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

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Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 7 part 1 By Herb I. Gross and Richard A. Medeiros next."— Presentation transcript:

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

2 Find a value of n for which 7 2n + 1 = 7 5. Problem #1 © 2007 Herbert I. Gross Answer: n = 2 next

3 Answer: 2 Solution: Clearly for any base, b, and any exponent, m, it's clear that b m = b m. Hence, one solution of the equation 7 2n+1 = 7 5 occurs if… 2n +1 = 5. The only value of n that satisfies that equation is n = 2. next © 2007 Herbert I. Gross

4 In some “programs” more than one input can produce the same output. For example, if we square 3 the answer is 9. However if the square of a number is 9, the number doesn't have to be 3 (it can be - 3). In more mathematical language… next Note 1 If x = 3 then x 2 = 9. However, if x 2 = 9, then either x = 3 or x = - 3. next © 2007 Herbert I. Gross

5 More generally in using an “undoing” program it's possible that undoing a step results in defining more than one input. Thus a very natural inquiry might be something like… next Note 1 “We know that if n = 2, then 7 2n+1 = 7 5. However, if we know that 7 2n+1 = 7 5, can we be sure that n has to be 2?” next © 2007 Herbert I. Gross

6 Rather than to give a formal mathematical proof that if m ≠ n, then 7 m ≠ 7 n, we can think in terms computing compound interest. Since the investment earns interest on the interest, the longer the money is invested the faster it grows. In other words if m < n then 7 m < 7 n and if m > n then 7 m > 7 n. next Note 1 next © 2007 Herbert I. Gross

7 next Find a value of n for which 8 15 = 2 n. Problem #2 © 2007 Herbert I. Gross Answer: n = 45 next

8 Answer: 45 Solution: The solution here involves a slight modification of what we did in Exercise 1. Namely in this exercise the two bases are different. So the key here is to recognize that 8 = 2 3. Therefore, starting with… next 8 15 = 2 n © 2007 Herbert I. Gross

9 Solution: We may replace 8 by 2 3 in the equation to obtain… next and because (2 3 ) 15 = = 2 n (2 3 ) 15 = 2 n 8 15 = 2 n From our discussion in Exercise 1, we see from the equation that the value of n has to be 45. n = 45 next © 2007 Herbert I. Gross

10 Notice that we multiplied the exponents when we raised a power to a power. However, when we multiply like bases we add the exponents. next Note 2 next If you're not sure about what to do, an excellent strategy is to go back to the basic definition. © 2007 Herbert I. Gross

11 For example, if you're not sure whether you multiply or whether you add the exponents in a problem such as 2 3 × 2 5, rewrite 2 3 as 2 × 2 × 2 and 2 5 as 2 × 2 × 2 × 2 × 2. next Note 2 next In this way … 2 3 × 2 5 becomes… © 2007 Herbert I. Gross 2 × 2 × 2× 2 × 2 × 2 × 2 × 2 next = 2 8

12 For what value of n does 2 15 × 2 13 × 2 11 = 2 n ? Problem #3 Answer: n = 39 next © 2007 Herbert I. Gross

13 Answer: 39 Solution: The property that we add the exponents when we multiply like bases was proven only in the case in which there were two factors. However, we can extend this result by use of the associative property of multiplication. More specifically, we may rewrite… next 2 15 × 2 13 × 2 11 © 2007 Herbert I. Gross as… ( ) next

14 Solution: We know that 2 15 × 2 13 = If we replace 2 15 × 2 13 by 2 28 the expression next © 2007 Herbert I. Gross becomes… × 2 11 (2 15 × 2 13 ) × 2 11 next which in turn becomes…

15 We actually used a generalization of this result in the previous exercise when we replaced (2 3 ) 5 by next Note 3 next (2 3 ) 5 © 2007 Herbert I. Gross = 2 15 = = 2 3 × 2 3 × 2 3 × 2 3 × 2 3 That is…

16 Again keep in mind that we can always revert back to our basic definition. For example, if we aren't sure that (2 3 ) 5 = 2 15, we may rewrite… next Note 3 © 2007 Herbert I. Gross 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 next 2 3 × 2 3 × 2 3 × 2 3 × 2 3 as…

17 and then count the number of times 2 appears as a factor… next Note 3 © 2007 Herbert I. Gross next 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × × 2 3 × 2 3 × 2 3 × There are 15 factors of 2. Hence the expression (2 3 ) 5 is equal to 2 15.

18 next Write (3 × 10) 3 + (4 × ) × as a place value numeral. Problem #4 Answer: n = 31,035 next © 2007 Herbert I. Gross

19 Answer: 39 Solution: Knowing that (b × c) n = b n × c n, we may rewrite (3 × 10) 3 as 3 3 × next © 2007 Herbert I. Gross next 27, × 1,000 3 × 3 × 3 × 10 × 10 × × 10 3

20 Solution: Therefore we may rewrite… next © 2007 Herbert I. Gross next 27,000 + (4 × ) × (3 × 10) 3 + (4 × ) × as… From our agreement concerning the order of operations (PEMDAS), we may rewrite (4 × ) as (4 × 100) + 3 which equals 403.

21 Solution: Therefore we may rewrite the expression as… next © 2007 Herbert I. Gross next 27, , ,000 + (4 × ) × Again using PEMDAS we may rewrite the expression… 27,000 + (403 × 10) , × ,000 + (4 × ) × = 31,305

22 next For what value of n is it true that … 8 n ÷ 2 5 ÷ 4 3 = 2? Problem #5 Answer: n = 4 next © 2007 Herbert I. Gross

23 Answer: n = 4 Solution: Since this exercise involves only division it would be nice to convert all numbers to the same base. So we may rewrite 8 as 2 3 and 4 as 2 2, in which case, the equation next (2 3 ) n ÷ 2 5 ÷ (2 2 ) 3 = 2 © 2007 Herbert I. Gross next may be written as… 8 n ÷ 2 5 ÷ 4 3 = 2

24 Solution: Since (2 3 ) n = 2 3n, (2 2 ) 3 = 2 6 and 2 = 2 1 we may rewrite the equation next © 2007 Herbert I. Gross next Using PEMDAS we may rewrite the equation as… (2 3n ÷ 2 5 ) ÷ 2 6 = n ÷ 2 5 ÷ 2 6 = 2 1 (2 3n ÷ 2 5 ) ÷ (2 2 ) 3 = 2 as…

25 next © 2007 Herbert I. Gross next Again using our rule for dividing like bases we may rewrite the equation as… or… 2 3n – 11 = 2 1 Solution: and by our rule for dividing like bases, the equation may be rewritten as… 2 3n – 5 – 6 = n – 5 ÷ 2 6 = 2 1 (2 3n ÷ 2 5 ) ÷ 2 6 = 2 1 next

26 Solution: Since the bases in the equation are the same, the exponents must be equal. next © 2007 Herbert I. Gross n = 4 3n = 12 3n – 11 = 1 2 3n – 11 = 2 1 Hence… next

27 As a check we replace n by 4 in… next Note 5 next 8 n ÷ 2 5 ÷ 4 3 © 2007 Herbert I. Gross 4 and verify that… = 2 We can use our rules for exponents to show that the equation is a true statement, but if we prefer, we may revert once again to our basic definitions and perform the calculations. next

28 For example (and it's okay to use a calculator if you wish)… next Note 5 next © 2007 Herbert I. Gross 8 4 ÷ 2 5 ÷ 4 3 =2 8 4 = 8 × 8 × 8 × 8 = 4, = 2 × 2 × 2 × 2 × 2 = = 4 × 4 × 4 = 64

29 Hence, we may rewrite the expression as… next Note 5 next © 2007 Herbert I. Gross 128 ÷ 64 and doing the divisions from left to right the expression becomes… = 2 4,096 ÷ 32 ÷ 64 () 8 4 ÷ 2 5 ÷ 4 3 = 2

30 If we are careless, we might look at 8 n ÷ 2 5 ÷ 4 3 and view it as if the grouping was 8 n ÷ (2 5 ÷ 4 3 ). However when there are no grouping symbols and only division is involved, PEMDAS requires us to perform the divisions from left to right. In other words we have to read 8 n ÷ 2 5 ÷ 4 3 as if it were written as (8 n ÷ 2 5 ) ÷ 4 3. next Note 5 next © 2007 Herbert I. Gross


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