1. Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 6 By Herb I. Gross and Richard A. Medeiros next

2. Which number is greater and by how much 2 10 or 10 3 ? Problem #1 © 2007 Herbert I. Gross Answer: 2 10 by 24 next

3. 2 10 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2; Hence 2 10 exceeds 10 3 by… 1,024 – 1,000, or 24. 48163264128256512 1,024 next 10 3 = 10 × 10 × 10 100 1,000 Answer: 2 10 by 24 Solution: – 24 © 2007 Herbert I. Gross

4. Be careful not to confuse 2 10 with 2 ×10. 2 ×10 is the sum of 2 tens or, equivalently the sum of ten 2's. On the other hand 2 10 is the product of ten 2's. next © 2007 Herbert I. Gross Note 1 It is important to understand the structural difference between multiplying two numbers and raising a number to a power. For example  While 2 ×10 is less than 10 × 3, 2 10 is greater than 10 3. next

5. Every time we increase the exponent by 1 we multiply by the base one more time. Thus, for example, if we increase the exponent in 2 10 by 1, we multiply 2 10 by 2. next © 2007 Herbert I. Gross Note 1 On the other hand if we increase the exponent in 10 3 by 1, we multiply 10 3 by 10. next So while 2 10 is greater than 10 3, 2 11 is less than 10 4. Hence 2 11 = 2,048. Hence 10 4 = 10,000

6. next Knowing that 2 3 = 8, find the value of n for which 8 4 = 2 n. Problem #2 © 2007 Herbert I. Gross Answer: n = 12 next

7. Answer: n = 12 Solution: © 2007 Herbert I. Gross 8 4 = 8 × 8 × 8 × 8 (2 × 2 × 2) 8 4 = × × × Since 2 3 = 8 we may replace each 8 in the equation by (2 × 2 × 2) to obtain…

8. next Solution: © 2007 Herbert I. Gross (2 × 2 × 2) 8 4 = × × × Since multiplication is associative, we may omit the parentheses, And since 2 12 is an abbreviation for twelve factors of 2, we see that 8 4 = 2 12. In summary, if 8 4 = 2 n, 1 2 3 45 67 8 910 1112 and we can count the number of 2’s. 8 4 = 2 12 n next 12 then n = 12.

9. If you are comfortable thinking in terms of exponents, a quick way to arrive at the answer is to notice that every 8 contains 3 factors of 2. next © 2007 Herbert I. Gross Note 2 next 3 factors 1 time2 times3 times4 times Therefore, if we have 8 as a factor four times, we also have three factors of 2 four times, or 12 factors of 2. 8 4 = 8 × 8 × 8 × 8 (2 × 2 × 2) 8 4 = × × ×

10. Among other things this problem is a segue for our next lesson in which we will talk about the rules of arithmetic for exponents. In this problem we showed that (2 3 ) 4 = 2 12. This is a special case of the fact that when we raise a power to a power we multiply the exponents. next © 2007 Herbert I. Gross Note 2

11. next © 2007 Herbert I. Gross Note 2 For example… (5 2 ) 3 = 5 6 because we are multiplying three factors of 5 2. That is we are multiplying two factors of 5 three times; which means that we are multiplying six factors of 5. next Stated more generally, if m and n are positive whole numbers then… (b n ) m = b nm (5 2 ) 3 = 5 2 × 5 2 × 5 2 5 × 55 × 55 × 5= 5 6

12. © 2007 Herbert I. Gross Note 2 On a more subtle note we are using the fact that no two different powers of 2 can be equal. More specifically we see that if n is any positive whole number, as n increases 2 n also increases. In other words when we concluded from 2 n = 2 12 that n = 12, we were assuming that no two different powers of 2 could be equal. Stated a bit differently, if n is not equal to 12, 2 n will not equal 2 12

13. next Knowing than 2 + 8 = 10 whenever 2, 8, and 10 modify the same noun, for what value of n does 2(10 5 ) + 8(10) 5 = 10 n? Problem #3 © 2007 Herbert I. Gross Answer: n = 6 next

14. © 2007 Herbert I. Gross Answer: n = 6 Solution: We may think of 10 5 as being a noun. That is, just as we talk about 8 thousands we can talk about 8 “ten-to-the-third power” 's. More generally we may view the powers of 10 as nouns; in which case 5(10) n means 5 ×10 n or five “ten to the n's”.

15. next © 2007 Herbert I. Gross Solution: With this in mind, we know that since 2 + 8 = 10 whenever 2, 8, and 10 modify the same noun, it follows that… “10 5 ’s”10 = 10 6 (2 + 8) “10 5 ’s” 2 “10 5 ’s” + 8 “10 5 ’s”2(10) 5 + 8(10) 5

16. next © 2007 Herbert I. Gross 2 (10) 5 + 8 (10) 5 × 10 5 Solution: With this in mind, we know that since 2 + 8 = 10 whenever 2, 8, and 10 modify the same noun, it follows that… 28+(2 + 8)10 We know that if we multiply 5 factors of 10 by 1 more factor of 10 we will have 6 factors of 10. That is, 10 1 × 10 5 = 10 6 Therefore we may rewrite the expression as … 2(10) 5 + 8(10) 5 = 10 6

17. next © 2007 Herbert I. Gross 2 (10) 3 + 8 (10) 3 = 10 n Somehow or other, we seem to be more comfortable working with smaller powers of 10 than with the larger powers. For example if the problem had been… (1,000) 2,000 8,000 = 10,000 4 We might have rewritten it in the place value form… and seeing that if 2(10) 3 + 8(10) 3 = 10 n, then n would equal 4. next Note 3

18. next © 2007 Herbert I. Gross Based on the above note it is often a good idea when numbers are too great and/or the abstract symbolism is confusing to work with lesser numbers. For example, with respect to the above note, once we saw what happened when n was 3, we might have decided to replace 3 by n and go through the same procedure. Note 3

19. next © 2007 Herbert I. Gross More specifically, if we start with… and replace 3 by n and 4 by n + 1 (that is n + 1 is 1 more than n, just as 4 is 1 more than 3), we obtain the more general result that for any positive integer n… Note 3 2 (10) 3 + 8 (10) 3 = 10 4 n+1n+1nn

20. next © 2007 Herbert I. Gross Note that there is a big difference when we add numbers that have the same power and when we multiply them. For example, 10 3 + 10 3 [that is, 1(10 3 )+ 1(10 3 )] = 2(10 3 ); While 10 3 × 10 3 = 10 6 ; that is 10 3 × 10 3 = (10 × 10 × 10) × (10 × 10 × 10). Note 3

21. next © 2007 Herbert I. Gross The fact that 10 3 × 10 3 = 10 6 follows from the fact that when we multiply 3 factors of 10 by 3 more factors of 10, we have multiplied 3 + 3 or 6 factors of 10. That is, even though this was a multiplication problem we added the exponents. This will be discussed in greater detail in the next lesson. Note 3

22. next a. Between what two consecutive whole number values for n is it true that (1.14) n = 2? Problem #4 © 2007 Herbert I. Gross Answer: Between 5 and 6 next

23. © 2007 Herbert I. Gross Answer: Between 5 and 6 Solution: The most primitive way to solve this problem is to successively multiply 1.14 by itself until the product exceeds 2.

24. next © 2007 Herbert I. Gross (1.14) 3 = 1.14 × (1.14) 2 = 1.14 × 1.2996 = 1.481544 (1.14) 4 = 1.14 × (1.14) 3 = 1.14 × 1.481544 = 1.68896016 (1.14) 5 = 1.14 × (1.14) 4 = 1.14 × 1.68896016 = 1.9254145024 (1.14) 6 = 1.14 × (1.14) 5 = 1.14 × 1.9254145024 =2.194972623936 This shows us that if n = 5, (1.14) n is less than 2, and if n = 6, (1.14) n is greater than 2. Therefore the required value for n is less than 6 but greater than 5. next Solution: That is… (1.14) 2 = 1.14 × 1.14 = 1.2996

25. next © 2007 Herbert I. Gross It is tedious, even with a calculator to compute the value of (1.14) 6 ; that is… 1.14 × 1.14 × 1.14 × 1.14 × 1.14 × 1.14. Note 4a However if you have a calculator with a x y key, the following sequence of strokes gives the value rather quickly; namely… 1.14xyxy 6= next

26. © 2007 Herbert I. Gross While the calculator will quite easily compute the value of (1.14) 6, it cannot tell you when you should be performing this operation. In this context, problem 4 (b) provides a practical application. Note 4a

27. next b. To the nearest year how long will it take for $30,000 to double in value if it is invested at an interest rate of 14% compounded annually? Problem #4 © 2007 Herbert I. Gross Answer: 6 next

28. © 2007 Herbert I. Gross Answer: 6 Solution: Compound interest means that you earn interest on your interest. Therefore every year the value of $1 increases by a factor of 1.14.

29. next © 2007 Herbert I. Gross Answer: 6 Solution: So at the end of the first year the $1 is worth… 1.14 × At the end of the second year the $1 is worth … 1.14 × $1(1.14) 2 That is, at the end of the second year the $1 is worth… At the end of the third year the $1 is worth... 1.14 × 1.14 × 1.14(1.14) 3 In other words, at the end of the third year each dollar is worth… next

30. © 2007 Herbert I. Gross Solution: Continuing in this manner we see that… At the end of the fourth year the $1 is worth... 1.14 × 1.14 × 1.14 × 1.14 × $1 next At the end of the fifth year the $1 is worth... (1.14) 4 1.14 × 1.14 × 1.14 × 1.14 × 1.14(1.14) 5 More generally, at the end of the n years the $1 is worth... n

31. next © 2007 Herbert I. Gross Solution: Since in part (a) we showed that (1.14) 5 =1.9254145024, we see that at the end of the fifth year each $1, rounded off to the nearest cent, will be worth $1.93. Hence, it will take more than five years for the value of the $1 to double In a similar way, the fact that (1.14) 6 =2.194972623936 means that rounded off to the nearest cent at the end of the sixth year each dollar is worth $2.19.

32. next © 2007 Herbert I. Gross Solution: Hence, the value of each $1 doubles during the sixth year. Since we may view a $30,000 investment as 30,000 investments of $1 each, the entire $30,000 doubles in value in the same amount of time that it takes for $1 to double in value.

33. next © 2007 Herbert I. Gross Keep in mind that a rate of 14% means a rate of 14 per hundred. So with respect to money 14% means that there is an increase of 14 cents per hundred; that is, 14 cents per dollar (100 cents). Thus an increase of 14% is equivalent to saying that after a year each $1 is worth $1.14. Note 4b

34. next © 2007 Herbert I. Gross The result of Part (b) can be generalized as follows. Since in each year the investment increases by a factor of 1.14, in n years the original investment increases by a factor of (1.14) n. Hence, at the end of n years an original investment of $30,000 will be worth (1.14) n × $30,000. Thus, for example, using the calculator we would compute the value, in dollars, of the $30,000 investment at the end of, say, 30 years by the sequence of key strokes… Note 4b 1.14xyxy 30×30,000 next

35. © 2007 Herbert I. Gross The value of (1.14) 30, to the nearest hundredth, is 50.95. So roughly speaking at the end of 30 years the original investment of $30,000 would be worth approximately 51 × $30,000 or $1,530,000. Note 4b

36. next © 2007 Herbert I. Gross Note 4b The above note shows the advantage of having interest paid on the interest (or the disadvantage if you owe money on your credit cards and don't pay the balance promptly). For example, if the 14% interest rate per year was applied only to the original $30,000 it wouldn't be until sometime during the eighth year of the investment that the original amount would double. That is at the end of the 7th year each $1 would be worth $1.98, and at the end of the 8th year it would be worth $2.12 (that is, 7 × 14% = 98% and 8 × 14% = 112%.

37. next a. You invest $10,000 at an interest rate of 8% compounded annually. To the nearest dollar, how much will the investment be worth in 5 years? Problem #5 © 2007 Herbert I. Gross Answer: $14,693 next

38. © 2007 Herbert I. Gross Answer: $14,693 Solution: % means cents per 1 dollar. Therefore at the end of each year the current investment increases by a factor of 1.08. Hence, at the end of 5 years the current investment increases by a factor of (1.08) 5. (1.08) 5 = 1.4693280768. Therefore, after 5 years the value of the original investment is (1.08) 5 × $10,000 = $14,693.280768; or rounded off to the nearest dollar the value is $14,693.

39. next © 2007 Herbert I. Gross If we were to compute the value of (1.08) 5 by multiplying five factors of 1.08, and if each time we rounded the amount off to the nearest dollar, we would be building errors upon errors. So one of the nice things about a calculator is that we can compute the value of (1.08) 5 without having to round off until the very last step. By way of review we use the sequence of key strokes… Note 5a 1.08xyxy 5 next

40. © 2007 Herbert I. Gross Note 5a next The procedure for solving this problem can be generalized quite easily. Namely (1) instead of talking about an initial investment of $10,000 we talk about an investment of P dollars; (2) instead of talking about an interest rate of 8% compounded annually, we talk about an interest rate of r% compounded annually and (3) instead of talking about a time period of 5 years we talk about a time period of n years.

41. next © 2007 Herbert I. Gross Note 5a next Since when written as a fraction r% means r/100, the general expression becomes… 10,000 × ( 1.08 ) 5 P 1 + 8/100 n 1 + r/100

42. next © 2007 Herbert I. Gross Note 5a next A = P × (1 + r/100) n If we now let A represent the value of an investment of P dollars after n years at an interest rate of r% compounded annually we obtain the general formula…

43. next © 2007 Herbert I. Gross Note 5a next In our problem, we were given the values of P, r, and n and were asked to find the value of A. In particular, this resulted in the direct computation … A = P × (1 + r/100) n 10,000(1 + 8/100) 5 (1.08) 5 next

44. b. To the nearest dollar, how much money must you invest now at a 8% rate of interest compounded annually if you want the value of the investment to be $10,000 five years from now? Problem #5 © 2007 Herbert I. Gross next Answer: $6,806

45. next © 2007 Herbert I. Gross Answer: $6,806 Solution: In part (a) we saw that the value (A) after n years of P dollars invested at a rate of 8% compounded annually is given by… A = P × (1.08) 5 In this case we want the value of the investment to be $10,000 at the end of 5 years. So in the equation we replace A by 10,000 to obtain the indirect computation… 10,000 next

46. © 2007 Herbert I. Gross Solution: We may convert the equation into a direct computation if we divide both sides of the equation by (1.08) 5. Doing this we obtain… 10,000 So rounded off to the nearest dollar it means that in order to have $10,000 five years from now we need to invest $6,806 if the interest rate is 8% interest compounded annually. next = (1.08) 5 P× ÷= 1.4693280768 = P 6,805.831970337... 6 = P

47. next © 2007 Herbert I. Gross Note 5b next In part (b), however, we were told that A = 10,000 and this led to the indirect computation (algebra). 10,000 ÷ (1.08) 5 = P As a reinforcement to our comments in Lesson 1, notice that the formula A = P ×(1.08) 5 is neither arithmetic nor algebra. It is a relationship between A and P. In part (a) we were told that P = 10,000 and this led to the direct computation (arithmetic). A = 10,000 × (1.08) 5

48. next © 2007 Herbert I. Gross Note 5b Notice the importance of reading comprehension. Namely it's important to determine whether it's A or P that is represented by 10,000. In part (a) the situation was that we had $10,000 now and wanted to know how much it would be worth in 5 years. In part (b), however we wanted to know how much money we had to invest now in order to have $10,000 in 5 years.