Exponents
Power
5
exponent
3
base
Example: 125  53 means that 53 is the exponential
form of the number 125.
53 means 3 factors of 5 or 5 x 5 x 5
The Laws of Exponents:
#1: Exponential form: The exponent of a power indicates
how many times the base multiplies itself.
x  x  x  x x  x  x  x
n
n times
n factors of x
Example: 5  5  5  5
3
#2: Multiplying Powers:
If you are multiplying Powers
with the same base, KEEP the BASE & ADD the EXPONENTS!
x x  x
m
So, I get it!
When you
multiply
Powers, you
add the
exponents!
n
m n
2 2  2
6
3
63
2
9
 512
#3: Dividing Powers: When dividing Powers with the
same base, KEEP the BASE & SUBTRACT the EXPONENTS!
m
x
m
n
mn

x

x

x
n
x
So, I get it!
When you
divide
Powers, you
subtract the
exponents!
6
2
6 2
4

2

2
22
 16
Try these:
12
1. 3  3 
2
2
7.
2. 52  54 
8.
3. a  a 
5
2
4. 2s  4s 
2
7
12 8
9.
5. (3)  (3) 
2
3
6. s t  s t 
2 4
7 3
s

4
s
9
3

5
3
s t

4 4
st
5 8
10.
36a b

4 5
4a b
SOLUTIONS
2 2
4
5 2
a
1. 3  3  3  3  81
2 4
6
2
4
2. 5  5  5  5
2
2
3. a  a  a
5
2
4. 2s  4s  2  4  s
2
7
5. (3)  (3)  (3)
2
3
6. s t  s t 
2 4
7 3
s
7
2 7
2 3
 8s
 (3)  243
2 7 43
t
9
5
s t
9 7
SOLUTIONS
12
7.
8.
9.
10.
s
12  4
8
s

s

4
s
9
3
9 5
4
3

3

81

5
3
12 8
s t
12  4 8 4
8 4
s
t

s
t

4 4
st
5 8
36a b
5  4 8 5
3
36

4

a
b

9
ab

4 5
4a b
#4: Power of a Power: If you are raising a Power to an
exponent, you multiply the exponents!
x 
n
m
So, when I
take a Power
to a power, I
multiply the
exponents
x
mn
32
(5 )  5
3
2
5
5
#5: Product Law of Exponents: If the product of the
bases is powered by the same exponent, then the result is a
multiplication of individual factors of the product, each powered
by the given exponent.
 xy 
So, when I take
a Power of a
Product, I apply
the exponent to
all factors of
the product.
n
 x y
n
n
(ab)  a b
2
2
2
#6: Quotient Law of Exponents: If the quotient of the
bases is powered by the same exponent, then the result is both
numerator and denominator , each powered by the given exponent.
n
 x
x
   n
y
 y
So, when I take a
Power of a
Quotient, I apply
the exponent to
all parts of the
quotient.
n
4
 2  2 16
   4 
81
3 3
4
Try these:
 
5
2 5
1. 3

 
3. 2a  
4. 2 a b 
2. a
s
7.   
t 2
 39 
8.  5  
3 
3 4
2 3
2
5 3 2

5. (3a ) 
2 2
 
2 4 3
6. s t

2
 st 
9.  4  
 rt 
5 8 2
 36a b 
 
10. 
4 5 
 4a b 
8
SOLUTIONS
 
2 5
1. 3
 
2. a
3 4

10
3

a12
 
 2a


3. 2a
2
2 3
3
5 3 2
4. 2 a b
23
 8a
6
22 52 32
4 10 6
10 6
2
a
b

2
a
b

16
a
b

5. (3a )  3  a 22  9a 4
2
2 2
 
2 4 3
6. s t
23 43
s t
s t
6 12
SOLUTIONS
5
s
7.   
t
5
s
5
t
2
3 
8.  5   34
3 
9
 
2
3
8
2
4 2
2 8
 st 


st
s
t
9.  4   
  2

r
 rt 
 r 
8
 36a b
10
4 5
4
a
b

5 8
2

 

9ab 
3 2
 92 a 2b32  81a 2b6
#7: Negative Law of Exponents: If the base is powered
by the negative exponent, then the base becomes reciprocal with the
positive exponent.
So, when I have a
Negative Exponent, I
switch the base to its
reciprocal with a
Positive Exponent.
Ha Ha!
If the base with the
negative exponent is in
the denominator, it
moves to the
numerator to lose its
negative sign!
x
m
1
 m
x
1
1
5  3 
5
125
and
3
1
2

3
9
2
3
#8: Zero Law of Exponents: Any base powered by zero
exponent equals one.
x 1
0
So zero
factors of a
base equals 1.
That makes
sense! Every
power has a
coefficient
of 1.
50  1
and
a0  1
and
(5 a ) 0  1
Try these:
1.
2a b
0
2

y 2  y 4 
2.
a 
5 1
3.
2

4. s  4s 

7
 
s t  
2
5. 3x y
6.
3 4
2 4 0
1
2 
7.   
 x 2
 39 
8.  5  
3 
2
2
s t 
9.  4 4  
s t 
2
5
 36a 
10.  4 5  
 4a b 
2 2
SOLUTIONS


0
1. 2a b  1
2
 
1
3. a
 5
a
5
2
7
4. s  4s  4s
5 1

2
5. 3x y
 
2 4 0
6. s t

3 4
 3 x y
 1
4
8
12

8
x

81y12
SOLUTIONS
1
2 
7.  
 x 
9 2
3 
8.  5 
3 
2
1
x
4
  
4
 x
 3
2

4 2

1
3  8
3
8

s t 
 2  2 2
4 4
s t
9.  4 4   s t
s t 
10
2
5
b

2

2
10
 36a 
9
a
b

2


10.  4 5  
81a
4
a
b


2 2