1. Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 7 part 2 By Herb I. Gross and Richard A. Medeiros next

2. 2 2 × 2 ‾3 = 2 n Problem #1a © 2007 Herbert I. Gross Solve for the value of n that makes the equation a true statement. next Answer: n = - 1

3. Solution: We chose our definition of a negative exponent in a way that ensured we could still add exponents when we multiplied like bases. Therefore… 2 2 × 2 ‾3 = 2 2+ ‾3 = 2 ‾1. Hence, if 2 2 × 2 ‾3 = 2 n, it follows that n = - 1 next © 2007 Herbert I. Gross

4. Among other things, this exercise is an example in which we have to use the arithmetic of signed numbers. That is, we had to know that 2 + - 3 = - 1. next Note 1a If we didn't remember the rule for multiplying like bases, we could have reverted to the basic definitions. next © 2007 Herbert I. Gross

5. next Note 1a For example, knowing that 2 -3 is the reciprocal of 2 3, we could have rewritten next © 2007 Herbert I. Gross as… 2 2 × 1 2 3 2 2 × 2 -3

6. Knowing that 2 2 = 2 × 2 and 2 3 = 2 × 2 × 2, we may rewrite 2 2 × 2 -3 in the form… next Note 1a and if we reduce it to lowest terms we see that… next © 2007 Herbert I. Gross 1212 = 2 2 × 2 -3 1212 = 2 × 2 2 × 2 × 2 next = 2 -1

7. One might argue that it might be easier just to convert the numbers into place value notation and then use “ordinary” arithmetic to answer the question. For example, we know that 2 × 2 = 4 and 2 × 2 × 2 = 8; and therefore… next Note 1a © 2007 Herbert I. Gross 2 2 × 2 -3 4848 = 2 × 2 2 × 2 × 2 = 1212 = = 2 -1

8. However when we are using exponents in our study of algebra and we are given an expression such x 2 (x -3 ) we have no choice but to use the properties of the arithmetic of exponents. next Note 1a © 2007 Herbert I. Gross

9. next Problem #1b © 2007 Herbert I. Gross Solve for the value of n that makes the equation a true statement. 2 2 ÷ 2 ‾3 = 2 n next Answer: n = 5

10. Solution: Keep in mind that our definition of negative exponents enables us to use the property b n ÷ b m = b n – m even when n and/or m is a negative number. Therefore… 2 2 ÷ 2 ‾3 = 2 2 – ‾3 = 2 2 + 3 = 2 5. Hence, if 2 2 ÷ 2 ‾3 = 2 n, it follows that n = 5 next © 2007 Herbert I. Gross

11. In this exercise we see the importance of knowing how to subtract signed numbers. That is, our solution required that we knew that 2 – - 3 = 2 + + 3 = 5. next Note 1b © 2007 Herbert I. Gross 22 22 ÷ 2 ‾3 means 4 ÷ 1/8 1/8 or 4 × 8 32 or 25.25. In terms of size, the result means that 22 22 is 32 times as great as 2 -3. next

12. However when we have to work with an expression such as… next Note 1b © 2007 Herbert I. Gross next x 2 + 3 there is little alternative but to rewrite the expression as… and then use the fact that… x 2 x 3 x 2 ÷ x -3 next

13. Problem #1c © 2007 Herbert I. Gross 2 4 × 3 4 = (2 × 3) n next Answer: n = 4 Solve for the value of n that makes the equation a true statement.

14. Answer: n = 4 Solution: In this case we use the property b n × c n = (b × c) n with b = 2, c = 3, and n = 4. next © 2007 Herbert I. Gross 2 4 × 3 4 = (2 × 3) 4 b n × c n = (b × c) n Therefore if 2 4 × 3 4 = (2 × 3) n, then n = 4 next

15. Note 1c © 2007 Herbert I. Gross A main purpose of this exercise is to show the inherent danger of what happens when memorization fails us. Namely in this case we are multiplying different bases that have the same exponent. This is sort of the “reverse” of multiplying like bases that have different exponents. In the latter case we keep the common base and add the exponents. In the former case we multiply the two bases and keep the common exponent. If all we do is memorize the rules it is possible that we may confuse one rule with another.

16. The “cure” for the above problem is to revert once again to our basic definitions. For example, we may rewrite… next Note 1c © 2007 Herbert I. Gross 2 × 2 × 2 × 2 2 4 × 3 4 next 3 × 3 × 3 × 3× in the form…

17. And because multiplication is both associative and commutative we may rearrange the terms and rewrite… next Note 1c © 2007 Herbert I. Gross 2 × 32 × 3 next (2 × 2 × 2 × 2)(3 × 3 × 3 × 3)× as… 2 × 32 × 32 × 32 × 32 × 32 × 32 × 32 × 3× next ×× ( ) 2 × 32 × 32 × 32 × 32 × 32 × 3 1234

18. (5 -1 ) 0 = n Problem #2a © 2007 Herbert I. Gross Solve for the value of n that makes each equation a true statement. next Answer: n = 1

19. Solution: Under the condition that we wanted the properties for exponent arithmetic to remain valid when the exponent was 0, we had to define b 0 to equal 1 for every non- zero number b. Therefore, if we let b = (5 -1 ), we see that (5 -1 ) 0 = 1 next © 2007 Herbert I. Gross

20. next Note 2a © 2007 Herbert I. Gross There is a tendency for students to want b 0 to equal 0. However if this were true, the property 2 3 × 2 0 = 2 3+0 would not be correct because 2 3 × 2 0 would be 2 3 × 0, or 0; while 2 3+0 would be 2 3 or 8. While it might seem a bit like nit picking, for the sake of completeness we should point out that 0 0 is indeterminate. To see why, look at an expression such as 0 3 × 0 0. Since 0 3 = 0, 0 3 × 0 0 will equal 0 regardless of the value we assign to 0 0. So just as 0 ÷ 0 is indeterminate, so also is 0 0. next

21. Problem #2b © 2007 Herbert I. Gross Solve for the value of n that makes each equation a true statement. (5 -1 ) -3 = 5 n next Answer: n = 3

22. Solution: By our definition of negative exponents, the property (b m ) n remains valid even if the exponents are negative. Hence… (5 -1 ) -3 = 5 -1×-3 = 5 3 Therefore if (5 -1 ) -3 = 5 n, then n=3 next © 2007 Herbert I. Gross

23. next Note 2b © 2007 Herbert I. Gross This exercise affords us a practical application for seeing why the product of two negative numbers is positive. As a check we can rewrite 5 -1 as 1 / 5. next -- Therefore we may rewrite (5 -1 ) -3 as ( 1 / 5 ) -3. -- In turn, we may rewrite ( 1 / 5 ) -3 as 1 ÷ ( 1 / 5 ) 3 or 5 3. -- Therefore, (5 -1 ) -3 = 5 3.

24. Note 2b © 2007 Herbert I. Gross However, notice how much less cumbersome it is if we know the rule for raising a power to a power. If we wanted to do the arithmetic using fractions and whole numbers, we could rewrite 5 -1 as 1 / 5. next (5 -1 ) -3 = next 1 1 / 5 × 1 / 5 × 1 / 5 1(1/5)31(1/5)3 = 1(1/53)1(1/53) = = 1 ÷ 1 / 5 3 = 5 3 = 125

25. next Note 2b © 2007 Herbert I. Gross However, in algebra if we were given an expression such as (x -1 ) -3 our only choice would be to use the rule for raising a power to a power and thus obtaining x 3 as our answer.

26. next 638.97 × 10 -4 = n Problem #3a © 2007 Herbert I. Gross Solve for the value of n that makes each equation a true statement. next Answer: n = 0.063897

27. Solution: By our definition of negative exponents,10 -4 is the reciprocal of 10 4. In other words, multiplying by 10 -4 is the same as dividing by 10 4. Every time we divide by 10 we move the decimal point 1 place to the left. next © 2007 Herbert I. Gross

28. Answer: n = 0.063897 Solution: Hence to divide by 10 4 we move the decimal point 4 places to the left. That is… next © 2007 Herbert I. Gross 638.97 × 10 -4 = 0 0 6 3 8 9 7..... In summary, if 638.97 × 10 -4 = n, then n = 0.063897 next

29. Problem #3b © 2007 Herbert I. Gross Solve for the value of n that makes each equation a true statement. next 0.000063897 = 6.3897 × 10 n Answer: n = - 5

30. Solution: To get from 0.000063897 to 6.3897 we have to move the decimal point 5 places to the right. That is, we have to multiply by 10 5. However we don't want the value of 0.000063897 to change. In other words, to “undo” multiplying by 10 5, we have to divide by 10 5 (or, equivalently to multiply by 10 -5 ). next © 2007 Herbert I. Gross

31. Answer: n = - 5 Solution: Therefore we rewrite 0.000063897 as… next © 2007 Herbert I. Gross (0.000063897 × 10 5 ) × 10 -5 = 6.3897 × 10 -5 In summary if 0.000063897 = 6.3897 × 10 n then n = - 5. next

32. Note 3 © 2007 Herbert I. Gross In describing very large or very small numbers, we make use of the integral powers of 10. Thus for example, instead of writing 2,000,000,000,000,000 we might write 2 × 10 15 ; and instead of writing 0.000000000000002 we might write 2 × 10 -15.

33. next Note 3 © 2007 Herbert I. Gross In our next lesson we will talk about scientific notation. To write a number in scientific notation, we write it in the form of an integral power of 10 multiplied by a number that is less than 10 but greater than or equal to 1.

34. next Note 3 © 2007 Herbert I. Gross For example, when we write 2,000,000,000,000,000, as 2 × 10 15 we are using scientific notation. Of course, there are other ways to express 2,000,000,000,000,000, as a whole number times a power of ten. For example, some other ways include 200 × 10 15, 2,000 × 10 14, but these two forms are not examples of scientific notation because neither 200 nor 2,000 is between 1 and 10. In the next lesson we will discuss why scientific notation is so helpful. next

35. 5 × 10 4 × 3 × 10 -6 × 4 × 10 -1 Problem #4a © 2007 Herbert I. Gross Write the following in decimal form. next Answer: n = 0.06

36. Answer: 0.06 Solution: Since the given expression involves only multiplication and since multiplication is both commutative and associative, we may regroup the factors and rewrite… next © 2007 Herbert I. Gross (5 × 3 × 4) × (10 4 × 10 -6 × 10 -1 )5 × 10 4 × 3 × 10 -6 × 4 × 10 -1 as… next

37. Solution: next © 2007 Herbert I. Gross next 60 × 10 -3 60 (10 4 + -6 + -1 )(5 × 3 × 4) × (10 4 × 10 -6 × 10 -1 ) ×

38. Solution: To multiply 60 by 10 -3, we move the decimal point three places to the left to obtain… next © 2007 Herbert I. Gross next 60 × 10 -3 6 000.... 123 next Therefore in place value notation… 5 × 10 4 × 3 × 10 -6 × 4 × 10 -1 = 0.06 next

39. Problem #4b © 2007 Herbert I. Gross Write the following in decimal form. next 5 × 10 4 × 3 × 10 -6 + 4 × 10 -1 Answer: n = 0.55

40. Answer: 0.55 Solution: At first glance Exercise 4b can be easily confused with Exercise 4a. The only difference is that in part (b) one of the time signs is replaced by a plus sign; and since plus signs separate terms we have to read… next © 2007 Herbert I. Gross (5 × 10 4 × 3 × 10 -6 ) + (4 × 10 -1 ) 5 × 10 4 × 3 × 10 -6 + 4 × 10 -1 as… next

41. Solution: To rewrite the expression below… next © 2007 Herbert I. Gross notice that… next 0.4 0.15 15 × 10 -2 15 × 10 4 + -6 (5 × 3) × (10 4 × 10 -6 ) (5 × 10 4 × 3 × 10 -6 ) + (4 × 10 -1 ) and… + = 0.55 Hence, we may rewrite it as 0.15 + 0.4 or 0.55. next

42. Note 4 © 2007 Herbert I. Gross When we use scientific notation we are often doing arithmetic with numbers that involve powers of 10. In such cases we often rewrite products so that the powers of 10 are grouped together. At the same time it is important for us to keep track of the order of operations. For example, it was important in part (b) to observe that there was a plus sign that separated one group of times signs from the other. next

43. Important Note 4 © 2007 Herbert I. Gross For example, while it's nice that we can rewrite 3 10 × 3 20 as 3 30, there is no simple way to rewrite 3 10 + 3 20.

44. next Suppose the cost of your buying groceries doubles every 10 years. If a particular grocery order costs $240 today, how much did the same order cost 30 years ago? Problem #5 © 2007 Herbert I. Gross next Answer: $30

45. Solution: If the cost doubles every ten years then ten years earlier the cost was half of the present cost. Hence, we can show by a chart such as the one shown below that 30 years ago the cost was $30. More specifically… next © 2007 Herbert I. Gross Present Cost Cost 10 years ago Cost 20 years ago Cost 30 years ago $240 $120 $60 $30 next

46. Note 5 © 2007 Herbert I. Gross In this exercise it is not very cumbersome to make a chart. However, this isn't always the case. For example, if we wanted to know what the cost would have been 100 years ago we would have had to add 7 more rows to our chart.

47. next Note 5 © 2007 Herbert I. Gross In such an event it's often more helpful to write the relationship in the form of a formula. In the present example, we could let d denote the number of decades ago that we wanted to know what the cost would have been (we use decades instead of years because the doubling takes place every ten years); we could use P to denote the present cost in dollars and we could use V to denote the cost in dollars t decades ago. In that case the formula would be… P = V × 2 d next

48. Note 5 © 2007 Herbert I. Gross To check that the formula P = V × 2 d is correct, we know from our chart that when d = 3 (that is, 30 years ago), V = 30 and P= 240; and in this case the formula is the true statement 240 = 30 × 2 3 ( = 30 × 8).

49. next Note 5 © 2007 Herbert I. Gross Since we know the value of P and want to find the value of V, it might be more convenient to undo formula V = P × 2 d by dividing both sides of the formula by 2 d (or equivalently by multiplying both sides by 2 -d ) to obtain the equivalent formula… V = P × 2 -d next

50. Note 5 © 2007 Herbert I. Gross and in the case of the present exercise, the formula V = P × 2 -d would become… V = 240 × 2 -3 ( = 240 ÷ 80) = 30 next More importantly, the formula above allows us to compute the cost any number of decades ago. For example, if it was 10 decades ago the cost would have been… V = 240 × 2 -10

51. next Note 5 © 2007 Herbert I. Gross And using a calculator we would enter the following sequence of key strokes… next and the answer would appear as 0.234375; that is, about 24 cents. 2xyxy 10 +/-+/- =×240= next

52. Note 5 © 2007 Herbert I. Gross This illustrates the power of exponential growth. In other words if you somehow invested 24¢ a hundred years ago and your investment doubled every 10 years, the present value would be about $240.

53. next Note 5 © 2007 Herbert I. Gross An important scientific application of this idea is when we talk about the half-life of a radioactive substance. The half-life is the amount of time it takes for the substance to lose half of its present size. Thus if the half life of a 240 gram radioactive substance was 200 years, 200 years from now its mass would be 120 grams; and 200 years after that its mass would be 60 grams; and 200 years after that its mass would be 30 grams and so on.