1. Keystone Problems… Keystone Problems… next Set 19 © 2007 Herbert I. Gross

2. You will soon be assigned problems to test whether you have internalized the material in Lesson 19 of our algebra course. The Keystone Illustrations below are prototypes of the problems you’ll be doing. Work out the problems on your own. Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problems next © 2007 Herbert I. Gross

3. As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

4. next © 2007 Herbert I. Gross Given that… 10 0.3010 = 2; what is the value of x if 10 4.3010 = x? next Keystone Problem #1 for Lesson 19

5. next © 2007 Herbert I. Gross When a number is expressed in the form 10 4.3010, the whole number part of the exponent (in this case, 4) tells us the order of magnitude of the number, and the part of the exponent that’s to the right of the whole number (in this case,.3010) tells us the sequence of digits in the number. Solution for Problem #1

6. next © 2007 Herbert I. Gross To see why this is true, we use the fact that 4.3010 = 4 + 0.3010 and rewrite10 4.3010 in the form… next Solution for Problem #1 By our rule for multiplying like bases, we know that …10 4 + 0.3010 = 10 4 × 10 0.3010. We also know that 10 4 = 10,000 and 10 0.3010 = 2. Therefore… 10 4.3010 = 10 4 + 0.3010 10 4.3010 = 10 4 × 10 0.3010 = 10,000 × 2 = 20,000

7. Even without knowing the exact value of 10 4.3010, the fact that 10 x increases as x increases tells us that since 4 < 4.3010 < 5, it means that 10 4 < 10 4.3010 < 10 5 Notes #1 And since 10 4 = 10,000 and 10 5 = 100,000 we see that… 10,000 < 10 4.3010 < 100,000 next © 2007 Herbert I. Gross

8. next © 2007 Herbert I. Gross Given that… 10 0.6990 = 5, 10 0.4771 = 3 and 10 0.3010 = 2; for what value of x will 10 x = 30? Keystone Problem #2 for Lesson 19

9. One of the nice things about working with exponents is that it allows us to replace multiplication problems by equivalent addition problems. This was very important in the days before the advent of calculators and computers. Preface to the Solution Preface to the Solution next © 2007 Herbert I. Gross

10. For example, using pencil and paper to compute such products as 3.14159 X 2.98764 X 5.12347 would be very tedious to say the least. On the other hand, with a hand calculator we can compute this product with ease. However, prior to the age of calculators, Napier’s invention of logarithms (arithmetic using exponents) was a way to replace such tedious multiplication problems by addition problems. Preface Preface next © 2007 Herbert I. Gross

11. Of course with the advent of calculators there is no longer a need to use logarithms in order to perform arithmetic calculations. However, it is still very important to know and to be able to use the rules for exponent arithmetic because so many things in “real life” involve exponential growth. Preface Preface next © 2007 Herbert I. Gross Using small whole numbers for simplicity, this exercise is designed to illustrate how we can replace a multiplication problem by an equivalent addition problem. next

12. © 2007 Herbert I. Gross Solution for Problem #2 We are given the following three pieces of information… 10 0.6990 = 5 We also know that… next 10 0.4771 = 3 10 0.3010 = 2 5 × 3 × 2 = 30 1 23

13. © 2007 Herbert I. Gross Solution for Problem #2 If we now replace 5, 3, and 2 in the equation 5 × 3 × 2 = 30 by their above values, we see that… next 5 × 3 × 2 = 30 10 0.6990 10 0.6990 = 5 10 0.4771 10 0.4771 = 3 10 0.3010 10 0.3010 = 2

14. next © 2007 Herbert I. Gross Solution for Problem #2 By the rule for multiplying like bases, the above equation can be rewritten in the equivalent form… next 10 0.6990 × 10 0.4771 × 10 0.3010 = 30 10 0.6990 + 0.4771 + 0.3010 = 30 Since 0.6990 + 0.4771 + 0.3010 = 1.4771, we may rewrite the above equation as… 10 1.4771 = 30

15. © 2007 Herbert I. Gross Solution for Problem #2 Since no two different powers of 10 can have the same output, we see from the above equation that the only way in which 10 x can equal 30 is if x = 1.4771. 10 1.4771 = 30

16. next © 2007 Herbert I. Gross Solution for Problem #2 To check our answer we may use a calculator with an x y key and then use the following sequence of key strokes… next Using my calculator the result is 10 1.4771 = 29.99853... The slight discrepancy between 29.99853… and 30 comes from the fact that the exponents in the equation 10 0.6990 × 10 0.4771 × 10 0.3010 = 30 have been rounded off to 4 decimal places. 10xyxy 1.4771=

17. Information such as 10 0.6990 = 5 was made available through tables that were developed by Napier. How he did this is well beyond the scope of this course. © 2007 Herbert I. Gross Notes #2 However, there seem to be a few interesting “coincidences” in our solution; one of which is that the sum of the two exponents 0.6990 and 0.3010 is exactly 1. next

18. To see why this is not a coincidence, observe that… © 2007 Herbert I. Gross Notes 2 5 × 2 = 10 0.6990 × 10 0.3010 = 10 0.6990 + 0.3010 = 10 1 = 10 next

19. Of course we would probably never have elected to compute 2 × 3 × 5 this way. Just imagine not having a calculator and having to compute the product 3.14159 × 2.98764 × 5.1234. However, from the exponential arithmetic (logarithm) tables developed by Napier, we could find that… © 2007 Herbert I. Gross Note #2 10 0.49715 = 3.14159 next 10 0.47533 = 2.98764 10 0.70956 = 5.12347

20. We could then proceed as we did in the previous solution by rewriting… next © 2007 Herbert I. Gross Note #2 n = 3.14159 × 2.98764 × 5.12347 next n = 10 0.49715 + 0.47533 + 0.70956 in the form… …and by the rules for multiplying like bases we see that… n = 10 0.49715 × 10 0.47533 × 10 0.70956 next

21. Since 0.49715 + 0.47533 + 0.70956 = 1.68204, we may rewrite our equation as… © 2007 Herbert I. Gross Note #2 next n = 10 0.49715 + 0.47533 + 0.70956 We may then use the x y key on the calculator to see that n = 48.0883637... 1.68204 next (or in the language of logarithms, n = log 1.68204).

22. © 2007 Herbert I. Gross Aside At the time of Napier, there were no calculators. His tables were designed to find the logarithm of a given number. Thus, if a person had been presented with the problem log n = 1.68204, he or she would have had to make an indirect computation. That is: starting with the logarithm of a number (1.68204), they would then have to work “backwards” in the table in order to find the number. next

23. © 2007 Herbert I. Gross Note #2 If we were to use “pencil and paper” to find the exact value of the product 3.14159 × 2.98764 × 5.12347, we would have found that its exact value was 48.088581743330172.

24. The discrepancy between the exact value and the value we got by using the log table, is that the numbers in the table were rounded off. As a result, only the first few decimal places in our estimated answer were significant. next © 2007 Herbert I. Gross Note #2 48.088363748.088581743330172 TableExact The above discussion serves to illustrate why the study of significant figures was so important before the advent of calculators. next

25. There is much more information that we can derive from the fact that 10 0.3010 = 2 © 2007 Herbert I. Gross Note #2 10 x = 8 next For example, suppose we wanted to find the value of x for which… We know that 8 = 2 3. Therefore, we may rewrite the above equation as… 10 x = (2) 3 next

26. We also know that 10 0.3010 = 2. Therefore, we may rewrite the equation 10 x = (2) 3 as… © 2007 Herbert I. Gross Note #2 10 x = (10 0.3010 ) 3 By our rules for the arithmetic of exponents we know that… next (10 0.3010 ) 3 = 10 (0.3010)3 = 10 0.9030

27. Thus, replacing 8 by 10 0.9030, the equation 10 x = 8 becomes… next © 2007 Herbert I. Gross Note #2 10 x = 10 0.9030 next Since no two different powers of 10 can be equal, we see from the above equation that x = 0.9030. As a check, we may use our calculator (or tables) to see that 10 0.9030 = 7.99... which tells us that our answer is at least plausible and probably correct.

28. © 2007 Herbert I. Gross Given that… 10 0.3010 = 2 and 10 1.3802 = 24; what is the value of x for which 2 x = 24? next Keystone Problem #3 for Lesson 19

29. Most calculators come equipped with a 10 x key (or, at least an x y key). However, they do not have other keys such as 2 x or 3 x (let alone a key such as 3.7 x etc.), but with the use of the 10 x key there is no need for having these other keys. Preface to the Solution Preface to the Solution next © 2007 Herbert I. Gross

30. With the rules for exponents all we would need is any one key of the form b x. However, b was chosen to be 10 primarily to take advantage of place value. Preface Preface next © 2007 Herbert I. Gross For example, when we see 10 5.1567 we know immediately that the number is between 10 5 and 10 6 ; that is: it is between 100,000 and 1,000,000. next

31. With the 10 x key, we can let x = 0.3010 and verify that 10 0.3010 = 2. However, other than for trial-and-error methods the 10 x key can’t help us find the value of x if we know the value of 10 x. Preface Preface next © 2007 Herbert I. Gross For example, suppose we were given a problem such as: Find the value of x for which 10 x = 7. next

32. In fact, to find the value of x for which 10 x = 7, we would want to use the “undoing” method. (That is: in the same way that we use subtraction to undo addition we would like to find a way to “undo” 10 x.) Preface Preface next © 2007 Herbert I. Gross In this regard, calculators that have a 10 x key also have another key that is labeled log. next

33. Let’s look at a rather simple illustration in terms of the equality 10 3 = 1,000. Namely, if we are told that 10 3 = x, to find the value of x using the calculator we would enter “3” and then press the 10 x key to obtain x = 1,000. Preface Preface next © 2007 Herbert I. Gross On the other hand, if we were told that 10 x = 1,000, we would enter “1,000” and then press the log key to obtain x = 3. next

34. It is not important for you to internalize, the full meaning of “logarithm”; just remember that it is simply an abbreviation for a specific exponent. Logarithms © 2007 Herbert I. Gross next For example, an equality such as log 56 = 1.748 is simply another way of saying that 10 1.748 = 56 That is: “log x” is an abbreviation for “the power to which 10 has to raised to yield x as the answer”.

35. More generally… An equivalent way of saying that 10 x = y is to say that log y = x. Logarithms © 2007 Herbert I. Gross In other words, if you understand the meaning of say, 10 4 = 10,000, but you don’t understand the meaning of log 10,000 = 4; it means that you are having a problem with the new vocabulary; not with mathematics. next

36. © 2007 Herbert I. Gross Logarithms Applying this idea to solving for x if 10 x = 7, we simply enter 7, press the log key and then the “=” key. That is, we enter… next …and 0.8450983… appears in the display. As a check, using the 10 x key and rounding off 0.8450983… to 0.845098, we see that 10 0.845098 = 6.9999993...; which validates that our answer is quite reasonable. 7log=

37. Exercise 3 illustrates how by using the 10 x and log keys, we can solve the problem 2 x = 24, Logarithms next © 2007 Herbert I. Gross

38. next © 2007 Herbert I. Gross Solution for Problem #3 We are told that 10 0.3010 = 2 and 10 1.3802 = 24 Hence, we may replace 2 by 10 0.3010 and 24 by 10 1.3802 in the equation 2 x = 24 to obtain the equivalent equation… next (10 0.3010 ) x = 10 1.3802. We know from our rules for exponents that (10 0.3010 ) x = 10 1.3010x. 10 0.3010x = 10 1.3802 Hence, we may rewrite the above equation as… next

39. © 2007 Herbert I. Gross Solution for Problem #3 Since no two different powers of x yield the same value for 10 x, we may conclude from the above equation that 0.3010x = 1.3802; or x = 1.3802 ÷ 0.3010; and using the calculator for convenience we see that… x = 4.58539… 10 0.3010x = 10 1.3802

40. next © 2007 Herbert I. Gross Solution for Problem #3 To check that our solution is correct we may round off 1,3802 ÷ 0.3010 to 4.58539 and replace x by 4.58539. Doing this we see that 2 4.58539 = 24.00711…; which indicates that the solution of the equation 2 x = 24 is approximately x = 4.58539.

41. Ordinarily in trying to solve the equation 2 x = 24, we would not have had the luxury of knowing in advance that: 10 0.3010 = 2 and 10 1.3802 = 24. However, we could have obtained this information by using the log key to see that log 2 = 0.301029995… and log 24 = 1.38021134… (which is even more accurate than the given information) next © 2007 Herbert I. Gross Note #3

42. So let’s go through the procedure one more time, and this time find the value of x for which… 11 x = 212 next © 2007 Herbert I. Gross Note #3 We use the log key to see that… log 11 = 1.04139… or 10 1.04139 = 11 and log 212 = 2.32633… or 10 2.32633 = 212 next

43. So if we now replace 11 by 10 1.04139 and 212 by 10 2.32633, the equation 11 x = 212 can be replaced by the equivalent equation… © 2007 Herbert I. Gross Note #3 …and we may therefore conclude that 1.04139x = 2.32633 (10 1.04139 ) x = 10 2.32633 Hence, x = 2.32633 ÷ 1.04139 = 2.23387… next

44. As a check that the solution of 11 x = 212 is x = 2.23387…, we approximate 2.23387… by 2.23387 and then replace x by 2.23387 in the equation 11 x = 212. We then see that 11 2.224267 = 212.0000081… and this indicates that we have found the correct answer, at least to several decimal place accuracy. © 2007 Herbert I. Gross Note #3

45. In solving the equation 11 x = 212, we found that x = 2.32633 ÷ 1.04139 = 2.23387. Let’s look at this quotient in more detail. next © 2007 Herbert I. Gross Namely, 2.32633 = log 212 and 1.04139 = log 11. Hence, what we have shown is that… If 11 x = 212, then x = log 212 ÷ log 11. next Enrichment Note

46. More generally, the log key allows us to solve any equation of the form b x = n where b and n are any positive numbers. next © 2007 Herbert I. Gross Namely if b x = n, then x = log n ÷ log b For example, in terms of 2 x = 24, we saw that x = 1.3802 ÷ 0.3010, and since log 24 = 1.3802 and log 2 = 0.3010, we see that x = log 24 ÷ log 2 next Enrichment Note