Lehman PTS-3 Summer 2020

1. 𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 3
Exponential Notation
"In mathematics the art of proposing
a question must be held of higher
value than solving it." – Georg Cantor
-

2. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
A product of identical numbers is usually written in
exponential notation. For example:
DEFINITION. Let 𝑎 be a real number and let 𝑛 ≥ 2 be a
positive integer, then the 𝑛-th power of 𝑎 is defined as:
The number 𝑎 is called the base, and 𝑛 is the exponent.
Example 1. Determine the following:
𝟓 ⋅ 𝟓 ⋅ 𝟓 = 𝟓 𝟑
𝒂 𝒏
= 𝒂 ⋅ 𝒂 ⋅ ⋯ ⋅ 𝒂
𝒏 𝐟𝐚𝐜𝐭𝐨𝐫𝒔
1
2
4
(a) (b) −2 3
(c) −23

3. Lehman College, Department of Mathematics
Integer Exponents (2 of 4)
Solution: Note the difference between (b) and (c):
Let us discover some of the rules for working with
exponential notation:
It appears that when we multiply two powers of the
same base, we add their exponents.
1
4
3
=(a)
(b) −2 4
=
(c) −24
=
1
4
1
4
1
4
=
1
64
−2 ⋅ −2 ⋅ −2 ⋅ −2 = 16
− 2 ⋅ 2 ⋅ 2 ⋅ 2 = −16
53 ⋅ 54 = 5 ⋅ 5 ⋅ 5 ⋅
3 factors
5 ⋅ 5 ⋅ 5 ⋅ 5 =
4 factors
57

4. Lehman College, Department of Mathematics
Integer Exponents (3 of 4)
In general, for any real number 𝑎 and positive integers 𝑛
and 𝑚 with 𝑛, 𝑚 ≥ 2:
Let us look at division of powers. Consider:
It appears that when we find the quotient of two powers
of the same base, we subtract their exponents.
Then, for any real number 𝑎 (𝑎 ≠ 0) and any positive
integers 𝑛 and 𝑚 with 𝑛, 𝑚 ≥ 2 and 𝑛 − 𝑚 ≥ 2, then:
𝒂 𝒎
𝒂 𝒏
= 𝒂 𝒎+𝒏
55
53
=
5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5
= 5 ⋅ 5 = 52
𝒂 𝒎
𝒂 𝒏
= 𝒂 𝒎−𝒏

5. Lehman College, Department of Mathematics
Integer Exponents (3 of 4)
The formulas we have discovered so far are true for
exponents 𝑛 with 𝑛 ≥ 2. Consider the following example
If we want the formula to still work, then we require that:
Similarly, consider the following example:
If we want the formula to still work, then we require that:
5
53−2 =
53
52
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5
=
515 =
1
53
53
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5
=
53−3 = 501 =

6. Lehman College, Department of Mathematics
Integer Exponents (3 of 4)
In general, for any nonzero real number 𝑎:
Let us look at negative exponents. Consider:
If we want the formula to still work, then we require that:
In general, for any nonzero real number 𝑎, and for any
integer 𝑛, we have:
𝒂 𝟏 = 𝒂
53
55
=
5 ⋅ 5 ⋅ 5
5 ⋅ 5 ⋅ 5 ⋅ 5 ⋅ 5
=
1
5 ⋅ 5
=
1
52
𝒂−𝒏 =
𝟏
𝒂 𝒏
𝒂 𝟎
= 𝟏and
1
52
= 53−5
= 5−2
𝒂 𝒏
=
𝟏
𝒂−𝒏
and

7. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
Example 2. Determine the following:
Solution.
Example 3. In Calculus I and II, we will solve problems
such as the following. Our second step should be:
4
7
0
(a) (b) −3 −2 (c) 𝑥−1
4
7
0
=(a) 1 (b) −3 −2 =
1
−3 2
=
1
9
(c) 𝑥−1
=
1
𝑥1
=
1
𝑥
(d)
1
𝑥−3
(d)
1
𝑥−3
= 𝑥−(−3)
= 𝑥3
𝑑
𝑑𝑥
1
𝑥5
=
𝑑
𝑑𝑥
𝑥−5(a)
1
2
1
𝑥3
𝑑𝑥 =(b)
1
2
𝑥−3 𝑑𝑥

8. Lehman College, Department of Mathematics
Helix Nebula in Aquarius (1 of 1)

9. Lehman College, Department of Mathematics
Integer Exponents (2 of 4)
Let us discover more rules for working with exponential
notation:
It appears that when we raise an exponential
expression to a power, we multiply the exponents.
In general, for any nonzero real number 𝑎 (base) and
integers 𝑛 and 𝑚 (exponents):
52 3
= 52
⋅ 52
⋅ 52
=
3 factors
5 ⋅ 5 5 ⋅ 5 5 ⋅ 5 =
6 factors
56
𝒂 𝒎 𝒏 = 𝒂 𝒏 𝒎 = 𝒂 𝒎𝒏
53 2 = 53 ⋅ 53 =
2 factors
5 ⋅ 5 ⋅ 5 5 ⋅ 5 ⋅ 5 =
6 factors
56

10. Lehman College, Department of Mathematics
Integer Exponents (2 of 4)
We will now look at the product and quotient of powers
of different bases, but the same exponent:
In general, for any nonzero real numbers 𝑎 and 𝑏 (base)
and integer 𝑛 (exponent):
We can similarly show that:
23
⋅ 53
=
= 2 ⋅ 5 2 ⋅ 5 2 ⋅ 5 =
3 factors
2 ⋅ 5 3
𝒂 𝒏 𝒃 𝒏 = 𝒂𝒃 𝒏
2 ⋅ 2 ⋅ 2 ⋅
3 factors
5 ⋅ 5 ⋅ 5 =
3 factors
𝒂
𝒃
𝒏
=
𝒂 𝒏
𝒃 𝒏

11. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
Example 3. Simplify the following expressions using
positive exponents only:
Solution. Group constants and factors with same base:
4𝑥2
𝑦
−3
(a) (b)
−35𝑥2
𝑦4
5𝑥6 𝑦−8
(c)(−7𝑥𝑦4)(−2𝑥5 𝑦6)
(a) −7𝑥𝑦4
−2𝑥5
𝑦6
= −7 −2)(𝑥 ⋅ 𝑥5
)(𝑦4
𝑦6
= 14𝑥6
𝑦10
𝑦
4𝑥2
3
=(b)
4𝑥2
𝑦
−3
=
𝑦3
4𝑥2 3
=
𝑦3
43 𝑥2 3
=
𝑦3
64𝑥6

12. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
Solution (cont’d). Group factors with same base:
−35𝑥2
𝑦4
5𝑥6 𝑦−8
=(c)
−35
5
𝑥2
𝑥6
𝑦4
𝑦−8
= −7 𝑥2−6 𝑦4−(−8)
= −
7𝑦12
𝑥4
= −7𝑥−4
𝑦12

13. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
Example 4. Simplify the following expression using
positive exponents only:
Solution 1. Raise factors to the required exponents,
then group constants and factors with the same base:
−4𝑎−2 𝑏3 −3 8𝑎𝑏 2
−4𝑎−2
𝑏3 −3
8𝑎𝑏 2
= −4 −3
𝑎−2 −3
𝑏3 −3
⋅ 82
𝑎2
𝑏2
=
1
−4 3
⋅ 𝑎−2 −3
𝑏3(−3)
⋅ 64𝑎2
𝑏2
=
64
−64
⋅ 𝑎6 𝑎2 𝑏−9 𝑏2
= −
𝑎8
𝑏7
= −𝑎8
𝑏−7

14. Lehman College, Department of Mathematics
Integer Exponents (1 of 4)
Example 4. Simplify the following expression using
positive exponents only:
Solution 2. Raise factors to the required exponents,
then group constants and factors with the same base:
−4𝑎−2 𝑏3 −3 8𝑎𝑏 2
−4𝑎−2
𝑏3 −3
8𝑎𝑏 2
=
1
−4𝑎−2 𝑏3 3
⋅ 64𝑎2
𝑏2
=
64𝑎2 𝑏2
−4 3 𝑎−2 3 𝑏3 3
=
64𝑎2 𝑏2
−64𝑎−6 𝑏9
= −
𝑎2− −6
𝑏9−2
= −
𝑎8
𝑏7

15. Lehman College, Department of Mathematics
Incorrect Correct
Common Errors to Avoid (4 of 9)
𝑏3
⋅ 𝑏4
= 𝑏12
32
⋅ 34
= 96
𝑏3
⋅ 𝑏4
= 𝑏7
32
⋅ 34
= 36
516
54
= 54
516
54
= 58
4𝑎 3
= 64𝑎3
4𝑎 3
= 4𝑎3
𝑏−𝑛 = 𝑏−𝑛
=
1
𝑏 𝑛
𝑎 + 𝑏 −1
=
1
𝑎
+
1
𝑏
𝑎 + 𝑏 −1
=
1
𝑎 + 𝑏
−
1
𝑏 𝑛