International projectionist (Nov-Dec 1933)

10 INTERNATIONAL PROJECTIONIST November 1933 equation may be stated in the form of numbers, letters or any other mathematical symbols. That part of the equation to the left of the equals sign is called the left-hand side of the equation; that part to the right, the right-hand side of the equation. The following examples will enable you to recognize and know just what an equation is: 5—3=2 a x+b y=7 10 (x+3)=4 In dealing with algebraic equations, there are six simple and fundamental rules which are self-apparent, once the definition of an equation and an understanding of that definition have become thoroughly familiar. To repeat: an equation is a statement of equality between two mathematical terms — that is, two terms, each one equal to the other and set up in such a form to show that equality — basically form an equation. In considering the six fundamental rules, let us take for our basic equation the example given above: 5 — 3=2. This is a true arithmetical statement, ard fulfills our definition of an equation. (a) . The same number may be added to each side of an equation without destroying the equality: 5 — 3 = 2 (Applying rule a) 5—3+7=2+7 or solving, 9=9 (b) . The same number may be subtracted from each side of an equation without destroying the equality: 5 — 3=2 (Applying rule b) 5—3—1 = 2—1 or solving, 1 = 1 (c). Both sides of an equation may be multiplied by the same number without destroying the equality: 5—3 = 2 9(5— 3) =9 2 (Applying rule c) or solving, 18 = 18 (d). Both sides of an equation may be divided by the same number without destroying the equality: 5—3 = 2 5—3=2 (Applying rule d) 2 2 or solving, 1=1 (e). Both sides of an equation may be raised to the same power without destroying the equality: 5—3=2 (5— 3)2=2a (Applying rule e) or solving, 4=4 (/) . The same root may be extracted of both sides of an equation without destroying the equality: 5—3=2 V5— 3=V2 (Applying rule f) Solving, by means of logarithms as explained in Lesson 2, 1.414=1.414 The important thing to remember in dealing with equations is that any operation performed on one side of the equation must also be performed on the other, in orded to maintain the equality. Should one side be operated upon without the same operation being performed on the other, the fundamental equality would be destroyed and the equation would no longer be a true equation. If we take our example equation, 5 — 3=2, and add, let us say, 6 to the right-hand side without adding 6 to the left, we Answers to Problems, Article II MATHEMATICS FOR THE PROJECTIONIST "R EFLECTING the growing interest within the craft in this series of articles is the greatly increased number of answers received to Article II, dealing with logarithms, which was printed last month. Into the swim this month plunged a total of 168 potential mathematicians, a majority of whom evidently will have to apply themselves more diligently, for of this number only 47, including most of last month's standbys, turned in perfect papers. But first let us consider the answers : Problem 5.— Explain in a few words the difference betzveen the Naperian system and the Common system of logarithms. Ans. — The Naperian system of logarithms is based upon the letter "e", a number of approximately the value 2.718; while the Common, or natural, system is based upon the number 10. Problem 6. — What is the characteristic of the logarithm of the following numbers: 5,579; 25; 611; 895,728; 323? Ans.— 3; 1; 2; 5; and 2. Problem 7.— Complete the folloiving multiplications by logarithms, showing each step in the calculations: 250-731 = 4282= Ans.— Log 250 equals 2.3979 Log 731 equals 2.8639 Adding. 5.2618 Referring to the log tables, the number whose logarithm is 2618 is 1827. From a consideration of the characteristic, we know that the product contains six (6) numbers to the left of the decimal point, hence the answer is 182,750. 4282 Log 428 equals 2.6314. Two (2) times 2.6314 equals 5.2628. The number whose log is 2628 is 1831. From a consideration of the characteristic we know that the answer must have six (6) places to the left of the decimal point, hence the answer is 183,100. Problem 8.— Divide 725 by 25 by logarithmic method, shoiving each step in the calculation. Ans.— Log 725 is 2.8603 Log 25 is 1.3979 Subtracting. .4624 Referring to the log tables, the number whose logarithm is 4624 is 29, Answer. Problem 9. — Extract the cube root of 512 by logarithmic method, showing each step in the calculation. Ans.— Log 512 equals 2.7092. The root index is 3, which divided into 2.7092 equals .9031, the log of the cube root of 512, or 8, Questions 6 and 7 evidently occasioned some difficulty, several papers listing the characteristic of the log of 5,579 as 4, whereas the correct answer is 3 ; and in question 7, several "students" let slip their minds the information given relative to "significant figures" and proceeded to worry unnecessarily about the last three digits in the answer. Including those who slipped up only on the first exercise of Problem 6, the complete list of those who submitted correct answers follows : G. L. Cummings, Edward Burke, Dale Danielson, H. L. MacKenzie, Joe Barnett, Alex Schaeffer, Harry Ginsburg, Lou Adams, Donald Taylor, Otis Clarke, A. Henderson, Ivar Olson, D. E. Weiss, Lloyd Garrity, John Butler, Patrick Callahan, Leo Sinoski, Sam Tillson, Charles Bederman. Also, Paul Gaeth, Jr. ; Elmore R. Towne, Carl Broening, Ellsworth Bauer, Jack Murray, Clyde Woods, Eddie Howson, James Edwards, Thos. L. McGuire, Herbert L. Armstrong, W. Hoy, L. N. Osgood, Frank Nerogin, James Murphy, M. Aquilino, Samuel Barr, Harry Gilman, Leonard Hudson, Clifford Newton, Saul Oberman, Willard Jarvis, Richard Egrey, John P. Fegan, O. Hollenbeck, Milton Krebs, F. Kassler, Elmer Jameson, and Mason Sperl ing . , Special mention for fine papers is due to Paul Gaeth, Jr., Elmore R. Towne, Herbert L. Armstrong, Eddie Howson, W. Hoy, and Dale Danielson. Just before press time there arrived a perfect set of answers to the problems in Article I from Max Cann, employed as a projectionist at the Opera House, Hawera, New Zealand. Comments Mr. Cann, in part: "Congratulations to you for the way every detail of this series is explained. Being only a beginner in projection I have had much benefit from your monthly issues, and I am taking the greatest interest in the series on mathematics. We should have more material of this sort." Elmore R. Towne, L. U. 225, Aberdeen, S. D., relates how the articles "chased" him into the library in search of additional information, as follows : "Your lesson on logarithms was so inspiring that I spent several hours digging into text books for additional information on the subject, which proves that the course has benefitted me already. Keep it up!" The author of this series of articles Mr. Gordon S. Mitchell, cites the probobility of many readers having questions to ask with respect to the various points developed in the lessons, as well as other questions which come to mind relative to different applications of the problems listed. Mr. Mitchell urges prompt submission _ of all such questions, the answers to which will appear in a special article to appear in the near future.