Laws of exponents

Exponents

{5

exponent

3

Power

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 =1
x ×4
x ×4
x4××2
××x4×x4×x43
×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

26 × 23 = 2 6+3 = 29
= 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
2
2
= 16


Try these:
12

7.

1. 3 × 3 =
2

2

2. 52 × 54 =

3.

8.

a ×a =
5

2

4. 2 s × 4 s =
2

7

12 8

9.

5. (−3) × (−3) =
2

6.

3

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+ 2

4

a ×a = a

5+ 2

=a

1. 3 × 3 = 3 = 3 = 81
2+ 4
6

2
4
=5
2. 5 × 5 = 5
2

3.

5

2

4. 2 s × 4 s = 2 × 4 × s
2

7

5. (−3) × (−3) = (−3)
2

6.

3

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

(5 ) = 5
3

2

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

4

16
2 2
  = 4 =
81
3 3


Try these:

( )

1. 3
2. ( a

5

2 5

)

3 4

=
=

( ) =
4. ( 2 a b )

3. 2 a

s
7.   =
t2
 39 
8.  5  =
3 

2 3

2

5 3 2

5. (−3a ) =
2 2

6. ( s t

)

2 4 3

=

=

2

 st 
9.  4  =
 rt 
5 8 2
 36a b 
 =
10. 
4 5 
 4a b 
8


SOLUTIONS

( )

1. 3

2 5

( )

2. a

3 4

=

(

2

3
a12

=

( )

3. 2a

10

2 3

3

= 2 a

)

5 3 2

4. 2 a b

2×3

= 8a

6

= 2 2×2 a 5×2b 3×2 = 2 4 a10b 6 = 16a10b 6

5. (−3a ) = ( − 3) × a

2×2

6. ( s t

6 12

2

2 2

)

2 4 3

2×3 4×3

=s t

=s t

= 9a

4


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
 4a b

5 8

2


 = 9ab 3


(

)

2

2 3×2

=9 a b
2

= 81a b

2 6


#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
a 0 =1
and
(5a ) 0 =1

Try these:

1.
2.

3.

( 2a b )
2

0

=

y 2 × y −4 =

(a )

=

−2

7

5 −1

5. ( 3 x y
−2

6.

(s t

)
)

3 −4

2 4 0

=

=

−2

s t 
9.  4 4  =
s t 
−2
5
 36a 
10.  4 5  =
 4a b 
2 2

4. s × 4 s =

−1

2 
7.   =
 x −2
 39 
8.  5  =
3 
2


SOLUTIONS

(

)

0

1. 2 a b = 1
1
−2
2
−4
2. y × y = y = 2
y
1
5 −1
3. a
= 5
a
−2

7
5
4. s × 4 s = 4s
2

( )

5. ( 3 x y
−2

6. ( s t

)

)

3 −4

2 4 0

= (3 x y

= 1

−4

8

−12

x
) = 81y12
8


SOLUTIONS
−1

2 
7.  
 x 
9 −2
3 
8.  5 
3 
2

−1

x
4
= x  = 4
 

( )

= 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  −2
10
5
b
−
2
−
2
10
 36a 
9
a
b
=
2



10.  4 5  =
81
a
 4a b 
2 2