Key Stone Problem…Key Stone Problem…
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Set 7 Part 2© 2007 Herbert I. Gross
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
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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(a) For what value of n is it
true that … 161/2 = 2n
Keystone Illustration for Lesson 7 Part 2
Answer: n = 2© 2007 Herbert I. Gross
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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.
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© 2007 Herbert I. Gross
Since the positive square root of 16 is 4, 161/2 = 4
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However, the problem wants the answer in the form 2 n; and since 4 = 22, we may
rewrite the equation in the equivalentform 161/2 = 22
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…
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© 2007 Herbert I. Gross
161/2 = (42)1/2
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= 42 × 1/2
= 41
= 4
= 22
Alternative Solution:
Using the result of (4) we may rewrite…
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© 2007 Herbert I. Gross
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2 = n
22 = 2n
4 = 2n161/2 = 2n
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© 2007 Herbert I. Gross
Special Note
• 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.
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© 2007 Herbert I. Gross
Special Note
• 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 thecase of exponents this is true. In other words, if b represents any base, and if
bn = bm , then n must be equal to m.
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© 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 thinkin terms of compound interest. Namely if
the rate of interest doesn't change the value of the investment continually
increases with time.
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© 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.
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© 2007 Herbert I. Gross
Note
• For example, starting with the propertybm × bn = bm+n
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we see that if we let b = 16, and m and n both equal = 1/2, the property becomes…
b m × b n =16 161/2 1/2
16 1/2 + 1/2 = 16 1 = 16
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© 2007 Herbert I. Gross
Special Note
• 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.
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© 2007 Herbert I. Gross
Special Note
• 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 1 = 16
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© 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 ofarithmetic 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.
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© 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
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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
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© 2007 Herbert I. Gross
Note
• 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 =
and we will see that 2 appears in the answer display.
2
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(b) For what value of n is it true that …
161/2 × 2-3 = 2n
Keystone Illustration for Lesson 7 Part 2
Answer: n = -1© 2007 Herbert I. Gross
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Solution for Part b:
From part (a) we already know that
161/2 = 4 = 22
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© 2007 Herbert I. Gross
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in the form…
22 × 2-3 = 2n
Hence, we may rewrite the equation…
161/2 × 2-3 = 2n
Solution for Part b:
We may now use our extended rules for the arithmetic of exponents to rewrite 22 × 2-3 as…
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© 2007 Herbert I. Gross
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If we now replace 22 × 2-3 by its value, we obtain…
2-1
22 + -322 × 2-3
And we see that… n = -1
= 2n
An Alternative Solution for Part b:
We could also have used the facts that…
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© 2007 Herbert I. Gross
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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
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© 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 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.
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© 2007 Herbert I. Gross
Note
• 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.
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© 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.
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(c) For what value of n is it true that …
161/2 × 2-3 ÷ 2-6 = 2n
Keystone Illustration for Lesson 7 Part 2
Answer: n = 5© 2007 Herbert I. Gross
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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.
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© 2007 Herbert I. Gross
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.
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© 2007 Herbert I. Gross
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…
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© 2007 Herbert I. Gross
(
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161/2 × 2-3 ÷ 2-6)
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.
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© 2007 Herbert I. Gross
(
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22
)
Therefore, we may rewrite the expression in the equivalent form…
× 2-3 × 26(161/2 × 2-3 ) ÷ 2-6
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Solution for Part c:
Since 22 × 2-3 = 22 + -3 = 2-1, we may rewrite the expression as…
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© 2007 Herbert I. Gross
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which is equal to…
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2-1+ 6
2-1 × 26(22 × 2-3 ) ÷ 2-6
= 25
Solution for Part c:
In summary,
(161/2 × 2-3 ) ÷ 2-6 = 25
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© 2007 Herbert I. Gross
If we now replace (161/2 × 2-3) ÷ 2-6 = 2n
by its value above, we see that 25 = 2n; so
n = 5.
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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 = 1/2
6 = 1/64.
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© 2007 Herbert I. Gross
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Therefore we may rewrite…
(4 × 1/8) ÷ 1/64(161/2 × 2-3 ) ÷ 2-6
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...
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© 2007 Herbert I. Gross
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as…
(1/2) × 64(4 × 1/8) ÷ 1/64
If we remember that 32 = 25, we see that n must be 5.
= 32
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© 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.
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© 2007 Herbert I. Gross
NoteMultiplying 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.
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© 2007 Herbert I. Gross
NoteMultiplying Like
Bases
• For example, we saw that 23 × 24 = 23+4. 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.
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© 2007 Herbert I. Gross
NoteMultiplying Like
Bases
• For example, using the definition of exponents we may write 23 × 24 × 25 in the form…
(2 × 2 × 2) × (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
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and verify that we have (3 + 4 + 5) factors of 2.
3 4 5
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© 2007 Herbert I. Gross
NoteMultiplying 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…
22 + -3 +6 = 25
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(d) For what value of n is it true that …
161/2 × 2-3 ÷ 2-6 + 23 = 2n
Keystone Illustration for Lesson 7 Part 2
Answer: n = 40© 2007 Herbert I. Gross
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Solution for Part d:
Using PEMDAS, we raise to powers, multiply and divide before we add. Therefore, we may rewrite…
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© 2007 Herbert I. Gross
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as …
32 + 8161/2 × 2-3 ÷ 2-6 + 23
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…
= 40( )
In summary, if 161/2 × 2-3 ÷ 2-6 + 23 = 2n, then n = 40.
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© 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.
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