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Key Stone Problem… Key Stone Problem… next Set 7 Part 2 © 2007 Herbert I. Gross.

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Presentation on theme: "Key Stone Problem… Key Stone Problem… next Set 7 Part 2 © 2007 Herbert I. Gross."— Presentation transcript:

1 Key Stone Problem… Key Stone Problem… next Set 7 Part 2 © 2007 Herbert I. Gross

2 You will soon be assigned five problems to test whether you have internalized the material in Lesson 7 part 2 of our algebra course. The Keystone Illustration below is a prototype of the problems you'll be doing. Work out the problem 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 the problem. Instructions for the Keystone Problem 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 (a) For what value of n is it true that … 16 1/2 = 2 n Keystone Illustration for Lesson 7 Part 2 Answer: n = 2 © 2007 Herbert I. Gross next

5 Solution for Part a: We saw in the lesson that raising a number to the one-half power means the same thing as taking the (positive) square root of the number. next © 2007 Herbert I. Gross Since the positive square root of 16 is 4, 16 1/2 = 4 next However, the problem wants the answer in the form 2 n ; and since 4 = 2 2, we may rewrite the equation in the equivalent form 16 1/2 = 2 2

6 Alternative Solution: Even though there may be only one correct answer. There are usually many paths that lead to it. For example, one might recognize that 16 = 4 2, whereupon… next © 2007 Herbert I. Gross 16 1/2 = (4 2 ) 1/2 next = 4 2 × 1/2 = 4 1 = 4 = 2 2

7 Alternative Solution: Using the result of (4) we may rewrite… next © 2007 Herbert I. Gross next 2 = n 2 2 = 2 n 4 = 2 n 16 1/2 = 2 n

8 © 2007 Herbert I. Gross Special Note The fact, for example, that 5 2 = 25 does not mean that if x 2 = 25, x then must equal 5. In fact since ( - 5) 2 is also 25, it means that if x 2 = 25, then x can be either + 5 or - 5. What this means in terms of our input/output model is that for certain recipes it is possible for two different inputs to yield the same output.

9 next © 2007 Herbert I. Gross Special Note In deducing from the equation (4 = 2 n ) that n = 2, we assumed that there was only one value of n for which 2 n = 4. In the case of exponents this is true. In other words, if b represents any base, and if b n = b m, then n must be equal to m.

10 next © 2007 Herbert I. Gross Special Note While a rigorous proof of this is beyond the scope of this course, an intuitive way to see that this is plausible is to think in terms of compound interest. Namely if the rate of interest doesn't change the value of the investment continually increases with time.

11 next © 2007 Herbert I. Gross Note There is no need to memorize the fact that raising a number to the one-half power means the same thing as taking the (positive) square root of the number. Rather we can derive the result using our basic rules for the arithmetic of exponents.

12 next © 2007 Herbert I. Gross Note For example, starting with the property b m × b n = b m+n next we see that if we let b = 16, and m and n both equal = 1/2, the property becomes… b m × b n =16 1/2 16 1/2 + 1/2 = 16 1 = 16 next

13 © 2007 Herbert I. Gross Special Note We see from 16 1/2 × 16 1/2 = 16 1 = 16 that 16 1/2 is that (positive) number which when multiplied by itself is 16; that is, it is the square root of 16.

14 next © 2007 Herbert I. Gross Special Note We could also have looked at the property (b m ) n = b m×n with b = 16, m = 1 / 2 and n = 2 to obtain… ( b m ) n = 16 1/22 16 1/2 × 2 = 16 1 = 16 next

15 © 2007 Herbert I. Gross Note In summary, knowing the basic properties and wanting to ensure that we preserve them allows us to derive the rules of arithmetic for fractional and/or negative exponents. We can generalize the result of what happens when we have unit fractions (that is, fractions whose numerator is 1) as exponents. Namely, for any non-zero value of n, 1/n × n = 1. next

16 © 2007 Herbert I. Gross Special Note So, for example, if we let n = 5, we see that … ( b 1/5 ) n = 5 b 1/5 × 5 = b 1 = b next In other words b 1/5 is that number which when raised to the 5 th power is equal to b. This number is known as the fifth root of b and is written as b. 5

17 next © 2007 Herbert I. Gross Note So, for example, the fact that 2 5 = 32 means that 2 is the 5 fifth root of 32. One way to check this is by using the calculator and showing that 32 1/5 = 2. Since 1 / 5 = 0.2, we can use the following sequence of key strokes… 32xyxy 0.2= and we will see that 2 appears in the answer display. 2 next

18 (b) For what value of n is it true that … 16 1/2 × 2 -3 = 2 n Keystone Illustration for Lesson 7 Part 2 Answer: n = -1 © 2007 Herbert I. Gross next

19 Solution for Part b: From part (a) we already know that 16 1/2 = 4 = 2 2 next © 2007 Herbert I. Gross next in the form… 2 2 × 2 -3 = 2 n Hence, we may rewrite the equation… 16 1/2 × 2 -3 = 2 n

20 Solution for Part b: We may now use our extended rules for the arithmetic of exponents to rewrite 2 2 × 2 -3 as… next © 2007 Herbert I. Gross next If we now replace 2 2 × 2 -3 by its value, we obtain… 2 -1 2 2 + -3 2 2 × 2 -3 And we see that… n = -1 = 2 n

21 An Alternative Solution for Part b: We could also have used the facts that… next © 2007 Herbert I. Gross next 2 -3 = 1 ÷ 2 3 = 1 ÷ 8 = 1 / 8, and 16 1/2 = 4 to rewrite 16 1/2 × 2 -3 in the form 4 ÷ 8 = 1 / 2 and then use the fact that… 1 / 2 = 1 ÷ 2 = 1 ÷ 2 1 = 2 -1

22 © 2007 Herbert I. Gross Note There is a tendency for beginning students to think that 2 -1 is negative. Keep in mind that if n is any positive number 2 n is positive. Hence, its reciprocal is also positive. Since raising a number to a negative power means taking the reciprocal of that number it means that 2 raised to any negative power is a positive number.

23 next © 2007 Herbert I. Gross Note Notice that 2 0 = 1. Therefore, if n is less than 0 (that is, if n is negative) 2 n will be less than 1. Moreover as a number increases its reciprocal decreases. next

24 © 2007 Herbert I. Gross Note The above discussion is not limited to having 2 as the base. It applies to any positive number b. The key point is that if n is a very large positive number, b -n is a positive number whose value is close to 0. next

25 (c) For what value of n is it true that … 16 1/2 × 2 -3 ÷ 2 -6 = 2 n Keystone Illustration for Lesson 7 Part 2 Answer: n = 5 © 2007 Herbert I. Gross next

26 Prelude to Solution for Part c: In showing why 2 n × 2 -n = 2 0 = 1, there is a tendency to think of n as denoting a positive number. However n can be negative, in which case -n (that is, the opposite of n) is positive. The point is that n + -n = 0 regardless of whether or not n is positive, and 2 0 = 1. next © 2007 Herbert I. Gross

27 Prelude to Solution for Part c: More generally if b is any non-zero number and n is any number, then b n and b -n are reciprocals of one another. In particular, dividing a number by b -n means the same thing as multiplying the number by b n. next © 2007 Herbert I. Gross

28 Solution for Part c: The expression 16 1/2 × 2 -3 ÷ 2 -6 contains only the operations of multiplication and division. Hence, by our PEMDAS agreement we perform the arithmetic from left to right. In other words we may read it as if it were… next © 2007 Herbert I. Gross ( next 16 1/2 × 2 -3 ÷ 2 -6 )

29 Solution for Part c: In earlier parts of this problem, we showed that 16 1/2 = 2 2. And as explained in the above prelude dividing by 2 -6 is equivalent to multiplying by 2 6. next © 2007 Herbert I. Gross ( next 2 ) Therefore, we may rewrite the expression in the equivalent form… × 2 -3 × 2 6 (16 1/2 × 2 -3 ) ÷ 2 -6 next

30 Solution for Part c: Since 2 2 × 2 -3 = 2 2 + -3 = 2 -1, we may rewrite the expression as… next © 2007 Herbert I. Gross next which is equal to… 2 -1+ 6 2 -1 × 2 6 (2 2 × 2 -3 ) ÷ 2 -6 = 2 5

31 Solution for Part c: In summary, (16 1/2 × 2 -3 ) ÷ 2 -6 = 2 5 next © 2007 Herbert I. Gross If we now replace (16 1/2 × 2 -3 ) ÷ 2 -6 = 2 n by its value above, we see that 2 5 = 2 n ; so n = 5. next

32 An Alternative Solution for Part c: When in doubt we can always resort to the basic definitions. For example, we already know that 16 1/2 = 4, 2 -3 – 1 / 8, and 2 -6 = 1 / 2 6 = 1 / 64. next © 2007 Herbert I. Gross next Therefore we may rewrite… (4 × 1 / 8 ) ÷ 1 / 64 (16 1/2 × 2 -3 ) ÷ 2 -6

33 An Alternative Solution for Part c: 4 × 1 / 8 = 1 / 2 and dividing by 1 / 64 is the same as multiplying by 64. Therefore, we may rewrite the expression... next © 2007 Herbert I. Gross next as… ( 1 / 2 ) × 64(4 × 1 / 8 ) ÷ 1 / 64 If we remember that 32 = 2 5, we see that n must be 5. = 32

34 © 2007 Herbert I. Gross Note Of course the above approach may be tedious at times, but (1) it always works and (2) by doing things the long way a few times may lead to a better internalization of the properties of the arithmetic of exponents.

35 next © 2007 Herbert I. Gross Note Multiplying Like Bases The fact that multiplication is associative and commutative (meaning that we can group the factors in any way that we wish) allows us to generalize some of the rules for exponents. For example, in demonstrating how to multiply like bases, we used only examples that involved two factors. next

36 © 2007 Herbert I. Gross Note Multiplying Like Bases For example, we saw that 2 3 × 2 4 = 2 3+4. However we did not look at an example of the form 2 3 × 2 4 × 2 5. Notice that the approach we used in the case of showing that 2 3 × 2 4 = 2 3+4 works in exactly the same way when there are more factors. next

37 © 2007 Herbert I. Gross Note Multiplying Like Bases For example, using the definition of exponents we may write 2 3 × 2 4 × 2 5 in the form… (2 × 2 × 2) × (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) next and verify that we have (3 + 4 + 5) factors of 2. 345

38 © 2007 Herbert I. Gross Note Multiplying Like Bases Therefore, in doing Part (c) when we were faced with the expression… 2 2 × 2 -3 ÷ 2 -6 We could have rewritten this expression in one step as… 2 2 + -3 +6 = 2 5 next

39 (d) For what value of n is it true that … 16 1/2 × 2 -3 ÷ 2 -6 + 2 3 = 2 n Keystone Illustration for Lesson 7 Part 2 Answer: n = 40 © 2007 Herbert I. Gross next

40 Solution for Part d: Using PEMDAS, we raise to powers, multiply and divide before we add. Therefore, we may rewrite… next © 2007 Herbert I. Gross next as … 32 + 816 1/2 × 2 -3 ÷ 2 -6 + 2 3 We know that 2 3 = 8, and from part (c) we know that (16 1/2 × 2 -3 ÷ 2 -6 ) = 32. Therefore, we may rewrite the expression as… = 40 ( ) In summary, if 16 1/2 × 2 -3 ÷ 2 -6 + 2 3 = 2 n, then n = 40.

41 © 2007 Herbert I. Gross Note Our rules for exponents apply to multiplication, division, and raising to powers. There are no nice rules for addition and/or subtraction. Therefore, in problems such as part (d) it is very important to pay attention to plus and minus signs because they separate terms. next


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