Laws of exponents

This aim to explain about different rules of exponents

1. Learning Targets:
•I can define exponent;
•I can do laws of exponents; and
•I can determined different
kinds of laws of exponents.

3. Exponents
3
5
Power
base
exponent33
means that is the exponential
form of t
Example:
he number
125 5 5
.125

5
3
means 3 factors of 5 or 5 x 5 x 5

4. 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

5. #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
2222
93636




6. #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
22
2
2
426
2
6




7. Try these:
22
33.1 
42
55.2 
25
.3 aa 
72
42.4 ss 
32
)3()3(.5 
3742
.6 tsts 
4
12
.7
s
s 
5
9
3
3
.8 
44
812
.9
ts
ts 
54
85
4
36
.10
ba
ba

8. 
22
33.1 
42
55.2 
25
.3 aa 
72
42.4 ss 
32
)3()3(.5 
3742
.6 tsts 8133
422

 725
aa
 972
842 ss
 SOLUTIONS642
55
 243)3()3(
532

 793472
tsts 


9. 
4
12
.7
s
s 
5
9
3
3
.8 
44
812
.9
ts
ts 
54
85
4
36
.10
ba
ba SOLUTIONS8412
ss
 8133
459

 4848412
tsts 
 35845
9436 abba 


10. #4: Power of a Power:If you are raising a Power to an
exponent, you multiply the exponents!
n
m mn
xx
So, when I
take a Power
to a power, I
multiply the
exponents52323
55)5( 


11. #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
nn
xy x y
So, when I take
a Power of a
Product, I apply
the exponent to
all factors of
the product.222
)( baab

12. #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
xx
yy




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







13. Try these:
5
2
3.1 
4
3
.2a 
3
2
2.3a  
2
352
2.4 ba 
22
)3(.5a 
3
42
.6ts 





5
.7
t
s 








2
5
9
3
3
.8 








2
4
8
.9
rt
st 








2
54
85
4
36
.10
ba
ba

14. 
5
2
3.1 
4
3
.2a 
3
2
2.3a  
2
352
2.4 ba 
22
)3(.5a 
3
42
.6ts SOLUTIONS10
3 12
a 6323
82 aa
 6106104232522
1622 bababa 
 
4222
93 aa
 1263432
tsts 


15. 





5
.7
t
s 








2
5
9
3
3
.8 








2
4
8
.9
rt
st 








2
54
85
4
36
10
ba
ba SOLUTIONS
622322
2
3
8199 babaab 
 2
82
2
4
r
ts
r
st








 
8
2
4
33 5
5
t
s

16. #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!93
3
1
125
1
5
1
5
2
2
3
3




and

17. #8: Zero Law of Exponents:Any base powered by zero
exponent equals one.0
1x 1)5(
1
15
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.

18. Try these:
0
2
2.1 ba 
42
.2 yy 
1
5
.3a 
 72
4.4 ss  


4
32
3.5 yx 
0
42
.6 ts 








1
2
2
.7
x 








2
5
9
3
3
.8 








2
44
22
.9
ts
ts 








2
54
5
4
36
.10
ba
a

19. SOLUTIONS
0
2
2.1ba 
1
5
.3a 
 72
4.4 ss  


4
32
3.5 yx 
0
42
.6ts 1 5
1
a 5
4s  
12
8
1284
81
3
y
x
yx 
 1

20. 








1
2
2
.7
x 








2
5
9
3
3
.8 








2
44
22
.9
ts
ts 








2
54
5
4
36
.10
ba
a SOLUTIONS4
4
1
x
x






 
8
8
2
4
3
1
33 

 
44
2
22
tsts 

 2
10
1022
81
9
a
b
ba 
