MATH REVIEW

2. Exponents
Exponents
 3
5
base
exponent
3 3
means that is the exponential
form of t
Example:
he number
125 5 5
.
125

53
means 3 factors of 5 or 5 x 5 x 5
Power

3. The Laws of Exponents:
The Laws of Exponents:
#1: Exponential form: The exponent of a power indicates
how many times the base multiplies itself.
n
n times
x x x x x x x x

   


   
      
3
Example: 5 5 5 5
  
n factors of x

4. #2: Multiplying Powers: If you are multiplying Powers
with the same base, KEEP the BASE & ADD the EXPONENTS!
m n m n
x x x 
 
So, I get it!
When you
multiply
Powers, you
add the
exponents!
512
2
2
2
2 9
3
6
3
6



 

5. #3: Dividing Powers: When dividing Powers with the
same base, KEEP the BASE & SUBTRACT the EXPONENTS!
m
m n m n
n
x
x x x
x

  
So, I get it!
When you
divide
Powers, you
subtract the
exponents!
16
2
2
2
2 4
2
6
2
6


 

6. Try these:

 2
2
3
3
.
1

 4
2
5
5
.
2

 2
5
.
3 a
a

 7
2
4
2
.
4 s
s



 3
2
)
3
(
)
3
(
.
5

 3
7
4
2
.
6 t
s
t
s

4
12
.
7
s
s

5
9
3
3
.
8

4
4
8
12
.
9
t
s
t
s

5
4
8
5
4
36
.
10
b
a
b
a

7. 
 2
2
3
3
.
1

 4
2
5
5
.
2

 2
5
.
3 a
a

 7
2
4
2
.
4 s
s



 3
2
)
3
(
)
3
(
.
5

 3
7
4
2
.
6 t
s
t
s
81
3
3 4
2
2



7
2
5
a
a 

9
7
2
8
4
2 s
s 

 
SOLUTIONS
6
4
2
5
5 

243
)
3
(
)
3
( 5
3
2




 
7
9
3
4
7
2
t
s
t
s 



8. 
4
12
.
7
s
s

5
9
3
3
.
8

4
4
8
12
.
9
t
s
t
s

5
4
8
5
4
36
.
10
b
a
b
a
SOLUTIONS
8
4
12
s
s 

81
3
3 4
5
9



4
8
4
8
4
12
t
s
t
s 


3
5
8
4
5
9
4
36 ab
b
a 

 


9. #4: Power of a Power: If you are raising a Power to an
exponent, you multiply the exponents!
 
n
m mn
x x

So, when I
take a Power
to a power, I
multiply the
exponents
= 15, 625
(5³)² = 5 ³ ² = 5
˟ ⁶

10. #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.
 
n n n
xy x y
 
So, when I take
a Power of a
Product, I apply
the exponent to
all factors of
the product.
2
2
2
)
( b
a
ab 

11. #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 n
n
x x
y y
 

 
 
So, when I take a
Power of a
Quotient, I apply
the exponent to
all parts of the
quotient.
81
16
3
2
3
2
4
4
4









12. Try these:
  
5
2
3
.
1
  
4
3
.
2 a
  
3
2
2
.
3 a
  
2
3
5
2
2
.
4 b
a

 2
2
)
3
(
.
5 a
  
3
4
2
.
6 t
s







5
.
7
t
s









2
5
9
3
3
.
8









2
4
8
.
9
rt
st









2
5
4
8
5
4
36
.
10
b
a
b
a

13.   
5
2
3
.
1
  
4
3
.
2 a
  
3
2
2
.
3 a
  
2
3
5
2
2
.
4 b
a

 2
2
)
3
(
.
5 a
  
3
4
2
.
6 t
s
SOLUTIONS
10
3
12
a
6
3
2
3
8
2 a
a 

6
10
6
10
4
2
3
2
5
2
2
16
2
2 b
a
b
a
b
a 




  4
2
2
2
9
3 a
a 

 
12
6
3
4
3
2
t
s
t
s 



14. 






5
.
7
t
s









2
5
9
3
3
.
8









2
4
8
.
9
rt
st









2
5
4
8
5
4
36
10
b
a
b
a
SOLUTIONS
  6
2
2
3
2
2
2
3
81
9
9 b
a
b
a
ab 
 
2
8
2
2
4
r
t
s
r
st









  8
2
4
3
3 
5
5
t
s

15. #7: Negative Law of Exponents: If the base is powered
by the negative exponent, then the base becomes reciprocal with the
positive exponent.
1
m
m
x
x


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!
9
3
3
1
125
1
5
1
5
2
2
3
3






and

16. #8: Zero Law of Exponents: Any base powered by zero
exponent equals one.
0
1
x 
1
)
5
(
1
1
5
0
0
0



a
and
a
and
So zero
factors of a
base equals 1.
That makes
sense! Every
power has a
coefficient
of 1.

17. Try these:
  
0
2
2
.
1 b
a

  4
2
.
2 y
y
  
 1
5
.
3 a


 7
2
4
.
4 s
s
  

 4
3
2
3
.
5 y
x
  
0
4
2
.
6 t
s









 1
2
2
.
7
x









 2
5
9
3
3
.
8









 2
4
4
2
2
.
9
t
s
t
s









 2
5
4
5
4
36
.
10
b
a
a

18. SOLUTIONS
  
0
2
2
.
1 b
a
  
 1
5
.
3 a


 7
2
4
.
4 s
s
  

 4
3
2
3
.
5 y
x
  
0
4
2
.
6 t
s
1
5
1
a
5
4s
  12
8
12
8
4
81
3
y
x
y
x 


1

19. 








 1
2
2
.
7
x









 2
5
9
3
3
.
8









 2
4
4
2
2
.
9
t
s
t
s









 2
5
4
5
4
36
.
10
b
a
a
SOLUTIONS
4
4
1
x
x








  8
8
2
4
3
1
3
3 
 

  4
4
2
2
2
t
s
t
s 



2
10
10
2
2
81
9
a
b
b
a 

