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ABOUT EXPONENTS
THE IDEA
One of the ideas for this section is to become very familiar with some of the most common and frequently used
properties for integer exponents. We would like to become fluent in their use and comfortable with the proofs where
appropriate. Specific cases for each of these can be proven and may be useful in gaining insight into why these are
true.
Below is a list of some of these famous exponent properties. The list is followed by some examples which make use
of these properties to simplify expressions. We then prove specific cases for some of these to shed some light into
why they hold.
About Z Exponents
Some common properties for Z Exponents, assume x, y ∈ C and x, y, n 6= 0 and exponents, n, m, k ∈ Z:
of
Natural
• Definition
xn = |x · x · {z
x · · · · x}
Exponents
• Definition of 0-Exponent [0-Expo] : x0 = 1
[N-Expo] :
• Power-to-Power [P2P] : (xn )m = xnm
`
´m
= xnm y km
• Distribute Power-to-Power [DP2P] : xn y k
n-times
• Definition of Integer Exponents [Z-Expo] : x−n =
1
xn
• Just Add Exponents [JAE] : xn xm = xn+m
• Just Subtract Exponents [JSE] :
xn
xm
= xn−m
EXAMPLES Simplifying Expressions
1. Simplify:
x−2 x5
Solution:
x−2 x5 = x−2+5
= x3
(JAE)
(BI)
2. Simplify:
y −12 y 5
Solution:
y −12 y 5 = y −12+5
= y −7
1
=
y7
(JAE)
(BI)
(Z-expo)
3. Simplify:
x12 y 5
Solution:
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x12 y 5
= x(12)(−3) y (5)(−3)
−3
= x−36 y −15
1
=
x36 y 15
4. Simplify:
x−12 y 5
Solution:
x−12 y 5
(DP2P)
3
(BI)
(Z-expo)
3
= x(−12)(3) y (5)(3)
(DP2P)
= x−36 y 15
=
(BI)
y 15
(Z-expo, BI)
x36
5. Simplify
x−12
x5
Solution:
1
x−12
= 5−(−12)
x5
x
1
=
x17
(Z-expo, BI)
(BI)
6. Simplify
x−12
x−2
Solution:
x−12
1
= −2−(−12)
x−2
x
1
=
x10
(Z-expo, BI)
(BI)
ABOUT RATIONAL EXPONENTS Definition & Caution
It should be noted that the above exponent properties also hold for fractional exponents, or rational exponents,
so long as the bases are positive real numbers. Even when the base is a negative real number under some special
conditions some of the properties still hold. Yet, a complete and honest discussion with rational exponents would
quickly lead us outside the scope of this class, thus we we limit our discussion of rational exponents to positive real
number bases. With that in mind we define:
About Q Exponents
Some common properties for Q Exponents, assume x, y ∈ R>0 and n, m 6= 0 and exponents, n, m ∈ Z:
Definition of Rational Exponents [Q-Expo] :
√
xm/n = n xm
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Where exponents are integers, assume all bases are non zero complex numbers, and where exponents are non-integer
rationals assume the base is a positive real number.
1. Simplify:
t2 t5
Solution:
t2 t5 = t2+5
= t7
(JAE)
(BI)
2. Simplify:
r2 r−5
Solution:
r2 r−5 = r2+−5
= r−3
1
=
r3
(JAE)
(BI)
(Z-expo)
3. Simplify:
z −2 z −5
Solution:
z −2 z −5 = z −2+−5
= z −7
1
=
z7
(JAE)
(BI)
(Z-expo)
4. Simplify:
t20 t5
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Solution:
t20 t5 = t20+5
= t25
(JAE)
(BI)
5. Simplify:
r2 r−15
Solution:
r2 r−15 = r2+−15
= r−13
1
=
r13
(JAE)
(BI)
(Z-expo)
6. Simplify:
z −20 z 5
Solution:
z −20 z 5 = z −20+5
= z −15
1
=
z 15
(JAE)
(BI)
(Z-expo)
7. Simplify:
r2 r15
Solution:
r2 r15 = r2+15
= r17
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(JAE)
(BI)
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8. Simplify:
z 30 z −5
Solution:
z 30 z −5 = z 30+−5
= z 25
(JAE)
(BI)
9. Simplify:
t−2 r−5
−3
Solution:
t−2 r−5
= t(−2)(−3) r(−5)(−3)
−3
= t6 r15
10. Simplify:
s−8 k −5
(DP2P)
(BI)
3
Solution:
s−8 k −5
3
= s(−8)(3) k (−5)(3)
= s−24 k −15
1
=
24 15
s k
(DP2P)
(BI)
(Z-expo)
11. Simplify:
p7 q −5
−3
Solution:
p7 q −5
−3
= p(7)(−3) q (−5)(−3)
= p−21 q 15
=
q 15
p21
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(DP2P)
(BI)
(Z-expo, BI)
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12. Simplify:
x−12 y 5
−3
Solution:
x−12 y 5
−3
= x(−12)(−3) y (5)(−3)
= x36 y −15
=
x36
y 15
(DP2P)
(BI)
(Z-expo, BI)
13. Simplify
y −12
y −5
Solution:
y −12
1
= −5−(−12)
−5
y
y
1
=
y7
(Z-expo, BI)
(BI)
14. Simplify
t−12
t−20
Solution:
t−12
= t−12−(−20)
t−20
= t8
15. Simplify
(Z-expo, BI)
(BI)
y 12
y −5
Solution:
y 12
= y 12−(−5)
y −5
= y 17
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(Z-expo, BI)
(BI)
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16. Simplify
t2
t−20
Solution:
t2
t−20
= t2−(−20)
= t22
17. Simplify
(Z-expo, BI)
(BI)
y 10
y5
Solution:
y 10
= y 10−(5)
y5
= y5
18. Simplify
(Z-expo, BI)
(BI)
t27
t20
Solution:
t27
= t27−(20)
t20
= t7
(Z-expo, BI)
(BI)
19. Simplify
t−27
t20
Solution:
t−27
1
= 20−(−27)
t20
t
1
=
t47
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(Z-expo, BI)
(BI)
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20. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
x7 · x5 = x12
Solution:
proof:
x7 · x5 = (x · x · x · x · x · x · x) (x · x · x · x · x)
= x·x·x·x·x·x·x·x·x·x·x·x
(N-expo)
(ALM)
= x12
(N-expo)
21. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
y2 · y3 = y5
Solution:
proof:
y 2 · y 3 = (y · y) (y · y · y)
=y·y·y·y·y
(N-expo)
(ALM)
= x5
(N-expo)
22. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
t5 · t4 = t9
Solution:
proof:
t5 · t4 = (t · t · t · t · t) (t · t · t · t)
= t·t·t·t·t·t·t·t·t
(N-expo)
(ALM)
= x9
(N-expo)
23. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
p2 · p2 = p4
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Solution:
proof:
p2 · p2 = (p · p) (p · p)
(N-expo)
=p·p·p·p
(ALM)
4
=x
(N-expo)
24. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
47 · 43 = 410
Solution:
proof:
47 · 43 = (4 · 4 · 4 · 4 · 4 · 4 · 4) (4 · 4 · 4)
(N-expo)
= 4·4·4·4·4·4·4·4·4·4
(ALM)
10
=x
(N-expo)
25. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
65 · 65 = 610
Solution:
proof:
65 · 65 = (6 · 6 · 6 · 6 · 6) (6 · 6 · 6 · 6 · 6)
(N-expo)
= 6·6·6·6·6·6·6·6·6·6
(ALM)
10
=x
(N-expo)
26. Explaining a small instance of JAE Use only the definition/s and/or axioms to explain why
123 · 122 = 125
Solution:
proof:
123 · 122 = (12 · 12 · 12) (12 · 12)
= 12 · 12 · 12 · 12 · 12
5
=x
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(N-expo)
(ALM)
(N-expo)
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27. Explaining a small instance of P2P Use only JAE, the definition/s and/or axioms to explain why
t5
4
= t20
Solution:
proof:
t5
4
= t5 · t5 · t5 · t5
=t
(N-expo)
20
(JAE)
28. Explaining a small instance of P2P Use only JAE, the definition/s and/or axioms to explain why
p2
3
= p6
Solution:
proof:
p2
3
= p2 · p2 · p2
=p
(N-expo)
6
(JAE)
29. Explaining a small instance of P2P Use only JAE, the definition/s and/or axioms to explain why
32
4
= 38
Solution:
proof:
32
4
= 32 · 32 · 32 · 32
=3
(N-expo)
8
(JAE)
30. Explaining a small instance of P2P Use only JAE, the definition/s and/or axioms to explain why
2
(−2)5 = (−2)10
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Solution:
proof:
5
(−2)
2
5
5
= (−2) · (−2)
(N-expo)
10
= (−2)
(JAE)
31. Explaining a small instance of P2P Use only JAE, the definition/s and/or axioms to explain why
5
(−2)2 = (−2)10
Solution:
proof:
5
(−2)2 = (−2)2 · (−2)2 · (−2)2 · (−2)2 · (−2)2
(N-expo)
10
= (−2)
(JAE)
32. Explaining a small instance of DP2P Use only JAE, the definition/s and/or axioms to explain why
32 x2
5
= 310 x10
Solution:
proof:
32 x2
5
= 32 x2 · 32 x2 · 32 x2 · 32 x2 · 32 x2
2
2
2
2
2 2
2
2
(N-expo)
2
2
= 3 ·3 ·3 ·3 ·3 x ·x ·x ·x ·x
(CoLM, ALM)
10 10
=3 x
(ALM, JAE)
33. Explaining a small instance of DP2P Use only JAE, the definition/s and/or axioms to explain why
32 52
5
= 310 510
Solution:
proof:
32 52
5
= 32 52 · 32 52 · 32 52 · 32 52 · 32 52
2
2
2
2
2 2
2
2
(N-expo)
2
= 3 ·3 ·3 ·3 ·3 5 ·5 ·5 ·5 ·5
10 10
=3 5
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(CoLM, ALM)
(ALM, JAE)
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34. Explaining a small instance of DP2P Use only JAE, the definition/s and/or axioms to explain why
y 5 x4
3
= y 15 x12
Solution:
proof:
y 5 x4
3
= y 5 x4 · y 5 x4 · y 5 x4
5
5
5 4
4
(N-expo)
4
=y ·y ·y x ·x ·x
15 12
=y x
(CoLM, ALM)
(ALM, JAE)
35. Using only definitions and/or axioms [i.e. do not use JAE], explain why for a positive real number, x:
x1/2 · x1/3 = x5/6
Solution: You will learn more on your quest for the solutions. To simply see the solution would abruptly and
prematurely end such quest.
36.
35x = 35
A. TRUE
B. FALSE
37.
x
5x
3t5x = (3t)
A. TRUE
B. FALSE
38.
35x = 32x 33x
A. TRUE
B. FALSE
39.
x1/3 + x1/4 = x1/12
A. TRUE
B. FALSE
40.
x1/3 + x1/4 = x7/12
A. TRUE
B. FALSE
41.
x1/3 + x1/4 = x4/12 (1 + x−1/12 )
A. TRUE
B. FALSE
42.
x1/3 + x1/4 = x3/12 (x1/12 + x0 )
A. TRUE
B. FALSE
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43.
x1/3 + x1/4 = x1/3 (x1/12 + 1)
A. TRUE
B. FALSE
44.
x1/3 · x1/4 = x7/12
A. TRUE
B. FALSE
45.
e1+5x = e · (ex )5
A. TRUE
B. FALSE
46.
e1+5x = e · (e5 )x
A. TRUE
B. FALSE
47.
e1+5x = (e · e5 )x
A. TRUE
B. FALSE
48.
e1+5x = e6x
A. TRUE
B. FALSE
49. Simplify
3x4 y −19
9x−3 y 12
Solution:
x7
3y 31
50. Simplify
Solution:
3x4 y −19
9x−3 y 12
5
x35
35 y 155
51. Simplify
Solution:
3x4 y −19
9x−3 y 12
−5
35 y 155
x35
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52. Simplify
−3 !−5
3x4 y −19
9x−6 y 12
Solution:
−3 !−5
3x4 y −19
9x−6 y 12
−3 −12 57 −5
3 x y
=
9x−6 y 12
45 −5
y
=
35 x6
y −225
= −25 −30
3 x
325 x30
= 225
y
53. Factor
x3 − y 3
Solution:
(x − y)(x2 + xy + y 2 )
54. Factor
x3 − 13
Solution:
(x − 1)(x2 + x · 1 + 12 ) = (x − 1)(x2 + x + 1)
55. Factor
R 3 − 13
Solution:
(R − 1)(R2 + R · 1 + 12 ) = (R − 1)(R2 + R + 1)
56. Factor
(x1/3 )3 − 13
Solution:
1/3
(x
2
2
1/3
2
1/3
+ x · 1 + 1 ) = (x1/3 − 1)( x1/3 + x1/3 + 1)
− 1) ( x
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57. Factor
x−1
(hint use the previous problems.. )
Solution:
(x1/3 − 1)
2
2
x1/3 + x1/3 · 1 + 12 = (x1/3 − 1) x1/3 + x1/3 + 1
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