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Multiplication and
Division of Exponents
Notes
Multiplication Rule
x x x
n
m
n m
In order to use this rule the base numbers
being multiplied must be the same.
Example:
3
x x
4
Written in multiplication form
x x x x x x x  x
Using Rule

x
34
x
7
7
Example 1
2 2
3
2*2*2
2
5
2*2*2*2*2
3 5
2
8

Example 2
x x x
354
3
5
x
x
12
4
Example 3
x x  y  y z
4
3
Remember

x
14
x

5
y
34
7
y z
4
xx
z
1
Example 4
3
5
3
2
(2x y )(3xy )( x y)
Multiply coefficients and add exponents of like bases
6x

(312) (531)
y
6x y
6

9
Example 5
4 x (2x y  5xy )
5
2
2
In order to simplify you must distribute. Since you are
multiplying when you distribute you must use the
multiplication rule for exponents

8x
52
y  20x
51
8x y  20x y
7
6
2
y
2

You Try
Simplify each expression
1.
2.
3,
4.
5.
y y
4
6
15x 3
2
(3x )(5x)
7
x
x x

5 6
3 
2
2 4
ab
(a b )(a b )
3
4
12y 3
2
(4 y )(3y)

5
6.
y
2
2
4
6
(x y )(x y )


9
x y
8


Simplify (remember when
adding only add coefficients of
like terms)
1.
2.
3x (2x  5x  2)
3
2
2x(4x 1)
4x(x  5)  (x 2  6x  8)


4.
4
3
8x 2  2x

3.
6x 15x  6x
5
4x 5 (2x 3  6x) 12x 6

4 x 2  20x  x 2  6x  8
5x 2  26x  8
8x 8  24 x 6 12x 6
8x 8  36x 6
Division Rule: If the bases are the
same subtract the exponents
m
x
mn
x
n
x
x 5 x x x x x
2


x
x3
x x x


5
OR
x
53
2

x

x
x3
Always do top
exponent minus
bottom exponent
Examples
1.
2.



3.
x2y6
xy 2
6a 7b 3
2ab 2

x 5  x8
x3

x
21
y
62
 xy
3a 71b 32  3a 6b
4
Divide coefficients,
subtract exponents of
the like bases.
x 58 x13
133
10
Use multiplication rule


x

x
3
3
x
x
on top, then use
division rule
Special Cases ( zero power):
any base raised to a power of
zero equals 1
x 1
0
Here is why, when the number in the numerator is the same as
the number in the denominator, the quotient is always 1.
2 42 * 2 * 2 * 2

1
4
2
2*2*2*2
So it makes
sense that
24
44
0

2

2
1
4
2
Examples
Simplify
3 4
1.
2.



ab
3
ab
a
10x 4 y 3
5x 4 y 2
2x 44 y 32  2x 0 y1  2 *1y  2y
33 41
b
 a b  1b  b
0 3
3

3.
m10 n 5

10 5
m n

1010
m
n
55
 m n  1*1  1
0
0
3
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