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

2. Instructions for the Keystone Problem
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Instructions for the Keystone Problem
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.
© Herbert I. Gross

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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.
© Herbert I. Gross

4. (a) For what value of n is it true that … 161/2 = 2n Answer: n = 2
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Keystone Illustration for Lesson 7 Part 2
(a) For what value of n is it true that … 161/2 = 2n
Answer: n = 2
© Herbert I. Gross

5. Since the positive square root of 16 is 4,
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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.
Since the positive square root of 16 is 4,
161/2 = 4
However, the problem wants the answer in the form 2 n; and since 4 = 22, we may rewrite the equation in the equivalent
form 161/2 = 22
© Herbert I. Gross

6. Alternative Solution:
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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 = 42, whereupon…
161/2 = (42)1/2
= 42 × 1/2
= 41
= 4
= 22
© Herbert I. Gross

7. Using the result of (4) we may rewrite…
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Alternative Solution:
Using the result of (4) we may rewrite…
161/2 = 2n
4 = 2n
22 = 2n
2 = n
© Herbert I. Gross

8. Special Note
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• The fact, for example, that 52 = 25 does not mean that if x2 = 25, x then must equal 5. In fact since ( -5)2 is also 25, it means that if x2 = 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.
© Herbert I. Gross

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

10. Special Note
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• 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.
© Herbert I. Gross

11. Note
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• 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.
© Herbert I. Gross

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

13. Special Note
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• We see from 161/2 × 161/2 = 161 = 16 that 161/2 is that (positive) number which when multiplied by itself is 16; that is, it is the square root of 16.
© Herbert I. Gross

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

15. arithmetic for fractional and/or negative exponents.
Note
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• 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.
© Herbert I. Gross

16. • So, for example, if we let n = 5, we see that …
Special Note
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• So, for example, if we let n = 5, we see that …
( b 1/5 ) n = 5
b 1/5 × 5 = b
b 1 =
In other words b1/5 is that number which when raised to the 5th power is equal to b. This number is known as the fifth root of b and is written as √b . 5
© Herbert I. Gross

17. and we will see that 2 appears in the answer display.
Note
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• So, for example, the fact that 25 = 32 means that 2 is the 5 fifth root of 32. One way to check this is by using the calculator and showing that 321/5 = 2. Since 1/5 = 0.2, we can use the following sequence of key strokes… 3 2 xy
0.2 = 2
and we will see that 2 appears in the answer display.
© Herbert I. Gross

18. 161/2 × 2-3 = 2n (b) For what value of n is it true that …
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Keystone Illustration for Lesson 7 Part 2
(b) For what value of n is it true that …
161/2 × 2-3 = 2n
Answer: n = -1
© Herbert I. Gross 18

19. From part (a) we already know that
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Solution for Part b:
From part (a) we already know that
161/2 = 4 = 22
Hence, we may rewrite the equation…
161/2 × 2-3 = 2n
22 × 2-3 = 2n
in the form…
© Herbert I. Gross

20. Solution for Part b: 22 × 2-3 22 + -3 2-1 = 2n
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Solution for Part b:
We may now use our extended rules for the arithmetic of exponents to rewrite 22 × 2-3 as…
If we now replace 22 × 2-3 by its value, we obtain…
22 × 2-3
2-1
= 2n
And we see that… n = -1
© Herbert I. Gross

21. An Alternative Solution for Part b:
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An Alternative Solution for Part b:
We could also have used the facts that…
2-3 = 1 ÷ 23 = 1 ÷ 8 = 1/8 ,and 161/2 = 4
to rewrite 161/2 × 2-3 in the form 4 ÷ 8 = 1/2
and then use the fact that…
1/2 = 1 ÷ 2 = 1 ÷ 21 = 2-1
© Herbert I. Gross

22. Note
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• There is a tendency for beginning students to think that 2-1 is negative. Keep in mind that if n is any positive number 2n 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.
© Herbert I. Gross

23. Note
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• Notice that 20 = 1. Therefore, if n is less than 0 (that is, if n is negative) 2n will be less than 1. Moreover as a number increases its reciprocal decreases.
© Herbert I. Gross

24. Note
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• 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.
© Herbert I. Gross

25. 161/2 × 2-3 ÷ 2-6 = 2n (c) For what value of n is it true that …
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Keystone Illustration for Lesson 7 Part 2
(c) For what value of n is it true that …
161/2 × 2-3 ÷ 2-6 = 2n
Answer: n = 5
© Herbert I. Gross 25

26. Prelude to Solution for Part c:
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Prelude to Solution for Part c:
In showing why 2n × 2-n = 20 = 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 20 = 1.
© Herbert I. Gross

27. Prelude to Solution for Part c:
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Prelude to Solution for Part c:
More generally if b is any non-zero number and n is any number, then bn 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 bn.
© Herbert I. Gross

28. Solution for Part c: ) ( 161/2 × 2-3 ÷ 2-6
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Solution for Part c:
The expression 161/2 × 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… (
161/2 × ÷ 2-6 )
© Herbert I. Gross

29. Solution for Part c: (161/2 × 2-3 ) ÷ 2-6 22 × 2-3 × 26 ( )
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Solution for Part c:
In earlier parts of this problem, we showed that 161/2 = 22. And as explained in the above prelude dividing by 2-6 is equivalent to multiplying by 26.
Therefore, we may rewrite the expression in the equivalent form…
(161/2 × ) ÷ 2-6 22
× 2-3
× 26 ( )
© Herbert I. Gross

30. Solution for Part c: (22 × 2-3 ) ÷ 2-6 2-1 × 26 2-1+ 6 = 25
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Solution for Part c:
Since 22 × 2-3 = = 2-1, we may rewrite the expression as…
which is equal to…
(22 × ) ÷ 2-6
2-1 × 26
2-1+ 6
= 25
© Herbert I. Gross

31. Solution for Part c: (161/2 × 2-3 ) ÷ 2-6 = 25 In summary,
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Solution for Part c:
In summary,
(161/2 × ) ÷ = 25
If we now replace (161/2 × 2-3) ÷ 2-6 = 2n by its value above, we see that 25 = 2n; so n = 5.
© Herbert I. Gross

32. An Alternative Solution for Part c:
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An Alternative Solution for Part c:
When in doubt we can always resort to the basic definitions. For example, we already know that 161/2 = 4, 2-3 – 1/8, and 2-6 = 1/26 = 1/64.
Therefore we may rewrite…
(161/2 × ) ÷ 2-6
(4 × 1/8) ÷ 1/64
© Herbert I. Gross

33. If we remember that 32 = 25, we see that
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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...
as…
(4 × 1/8) ÷ 1/64
(1/2) × 64
= 32
If we remember that 32 = 25, we see that
n must be 5.
© Herbert I. Gross

34. Note
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• 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.
© Herbert I. Gross

35. Multiplying Like Bases
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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.
© Herbert I. Gross

36. Multiplying Like Bases
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Note
Multiplying Like Bases
• For example, we saw that 23 × 24 = However we did not look at an example of the form 23 × 24 × 25.
Notice that the approach we used in the case of showing that 23 × 24 = 23+4 works in exactly the same way when there are more factors.
© Herbert I. Gross

37. Multiplying Like Bases
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Multiplying Like Bases
• For example, using the definition of exponents we may write 23 × 24 × 25 in the form…
and verify that we have ( ) factors of 2.
(2 × 2 × 2) × (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) 3 4 5
© Herbert I. Gross

38. • Therefore, in doing Part (c) when we were faced with the expression…
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Note
Multiplying Like Bases
• Therefore, in doing Part (c) when we were faced with the expression…
22 × 2-3 ÷ 2-6
We could have rewritten this expression in one step as…
= 25
© Herbert I. Gross

39. 161/2 × 2-3 ÷ 2-6 + 23 = 2n (d) For what value of n is it true that …
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Keystone Illustration for Lesson 7 Part 2
(d) For what value of n is it true that …
161/2 × 2-3 ÷ = 2n
Answer: n = 40
© Herbert I. Gross 39

40. Solution for Part d: ( 161/2 × 2-3 ÷ 2-6 + 23 32 + 8 ) = 40
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Solution for Part d:
Using PEMDAS, we raise to powers, multiply and divide before we add. Therefore, we may rewrite…
We know that 23 = 8, and from part (c) we know that (161/2 × 2-3 ÷ 2-6) = 32. Therefore, we may rewrite the expression as…
as … (
161/2 × ÷ )
= 40
In summary, if 161/2 × 2-3 ÷ = 2n,
then n = 40.
© Herbert I. Gross

41. Note
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• 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.
© Herbert I. Gross