the labor of using a sixplace table is much greater than that of using a five-place one. Seven-place tables contain the mantissas of the numbers from 1 to 99,999. Such tables are needed in certain kinds of engineering work and are of constant use in astronomy. In place of a table of logarithms engineers often use an instrument called a slide rule. This is really a mechanical table of logarithms arranged ingeniously for rapid practical use. Results can be obtained with such an instrument far more quickly than with an ordinary table of logarithms, and that without recording or even thinking of a single logarithm. A slide rule ten inches long usually gives results correct to three figures. For work requiring greater precision larger and more elaborate instruments which give a ten-figure accuracy are used.
131. Exponential equations. An exponential equation is an equation in which the unknown occurs in an exponent. Many exponential equations are readily solved by means of logarithms, since log at = x log a.
Thus let ar = c. Then x log a = log c. Whence x = log c = log a. EXAMPLE
Solve for x : 7* = 283. Solution. log 7* = log 283 x log 7 log 283. Whence log 283 2.4518 X = log 7 .8451
1. 4= 3. 182x 4. 92+1 43. 6. 52r-1 5. 3 =1.05. 7. (.75)* 52. Solution. x log .75 = log 52. log 52 1.7160 1.7160 1.7160 log .75 1.8751 - 1 +.8751 .1249
8. (.0863) + = .284. 9. (242)-- = 500.
10. (.0073)2+1 = (.084). 11. (72)22–3 = (28)-.
MISCELLANEOUS EXERCISES 1. Prove log ab log a + log b. 2. Prove log 7 = log a – log b. 3. Prove log (a)" = n log a. 4. Prove log Va=1 log a. 5. Does the logarithm of a negative number to a positive base exist? Explain.
Find, without using the table, the numeric values of : 6. log, 4. 9. log, 27. 7. logs 25. 10. 4 logs 125. 11. log: 8 + log: 4. 14. 4 log 125 – 3 log25 5 + 2 log, 25. 15. 5 log2 (1) + 3 log4 (64) + 7 log: 4.
Simplify: 16. log f + log 7. 17. log 17 – log 3. 18. log 3,5 + log 1. - log 25. 19. 3 log 5 + 5 log 3. 20. 2 log 9 + 3 log 5 – 7 log 15.
21. log v a2 - 1 log (a + x) + log (a – x). a3 + 23 22. log = log (a + x) + log (a? - ax + x2) - log a.
In the following obtain results to four figures: 24. The circumference of a circle is 2 TR. (1 = 3.1416.) Find the circumference of a circle whose radius is 23.15. 25. The area of a circle is aR?. Find the area of a circle whose diameter is 18.76 inches. Page 10
132. Introduction. In surveying, astronomy, and in countless other applications of science it is necessary to solve triangles. A triangle has six parts, three sides and three angles. When three of these parts, of which one at least must be a side, have been given or have been measured, the other three can be computed by the methods of trigonometry. In developing the principles of trigonometry it is simplest to begin with the right triangle. In order to attack the prob B lem stated above it is necessary to denote the ratios of the sides of a right triangle by certain names. Let ABC be any right triangle with the right angle at C. a side opposite The sine of the angle A hypotenuse 6 side adjacent The cosine of the angle A = hypotenuse side opposite The tangent of the angle A = b side adjacent b. side adjacent The cotangent of angle A side opposite Page 11
Find the missing parts in the following right triangles. Perform the multiplications and divisions as in Exercise 3 above, or by logarithms if the teacher prefers. a b с B 4. ? ? 300 18° ? 5. 120 ? ? ? 56° 6. ? 225 ? 67° ? 7. ? 86.6 100 ? ? 8. 707.1 ? 1000 ? ? 9. 40 20.38 ? ? ? 10. 376.8 500 ? ? 11. 17.05 ? 25 ? 12. ? 73.08 80? ? 13. 606 1500 ? ? 14. 125 50.5 ? ? 15. ? 67.84 80? ? 16. 1386 ? ? 78° ? ? 214 ? ? 53° 18. 10.45 99.45 ? ? ?
133. Interpolation. The word “interpolation' literally a placing between. The process consists in finding from the numbers in a given table a number not listed there. It applies to any table, one of square roots, cube roots, or logarithms, or a trigonometric table, such as that given on page 265. There are two problems of interpolation in a trigonometric table : (a) Given an angle not in the table, to find the corresponding sine, cosine, tangent, or cotangent. (6) Given the value of a sine, cosine, tangent, or cotangent not in the table, to find the corresponding angle. Page 12
The foregoing can be reduced to one simple statement : Minus 10 is understood after all logarithms in the table except those in the column at the top of which is the word "cotangent." EXERCISES Solve the following, using the logarithms tables on page 270: 1. In the right triangle ABC side AC = 9.86 and ZA = 34° 20'. Solve the triangle,
Solution. tan A a. b a = b tan A. Hence log a = log b + log tan A log 9.86 + log tan 34° 20'. log 9.86 .9939 log tan 34°20' = 9.8344 - 10 log a = .8283 6.734.
Compute the missing parts in the following right triangles. a b с A B 2. 187 ? ? 41° 40' ? 3. ? 64.9 ? 19° 20' ? 4. ? ? .0785 ? 35° 20' 5. 154 ? 200 ? 6. 609 481 ? ? Page 13
18. What does the expression in Problem 17 become when n = 1? 2? 3? Are these three results in arithmetic progression ? 19. A body falls 16 feet the first second, 48 feet the next, 80 feet the next, and so on. How far does it fall during the tenth second ? the twentieth? the nth ? 20. The digits of a certain three-digit number are in arithmetic progression. The sum of the digits is 15 and the sum of their squares is 93. Find the number. 21. If an arithmetic progression has an odd number of terms, show that the middle term is half the sum of the first and last terms. 22. In an arithmetic progression find the 9th term from the beginning; the 9th term from the end. 23. The pairs of numbers (1, 1), (2, 3), (3, ), (4, 3), (5, ), (6, – ) are respectively the x and y of coördinates of six points. Show that the distances of the points from the origin are in arithmetic progression. 24. Find the rth term from the beginning of an arithmetic progression and the rth term from the end. Show that the arithmetic mean of these two terms is the middle term of the progression if it has an odd number of terms. 25. The velocity of a body falling from rest increases uniformly 32 feet per second each second. Find the velocity at the end of the 10th second. Find the average velocity during the first second, during the third second, during the nth second.
138. Arithmetic means. The arithmetic mean between two numbers is a number which forms with the two given ones as the first and last terms an arithmetical progression. Page 14
18. Find the sum of the first n integers which are exact multiples of 13. 19. How many terms of 1 +5 + 9 + ... will amount to 630 ? HINT. We have here S, a, and d in S = 112 a + (n – 1)d), to solve the equation for n. 20. How many terms of 15 + 121 + 10 + ... will amount to 15 ? 21. How many terms of terms of - 12 - 102 - 9 ... 9... will amount to 54? 22. How many terms of 16 + 153 + 154 + will amount to 216? 23. The first term of an arithmetic progression is 7 and the 6th term is 12. Find the sum of the first 100 terms. 24. Assuming that a heavy iron ball is not retarded by the air, determine the number of seconds it would take to fall if it is dropped from a point on the Eiffel Tower 980 feet above the ground. (S = 16 t2.) 25. A ball shot vertically upward returned to the point it started from 18 seconds later. How high did it rise? 26. By Exercise 25, page 277, it is seen that a falling body obeys the law of an arithmetic progression. Show from the data of that exercise that the general formula S 12 12 a + (n 1)d] becomes the special one, S used in physics for such 2 problems. 27. A passenger in an aëroplane drops a rock and observes that it strikes the ground 11 seconds later. How high was the plane at the time? Page 15
13. Find the amount of 100 dollars at 52% compounded annually for 3 years. 14. What is the (n − 1)st term of a geometric progression ? the (n − 2)d? the (n − 3)d? the (n + 1)st? 15. The sum of the digits of a three-digit number is 14 and the digits are in geometric progression. If 594 is added to the number, the result is expressed by the digits in reverse order. Find the numbers. 16. Show that the distance of the following points from the origin are in geometric progression: (1,3), (V2,3V2), (2, 6), (2V2, 6V2). 17. A house worth $8000 depreciates 10% in value each year. What will it be worth at the end of 7 years? 18. An automobile selling at $800 depreciates in use yearly 30% of its cost. What is it worth at the end of 3 years ? 19. Show that the product of any two terms of a geometric progression equidistant from a given term is always the same. 20. Show that if each term of a geometric progression is subtracted from the following one, the successive remainders form a geometric progression. 21. One hundred dollars is deposited annually for six years in a bank which pays four per cent compound interest. What sum is due the depositor at the end of six years if he has withdrawn no money from the bank during that time? 22. A father deposits a certain sum in a bank at the beginning of the school year each of the four years his son is in the high school. If this amounts to sixteen hundred dollars when the boy enters college after his high school Page 16
was done five times. What portion of the original contents was then in the vessel ? 20. At each stroke an air pump withdraws 20 cubic inches of air from a vessel containing 200 cubic inches. After every stroke the air remaining in the vessel expands and completely fills it. What fraction of the original quantity of air remains in the vessel at the end of the eighth stroke? 144. Infinite geometric series. If the number of terms of a geometric series is unlimited, it is called an infinite geometric series. In the progression 3, 6, 12, the ratio is positive and greater than 1 and each term is greater than the preceding one. Such a progression is an increasing one. It is obvious that the sum of an unlimited number of terms in an increasing geometric progression is unlimited. If r> 1 (read" is greater than 1 "), the sum can be made as large as we please by taking a sufficient number of terms. If the geometric progression is a decreasing one, however, the result is very different. In the decreasing progression 4, 2, 1, 1, 1, , - the ratio is positive and less than 1, and each term is less than the preceding one. Here the sum of the first three terms is 7, the first four 71, the first five 73, the first six 73, etc. Obviously, the more terms taken the greater is the sum. But no matter how large the number of terms taken their sum is less than 8. This number 8 is called the limit of the series. Sn = 4 + 2 +1 + + This series and its limit can be illustrated by the motion of a point 'along a line in the diagram on the following page. Suppose line AX is eight inches long and a point starts from A and moves in one minute to point B which is į the Page 17
19. ABC is an equilateral triangle with side AB equal to 4. The triangle DEF is formed by joining the midpoints of the adjacent sides of ABC. Triangle IGH is formed similarly, and so on. Determine the series formed by the ML-L areas of the successive triangles. G KH Find the sum of the areas of all the А B triangles drawn as there supposed. D 20. A flywheel whose circumference is 12 feet makes 2400 revolutions per minute. If it slows down so that each second it makes ninety-eight per cent as many revolutions as it did the preceding second, how far will a point on its rim have traveled by the time it has stopped ? NOTE. In the study of geometric progressions we have seen that the sum of the infinite series 1 + x + x2 + 2a +. is a definite number when x has any value less than 1. But it has no finite value when x is equal to or greater than 1; that is, we have an expression which we cannot use arithmetically unless x has a properly chosen value. If we were studying a problem which involved such a series, it would be a matter of the most vital importance to know whether the values of x under discussion were such as to make the series meaningless. This question of distinguishing between expressions which converge (that is, the sum of whose terms approaches a limit) and those which do not has an interesting history. Newton and his followers in the seventeenth century dealt with infinite series and always assumed that they converged, as, in fact, most of them did. But as more complicated series came into use it became more difficult to tell from inspection whether the series converged or not for a given value of the variable. It was not until the beginning of the nineteenth century that Gauss, Abel, and Cauchy, in Germany, Norway, and France respectively, began to teach this subject effectively and to devise far-reaching tests to determine the values of x for which certain series converge to a finite limit. It is said that on hearing a discussion by Cauchy in regard to series which do not always converge, the astronomer La Place became greatly alarmed lest he had made use of some such series in Page 18
7. Which is the greater ratio 17: 25 or 23:33 ? 8. In an alloy of 52 ounces of zinc and copper there are 12 ounces of zinc. Find the ratio of zinc to copper. 9. Divide $1800 into two parts which are in the ratio of 5: 7. 10. Separate 360 into three parts in the ratio of 3:5:7. 11. Show that X X + 4 < if x is positive. x + 3 X + 7 HINT. Reduce the fractions to respectively equivalent fractions having a common denominator and then compare the numerators. 12. Show 2 + 3 < if x is positive. X + 5 X + 8 13. Show X + 4 < if x is positive. x + 3 X + 5 14. Arrange the ratios 3:4; 7:8; 13:16 in decreasing order of magnitude. 9 3 (1
2 a? – 5 a +3. a? – 2a + 1 16. a2 9 a’ – 5a +6 a(x2 - 3 a) - ax (2x - 3) ax 17. (x2 – 3x)2 x2 3 x 18. Which is the greater ratio if a is positive: 7 + 2 a 7 + 3 a 7 +5 a 19. If a positive number is added to or subtracted from both terms of a proper fraction, what change is produced in the value of the fraction? 20. Separate 660 into three parts such that the ratio of the first to the second is 5:6 and of the second to the third is 9:11. Page 19
The proportions given in Exercises 16, 18, and 20 are said to be derived from a: b = c:d by addition, subtraction, and addition and subtraction, respectively. a If ģ show that the following equalities are true: b 5 a 5 c a2 62 c2 b d 62 d2 2 a 762 c2 – 7 da 26. 76 62 d2 a2 ac a2 - 52 c2 d2 27. 62 bd 2 ab 2 cd 7 a2 -3 62 7 c2-3 d2 5α b 5 c d 5 ab 5 cd a2 + ab + 62 ab + 62 c2 + cd + d2 c2 cd + d2
30. In the proof which follows give the reason for each step and state the result verbally. If a to te b å d b + d + b d Page 20
9. Compare the volumes of two spheres whose radii are in the ratio 5:7. 10. Combining the results of Problems 5 and 8 show that
V2 11. The radius of the sun is to the radius of the earth as 109: 1. Find the ratio of their volumes and the ratio of their surfaces. 12. If ABC is any triangle and LM is a line parallel to BC, meeting AB at L and AC at M, it is proved in geometry that area ABC AB? AC2 BC2 area ALM AL AM LM If in the adjacent figure the area of triangle ABC is 196 square inches and AL is 15 inches and AB is 21 inches, find the area of the triangle ALM. What is the area of the trapezoid BCML? M 13. If in the figure of Exercise 12, AL is 18 and AB B is 28 and ALM equals 300, find the area of ABC. 14. If in the figure of Exercise 12, AL is 12 feet and AB is 20 feet, and area ABC equals 400 square feet, find the area of the trapezoid BCML. 15. If in the figure of Exercise 12, AL is 60 feet and area ALM : area BCML = 4:5, find AB. 16. If in the figure of Exercise 12, ABC equals BCML and AB = 100, find AL to two decimals. Page 21
8. One quantity may vary as the cube of another. Express this relation in symbols: (a) as a variation, (6) as a proportion.
154. Inverse variation. If two variables are so related that one increases as the other decreases and vice versa, either variable is said to vary inversely as the other. For example, if the area of a triangle, A, is constant while the base, b, and altitude, a, vary, the base varies inversely as the altitude. This is apparent from the equation A = 1 ab. Similarly if d is the distance covered by a motor bus each trip, and r its speed, and t the time, d = rt. Here for a fixed route of length d, t varies inversely as r.
1 Therefore ti : tz (5) 11 12 1 In the form of a variation (5) becomes t varies as In general, x varies inversely as y when x varies as the reciprocal of y; that is, X varies as 1 ข And if x and y denote any two corresponding values of the variable, and x, and yı a particular pair of corresponding values, 1:1 X : X1 (7) y Yu Whence X 1 (8) Yi ข
But xıyı is a constant, being the product of two definite numbers. Call this constant K. Then (8) becomes ху K. Page 22
9. The time of vibration of a pendulum varies directly as the square root of its length and inversely as the force of gravitation. Express the relation between t, 1, and g by a proportion. 10. If a pendulum 100 centimeters long vibrates once in 1 second, find the period of a pendulum 121 centimeters long. 11. If a pendulum 100 centimeters long vibrates once in one second at sea level where the force of gravitation is 982, find the time of one vibration of a pendulum 144 centimeters long on top of a mountain where the force of gravitation is 970. 12. The weight of an object above the surface of the earth varies inversely as the square of its distance from the center of the earth. Express this relation as a variation and also as a proportion. 13. An object weighing 160 pounds at the earth's surface would weigh how much 800 miles above the surface? 2000 miles above the earth's surface? (Radius of earth equals 4000 miles.) 14. A meteorite weighs one ton. How far from the earth's surface was it when its weight was half as much? one tenth as much ? 15. The weight of any object below the surface of the earth varies directly as its distance from the center. An object weighs 100 pounds at the earth's surface. Find its weight (a) 800 miles below the surface, (b) 2000 miles below the surface, (c) at the center. (Radius of earth = 4000 miles.) Page 23
159. The general linear system in three variables. The method of addition and subtraction applied to the system a12 + by + C12 di, (1) A22 + b2y + C2z = (2) 23.C + b3y + C3z = d3, (3) gives a d1b2c3+d2b3c1 +d3b1c2 – d3b2c1 – dıb3C2 — d2b1C3), (4) a1b2c3 +azb3cı +azbı22 - azb2cı - Qjb3C2 — azbic3 ajd2c3+Q2d3c1 +azdıcı – azdıcı – ajd3C2 — Azdıc3 Y (5) a1b2c3+azbzcı +azbica - a3b2cı - a_bzC2 - azbicz
By grouping and factoring as in § 158 the values (4), (5), and (6) become Page 24
Solution. Rewriting in standard form, (4) 2x – y + z = 1, (5) 33 + 5y + 72 = - 10. (6) From I and II, page 326, 2 3 5 1 1 1 10 5 7 100 5 1 3 5 20 1 2 5 2 1 1 3 10 7 120 20 The value of 2 can now be more easily obtained by substituting the values of x and y already found in (1), (2), or (3) than by determinants. Substituting in (2), – 10 +2 = -6 + 1. Hence, 5. Check as usual. EXERCISES Solve for x, y, and z as in the preceding example and check results. 1. x + y + 2 = 2. 2 x + y - 2 = 3, x – 3y + z = 2, 30 + y - 5z = - 8. x + y - 32 = -7, 2 + 32 + y = - 11. Hint. The first equation is 3 x + y +0z = 22. The coefficient zero must never be omitted in using determinants. Page 25
If the student meets the statement = 0, he should 0 regard it as a loose use of the statement that becomes n infinitely large as n becomes infinitesimally small. 0 0 becomes 2c2 9 when x = 3. For any value of x other than the critical value 3, the fraction equals a definite number. Usually we are concerned with the limit of such expressions as the variable approaches a critical value. The limit for the X 3 fraction is easily found. We assign to x successively 22 9 the values 2.9, 2.99, 2.999, 2.9999, etc. The corresponding values of the fractions are 39, 399, 1999, 19999, etc. Obviously these numbers approach the limit . We may arrive at this result more easily as follows: for X - 3 all values of x except 3 the terms of the fraction 22 -9 may 1 be divided by (x-3), obtaining • This result is true, however little x may differ from 3. Now if, without giving 1 x the value 3, we make it approach 3 as a limit, will approach į as a limit, and this is the limit of the original 3 fraction as well. 9 By either of the preceding methods it can be shown that x2 - 16 0 - 4 These two fractions are simple illustrations of the impor 0 tant fact that the symbol ő is not a definite number. The truth of this can be seen more clearly from a study of the following graph.
, which becomes ; for x = 4, has 8 as its limit. a Page 26
knowns, the system has an infinite number of sets of roots; that is, the system is indeterminate. 0
The symbol, then is a symbol of indetermination.
Solve by determinants and interpret the results: 1. 2 x - y = 1, 3. x+y+z= 3, 4 x – 2y = 2. 2 – 5 y +22=0, 2 x + 2 y +22= 3. 2. x - y = 1, 4. x + y +z=2, Y – X = 0. 3x – 5 y +2=8, 0x +0y+02= 0. 5. From the results obtained in Exercise 4, what conclusion is warranted regarding the number of sets of roots belonging to a system of two linear equations in three variables ? 6. 2 + y + z = 1, 0 x + y + 0%= 0, 0 x +0y+02= 0. 7. What do the results obtained in Exercise 6 show regarding the number of sets of roots belonging to one equation in three variables ? What limit does each of the following expressions approach as n approaches oo? 1 n 8 15. ī n 9. 13. n n(n + 4). Hint. This may be written 16. n? . 10. 1 n 1+ n(n + 1)(n + 2) 17. n 11. 14. |
Sabendo que x 2 y 2 30 e que x-y 8 dê o valor de left x y right 2
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