Download Lesson 7 part 2 Solutions

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Algebra
Problems…
Solutions
Set 7 part 2
© 2007 Herbert I. Gross
By Herb I. Gross and Richard A. Medeiros
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Problem #1a
Solve for the value of n that makes
the equation a true statement.
2
2 ×
‾3
2
=
n
2
Answer: n = -1
© 2007 Herbert I. Gross
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Answer: n = -1
Solution:
We chose our definition of a negative
exponent in a way that ensured we could
still add exponents when we multiplied like
bases. Therefore…
22 × 2‾3 = 22+ ‾3 = 2‾1.
Hence, if 22 × 2‾3 = 2n, it follows that n = -1
© 2007 Herbert I. Gross
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Note 1a
• Among other things, this exercise is an
example in which we have to use the
arithmetic of signed numbers. That is, we
had to know that 2 + -3 = -1.
If we didn't remember the rule for
multiplying like bases, we could have
reverted to the basic definitions.
© 2007 Herbert I. Gross
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Note 1a
For example, knowing that 2-3 is the
reciprocal of 23 , we could have rewritten
as…
2
2 ×
© 2007 Herbert I. Gross
1
-3
23
2
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Note 1a
• Knowing that 2
= 2 × 2 and 23 = 2 × 2 × 2,
we may rewrite 22 × 2-3 in the form… and if
we reduce it to lowest terms we see that…
2
2
×
2
1
2
-3
2 ×2 =
2×2×2
2
= 2-1
© 2007 Herbert I. Gross
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Note 1a
• One might argue that it might be easier
just to convert the numbers into place value
notation and then use “ordinary” arithmetic
to answer the question. For example, we
know that 2 × 2 = 4 and 2 × 2 × 2 = 8; and
therefore…
2
2 ×
© 2007 Herbert I. Gross
-3
2
2
×
2
1
4
-1
=
2
=
= =
2×2×2 8 2
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Note 1a
• However when we are using exponents
in our study of algebra and we are given
an expression such x2(x-3) we have no
choice but to use the properties of the
arithmetic of exponents.
© 2007 Herbert I. Gross
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Problem #1b
Solve for the value of n that makes
the equation a true statement.
2
2 ÷
‾3
2
=
n
2
Answer: n = 5
© 2007 Herbert I. Gross
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Answer: n = 5
Solution:
Keep in mind that our definition of negative
exponents enables us to use the property
bn÷ bm = bn – m even when n and/or m is a
negative number. Therefore…
22 ÷ 2‾3 = 22 – ‾3 = 22 + 3 = 25.
Hence, if 22 ÷ 2‾3 = 2n, it follows that n = 5
© 2007 Herbert I. Gross
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Note 1b
• In this exercise we see the importance
of knowing how to subtract signed
numbers. That is, our solution required
that we knew that 2 – -3 = 2 + +3 = 5.
•2
÷ 2‾3 means 4 ÷ 1/8 or 4 × 8 or 32 or 25.
In terms of size, the result means that 22 is
32 times as great as 2-3.
2
© 2007 Herbert I. Gross
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Note 1b
• However when we have to work with an
expression such as… there is little
alternative but to rewrite the expression
as… and then use the fact that…
2
x ֥
3
-3
x
2
+
3
x
© 2007 Herbert I. Gross
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Problem #1c
Solve for the value of n that makes
the equation a true statement.
4
2 ×
4
3
= (2 ×
n
3)
Answer: n = 4
© 2007 Herbert I. Gross
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Answer: n = 4
Solution:
In this case we use the property
bn × cn = (b × c)n with b = 2, c = 3, and n = 4.
n
4
b
2
×
n
4
c
3 =
(2
(b ×
4
n
3)
c)
Therefore if 24 × 34 = (2 × 3)n , then n = 4
© 2007 Herbert I. Gross
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Note 1c
• A main purpose of this exercise is to show
the inherent danger of what happens when
memorization fails us. Namely in this case we
are multiplying different bases that have the
same exponent. This is sort of the “reverse” of
multiplying like bases that have different
exponents. In the latter case we keep the
common base and add the exponents. In the
former case we multiply the two bases and
keep the common exponent. If all we do is
memorize the rules it is possible that we may
confuse one rule with another.
© 2007 Herbert I. Gross
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Note 1c
• The “cure” for the above problem is to
revert once again to our basic definitions.
For example, we may rewrite… in the form…
4
2 ×
4
3
2×2×2×2×3×3×3×3
© 2007 Herbert I. Gross
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Note 1c
• And because multiplication is both
associative and commutative we may
rearrange the terms and rewrite… as…
(2 × 2 × 2 × 2) × (3 × 3 × 3 × 3)
2×3 × 2×3 × 2×3 × 2×3
4
23
1
(
)
© 2007 Herbert I. Gross
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Problem #2a
Solve for the value of n that makes
each equation a true statement.
-1
0
(5 )
=n
Answer: n = 1
© 2007 Herbert I. Gross
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Answer: n = 1
Solution:
Under the condition that we wanted the
properties for exponent arithmetic to
remain valid when the exponent was 0, we
had to define b0 to equal 1 for every nonzero number b. Therefore, if we let
b = (5-1), we see that (5-1)0 = 1
© 2007 Herbert I. Gross
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Note 2a
• There is a tendency for students to want b
0
to
equal 0. However if this were true, the property
23 × 20 = 23+0 would not be correct because
23 × 20 would be 23 × 0, or 0; while 23+0 would be
23 or 8.
While it might seem a bit like nit picking, for
the sake of completeness we should point out
that 00 is indeterminate. To see why, look at an
expression such as 03 × 00. Since 03= 0, 03 × 00
will equal 0 regardless of the value we assign
to 00. So just as 0 ÷ 0 is indeterminate,
so also is 00.
•
© 2007 Herbert I. Gross
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Problem #2b
Solve for the value of n that makes
each equation a true statement.
-1
-3
(5 )
=
n
5
Answer: n = 3
© 2007 Herbert I. Gross
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Answer: n = 3
Solution:
By our definition of negative exponents, the
property (bm)n remains valid even if the
exponents are negative. Hence…
(5-1)-3 = 5-1×-3 = 53
Therefore if (5-1)-3 = 5n, then n=3
© 2007 Herbert I. Gross
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Note 2b
• This exercise affords us a practical
application for seeing why the product of
two negative numbers is positive.
• As a check we can rewrite 5
-1
as 1/5.
-- Therefore we may rewrite (5 ) as ( / ) .
-- In turn, we may rewrite ( / ) as 1 ÷ ( / )
-1 -3
1
--
or 53.
Therefore, (5-1)-3 = 53.
© 2007 Herbert I. Gross
5
-3
1
5
1
-3
5
3
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Note 2b
• However, notice how much less
cumbersome it is if we know the rule for
raising a power to a power.
• If we wanted to do the arithmetic using
fractions and whole numbers, we could
rewrite 5-1 as 1/5.
1
1
1
-1
-3
(5 ) = 1
= 1 3 = 1 3
1
1
/5 × /5 × /5
( /5)
( /5 )
= 1 ÷ 1/53 = 53
© 2007 Herbert I. Gross
= 125
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Note 2b
• However, in algebra if we were given an
expression such as (x-1)-3 our only choice
would be to use the rule for raising a
power to a power and thus obtaining x3 as
our answer.
© 2007 Herbert I. Gross
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Problem #3a
Solve for the value of n that makes
each equation a true statement.
638.97 ×
-4
10
=n
Answer: n = 0.063897
© 2007 Herbert I. Gross
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Answer: n = 0.063897
Solution:
By our definition of negative exponents,10-4
is the reciprocal of 104 . In other words,
multiplying by 10-4 is the same as dividing
by 104. Every time we divide by 10 we move
the decimal point 1 place to the left.
© 2007 Herbert I. Gross
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Answer: n = 0.063897
Solution:
Hence to divide by 104 we move the decimal
point 4 places to the left. That is…
638.97 × 10-4 = 0.0. 6. 3. 8.9 7
In summary, if 638.97 × 10-4 = n, then
n = 0.063897
© 2007 Herbert I. Gross
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Problem #3b
Solve for the value of n that makes
each equation a true statement.
0.000063897 = 6.3897 ×
n
10
Answer: n = -5
© 2007 Herbert I. Gross
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Answer: n = -5
Solution:
To get from 0.000063897 to 6.3897 we have
to move the decimal point 5 places to the
right. That is, we have to multiply by 105 .
However we don't want the value of
0.000063897 to change. In other words, to
“undo” multiplying by 105, we have to divide
by 105 (or, equivalently to multiply by 10-5).
© 2007 Herbert I. Gross
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Answer: n = -5
Solution:
Therefore we rewrite 0.000063897 as…
(0.000063897 × 105 ) × 10-5 = 6.3897 × 10-5
In summary if 0.000063897 = 6.3897 × 10n
then n = -5.
© 2007 Herbert I. Gross
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Note 3
• In describing very large or very small
numbers, we make use of the integral
powers of 10. Thus for example, instead
of writing 2,000,000,000,000,000 we might
write 2 × 1015; and instead of writing
0.000000000000002 we might write
2 × 10-15.
© 2007 Herbert I. Gross
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Note 3
• In our next lesson we will talk about
scientific notation. To write a number in
scientific notation, we write it in the form
of an integral power of 10 multiplied by a
number that is less than 10 but greater
than or equal to 1.
© 2007 Herbert I. Gross
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Note 3
• For example, when we write
2,000,000,000,000,000, as 2 × 1015 we are using
scientific notation. Of course, there are other
ways to express 2,000,000,000,000,000, as a
whole number times a power of ten. For
example, some other ways include 200 × 1015,
2,000 × 10 14, but these two forms are not
examples of scientific notation because
neither 200 nor 2,000 is between 1 and 10.
In the next lesson we will discuss why
scientific notation is so helpful.
•
© 2007 Herbert I. Gross
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Problem #4a
Write the following in decimal form.
5×
4
10
×3×
-6
10
×4×
-1
10
Answer: n = 0.06
© 2007 Herbert I. Gross
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Answer: 0.06
Solution:
Since the given expression involves only
multiplication and since multiplication is
both commutative and associative, we
may regroup the factors and rewrite… as…
4
4
-6
-6
-1
-1
(55 ×× 310× 4)
××
3 (10
× 10× 10
× 4 ×× 10
10 )
© 2007 Herbert I. Gross
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Solution:
(5 ×60
3×
4
4
+
-6
-6
+
-1
-1
4) × (10 (10
× 10 × 10) )
-3
60 × 10
© 2007 Herbert I. Gross
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Solution:
To multiply 60 by 10-3, we move the decimal
point three places to the left to obtain…
-3
60 × 10
0. 0.6.0.
3 2 1
Therefore in place value notation…
5 × 104 × 3 × 10-6 × 4 × 10-1 = 0.06
© 2007 Herbert I. Gross
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Problem #4b
Write the following in decimal form.
5×
4
10
×3×
-6
10
+4×
-1
10
Answer: n = 0.55
© 2007 Herbert I. Gross
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Answer: 0.55
Solution:
At first glance Exercise 4b can be easily
confused with Exercise 4a. The only
difference is that in part (b) one of the
time signs is replaced by a plus sign; and
since plus signs separate terms we have to
read… as…
1044××33××10
10-6-6) + (4
4 ×× 10
(55××10
10-1-1)
© 2007 Herbert I. Gross
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Solution:
To rewrite the expression below… notice
that… and…
(5(5××3)10×4 (10
10-6-6)) + (4 ×0.4
× 34 ×× 10
10-1)
15 × 104 + -6
15 × 10-2
0.15
= 0.55
+
Hence, we may rewrite it as 0.15 + 0.4 or 0.55.
© 2007 Herbert I. Gross
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Note 4
• When we use scientific notation we are
often doing arithmetic with numbers that
involve powers of 10. In such cases we
often rewrite products so that the powers of
10 are grouped together.
At the same time it is important for us to
keep track of the order of operations. For
example, it was important in part (b)
to observe that there was a plus sign that
separated one group of times signs from
the other.
•
© 2007 Herbert I. Gross
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Important Note 4
• For example, while it's nice that we can
rewrite 310 × 320 as 330, there is no simple
way to rewrite 310 + 320.
© 2007 Herbert I. Gross
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Problem #5
Suppose the cost of your buying
groceries doubles every 10 years. If a
particular grocery order costs $240
today, how much did the same order
cost 30 years ago?
Answer: $30
© 2007 Herbert I. Gross
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Answer: $30
Solution:
If the cost doubles every ten years then ten
years earlier the cost was half of the present
cost. Hence, we can show by a
chart such as the one shown below that 30
years ago the cost was $30. More
specifically…
Present Cost
$240
Cost 10 years ago
Cost 20 years ago
Cost 30 years ago
© 2007 Herbert I. Gross
$120
$60
$30
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Note 5
• In this exercise it is not very
cumbersome to make a chart. However,
this isn't always the case. For example, if
we wanted to know what the cost would
have been 100 years ago we would have
had to add 7 more rows to our chart.
© 2007 Herbert I. Gross
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Note 5
• In such an event it's often more helpful to
write the relationship in the form of a formula.
In the present example, we could let d denote
the number of decades ago that we wanted to
know what the cost would have been (we use
decades instead of years because the doubling
takes place every ten years); we could use P to
denote the present cost in dollars and we could
use V to denote the cost in dollars t decades
ago. In that case the formula would be…
© 2007 Herbert I. Gross
P = V × 2d
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Note 5
• To check that the formula P = V × 2
d
is
correct, we know from our chart that when
d = 3 (that is, 30 years ago), V = 30 and
P= 240; and in this case the formula is the
true statement 240 = 30 × 23 ( = 30 × 8).
© 2007 Herbert I. Gross
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Note 5
• Since we know the value of P and want to
find the value of V, it might be more
convenient to undo formula V = P × 2d by
dividing both sides of the formula by 2d (or
equivalently by multiplying both sides by
2-d) to obtain the equivalent formula…
V = P × 2-d
© 2007 Herbert I. Gross
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Note 5
• and in the case of the present exercise,
the formula V = P × 2-d would become…
V = 240 × 2-3 ( = 240 ÷ 80) = 30
• More importantly, the formula above allows
us to compute the cost any number of
decades ago. For example, if it was 10
decades ago the cost would have been…
© 2007 Herbert I. Gross
V = 240 ×
-10
2
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Note 5
• And using a calculator we would enter the
following sequence of key strokes…
2
xy
1
0
+/
-
=
×
2
4
0
=
• and the answer would appear as 0.234375;
that is, about 24 cents.
© 2007 Herbert I. Gross
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Note 5
• This illustrates the power of exponential
growth. In other words if you somehow
invested 24¢ a hundred years ago and
your investment doubled every 10 years,
the present value would be about $240.
© 2007 Herbert I. Gross
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Note 5
• An important scientific application of this
idea is when we talk about the half-life of a
radioactive substance. The half-life is the
amount of time it takes for the substance to
lose half of its present size. Thus if the half
life of a 240 gram radioactive substance
was 200 years, 200 years from now its mass
would be 120 grams; and 200 years after
that its mass would be 60 grams; and 200
years after that its mass would be 30 grams
and so on.
© 2007 Herbert I. Gross