Integer Exponents
Lecture Notes
Simplify each of the following.
without negative exponents.
2.
Assume that all variables represent positive numbers.
2
1. 3
15.
1
4
3. m
1
4.
x 5
5. a8 a
6. p3 p
7.
8.
17.
1
7
18.
11.
28.
18q 3
6q 3
2a
3
3p3 q 5
(2q 0 p3 )
2a
29.
2
1
1 b) 3
22 (a
x3 y 0 x
y 3
5
30.
5
31.
x3 y 7 x
y 3
3
20. (2y)
9
50a12
10a 3
2
32.
a3 b 5
22.
a 2 b3
x0
x0
13. b
3
3
5
21.
23. 3m3
2
24.
3
5
b2
b
1
5
(b ) (b2 ) (b
1
1)
c copyright Hidegkuti, Powell, 2011
25.
26.
2ab
k3
(k
2
2 b3
4
x
x
4
2b
2a
3
2
3
19. 2y
12. ( x)0
14.
27.
Present your answer
5
p8
t 3
9. 4
t
10.
2
m
m
x3 y 5
16.
z 4
3
2
page 1
1
x
x
+y
2
y
2a
33.
2
2
0
1
2
2 3 0
b a
2a2 ( 2a
3
a2 b
34.
3
5
1a
b7 ( ab2 )
3
aba
2b 2
3
2 b) 2 ab0
!
2
5 )2
2a 3 b5
3a3 b 2
2
a3 b
5
3
35.
x
2
2
yx5 (y
y 3 x0
2yx0 y
2x 2 0
2 x) 3 (2x 1 yx3 ) 1
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
page 2
Answers
1.
1
9
2. 8
3.
1
m4
4. x5
5.
a7
6.
p4
7. x5
8.
5a15
9.
1
t7
10. 1
11.
17. 3q 6
23.
18.
27
8
24.
19.
2
y3
1
12. 1
13.
1
b4
14. b4
15. m3
16.
x3 z 4
y5
c copyright Hidegkuti, Powell, 2011
20.
1
8y 3
22.
a5
b8
b9
8a3
25. k
30.
32.
a5
33.
xy
y
x
b8
2
8b4
28. 18p9 q 10
29.
x4
y6
31. 1
9
26. a3 b
4
27.
25
21.
9
1
9m6
4a10
b12
34.
1
b8
35.
2x4
y3
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
page 3
Solutions
Simplify each of the following.
without negative exponents.
Assume that all variables represent positive numbers.
2
1. 3
Solution: We just apply the rule a
n
=
1
.
an
2
3
2.
Present your answer
=
1
1
=
2
3
9
1
2
3
Solution: We apply the rule a
n
=
1
.
an
1
2
3
1
1
23
=
=
1
1
8
To divide is to multiply by the reciprocal:
1
1
8
This is true in general:
1
a
n
1
n
=
1
1
an
=1
an
= an
1
4
Solution: We apply the rule a
n
=
1
.
an
m
4.
8
=8
1
= an
a
3. m
=1
4
=
1
m4
1
x 5
Solution: We have already proven that
1
a
n
= an
1
= x5
x 5
5. a8 a
1
Solution 1: We can apply the rule an am = an+m
Solution 2: We can apply the rule a
a8 a
c copyright Hidegkuti, Powell, 2011
1
n
a8 a 1 = a8+( 1) = a7
1
an
= n and then the rule m = an
a
a
= a8
1
a8 1
a8
a8
=
=
=
= a8
a1
1 a
a
a1
1
m.
= a7
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
6. p3 p
7
page 4
p8
Solution 1: We can apply the rule an am = an+m
p3 p
Solution 2: We can apply the rules a
p3 p
7.
x
x
7
p8 = p 3
n
=
7
7)+8
= p4
n
1
n am = an+m and a = an
and
a
an
am
m.
1
p3 1 p8
p3 p 8
p3+8
p11
8
p
=
=
=
=
= p11
p7
1 p7 1
p7
p7
p7
7
= p4
4
9
Solution 1: We can apply the rule
an
= an
am
Solution 2: We can apply the rules a
x
x
4
n
=
m.
=x
9
4 ( 9)
4+9
=x
an
1
and
= an
an
am
4
x
x
8.
p8 = p3+(
9
x9
= x9
x4
=
4
= x5
m.
= x5
50a12
10a 3
Solution 1: We can apply the rule
an
= an
am
m.
50a12
= 5a12
10a 3
Solution 2: We can apply the rules a
n
=
( 3)
= 5a12+3 = 5a15
1
an
and
= an
an
am
m.
50a12
50a12 a3
=
= 5a12+3 = 5a15
10a 3
10
9.
t 3
t4
Solution 1: We can apply the rules
an
= an
am
m
and then a
t 3
=t
t4
Solution 2: We can apply the rule a
n
=
3 4
=t
7
n
=
=
1
.
an
1
t7
1
and then an am = an+m .
an
t 3
1
1
= 4 3 = 7
4
t
t t
t
c copyright Hidegkuti, Powell, 2011
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
page 5
10. x0
Solution: There is a separate rule stating that as long as x is not zero, then x0 = 1. So the answer is 1.
11.
x0
Solution: This is the opposite of x0 and so the answer is
x0 =
1 x0 =
1.
1 1=
1
12. ( x)0
Solution: This is again 1 because any non-zero riased to the power zero is 1.
13. b
5
b2
b
1
Solution 1: We can apply the rules an am = an+m and then a
5
b
Solution 2: We can apply the rule a
b
14.
5
b2
b
1
b2
n
=
b
1
=b
5+2+( 1)
1
.
an
=
4
=b
=
1
b4
1
and then just cancel.
an
1 2 1
1 b2 1
b2
b/ b/
1
b
=
=
=
= 4
b5
b1
b5 1 b1
b6
b/ b/ b b b b
b
=
1
(b
5 ) (b2 ) (b 1 )
Solution 1: We can apply the rules an am = an+m and then a
1
5
(b ) (b2 ) (b
1)
Solution 2: We can apply the rule a
n
=
=
1
5+2+( 1)
b
(b
m
m
=
1
b
4
n
=
an
1
and
then
= an
an
am
1
15.
n
5 ) (b2 ) (b 1 )
=
1
.
an
=
1
1
b4
=1
b4
= b4
1
m.
b5 b1
b6
= 2 = b6
2
b
b
2
= b4
2
5
Solution 1: We can apply the rules
an
= an
am
Solution 2: We can apply the rule a
m
m
2
n
=
5
=m
and then a
n
2 ( 5)
2+5
=m
=
1
an
and
then
= an
an
am
m
m
c copyright Hidegkuti, Powell, 2011
m
2
5
=
m5
= m5
m2
2
1
.
an
= m3
m.
= m3
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
16.
page 6
x3 y 5
z 4
Solution: Each variable occurs only once and so this problem is just about bringing it to the form required.
1
1
We can apply the rule a n = n . We hve alread shown that
= an .
a
a n
x3 y 5
x3 z 4
=
z 4
y5
17.
18q 3
6q 3
an
= an
am
Solution 1: We can apply the rule
m.
18q 3
6/ 3q 3 (
=
6q 3
6/ 1
n
Solution 2: We can apply the rules a
3)
=
3q 3+3
= 3q 6
1
1
and then an am = an+m .
an
=
18q 3
6/ 3q 3 q 3
=
= 3q 6
6q 3
6/ 1
18.
3
2
3
n
Solution: We can apply the rule a
2
3
=
3
1
.
an
1
=
2
3
a
b
Note that we basically proved here that
19. 2y
=
3
n
1
=
2 2 2
3 3 3
=
b
a
1
8
27
=1
27
27
=
8
8
n
.
3
Solution: We can apply the rule a
and not 2y.
n
=
1
. It is important to note that the base of exponentiation is y
an
2y
20. (2y)
3
=2
1
2 1
2
=
= 3
3
3
y
1 y
y
3
Solution: We can apply the rule a
the rule (ab)n = an bn .
n
=
1
. This time the base of exponentiation is 2y. So we will apply
an
(2y)
c copyright Hidegkuti, Powell, 2011
3
=
1
1
1
3 = 23 y 3 = 8y 3
(2y)
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
2
3
5
21.
n
Solution 1: We can apply the rule a
3
5
2
1
=
2
3
5
=
1
.
an
1
=
a
b
n
2
3
5
=
3
5
3
5
Solution 2: We proved previously that
22.
page 7
=
5
3
=
b
a
1
3
5
1
9
25
=
3
5
25
25
=
9
9
=1
n
. Using that,
2
5
3
=
5
3
=
25
9
a3 b 5
a 2 b3
Solution 1: We can apply the rule
a3 b 5
= a3
a 2 b3
( 2)
an
= an
am
5 3
b
Solution 2: We can apply the rules a
n
m
and then a
= a3+2 b
=
5 3
n
= a5 b
=
8
1
.
an
1
a5 1
a5
=
=
b8
1 b8
b8
= a5
1
and an am = an+m .
an
a3 b 5
a3 a2
a5
=
=
a 2 b3
b3 b 5
b8
2
23. 3m3
Solution: We can apply the rule a
3m3
24.
2ab
3
n
2
1
and then (ab)n = an bn and also (an )m = anm .
an
1
1
1
1
=
2 = 2
2 = 9m3 2 = 9m6
3
3
(3m )
3 (m )
=
3
Solution: We can apply the rule (ab)n = an bn and then (an )m = anm .
2ab
We now apply a
n
=
3
3
k3
(k
3
a
3
b
3
3
= ( 2)
3
a
3
b
3( 3)
3
= ( 2)
a
3 9
b
1
.
an
( 2)
25.
= ( 2)
3
3 9
a
b =
1
1 9
1
1 b9
=
3 a3 b =
8 a3 1
( 2)
b9
=
8a3
b9
8a3
3
5 )2
Solution: We can apply the rule (an )m = anm and then
k3
(k
c copyright Hidegkuti, Powell, 2011
3
5 )2
=
k
k 3( 3)
=
5
2
k
k
9
10
=k
an
= an
am
9 ( 10)
=k
m.
9+10
= k1 = k
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
26.
2
2a 3 b5
3a3 b 2
a3 b
page 8
3
5
Solution:
2a 3 b5
3a3 b 2
2
2a
6 b5+2
2
2a
6 b7
E =
=
=
=
=
=
=
=
=
=
=
27.
2a
3
2a
3
a b
5
a3 b
5
3
2
a3 b
3
a3 b
( 3)2
(2a
6 b7 )2
a3 b
3
5
a
b
n
3
apply
n
=
an
= an
am
m
n
b
a
an
bn
=
apply (ab)n = an bn and a
1
n
=
1
an
apply (an )m = anm and (ab)n = an bn
1
n
=
apply a
n
apply a
(a3 )3 (b 5 )3
9a12
1
4b14 a3 3 b( 5)3
9a12
1
4b14 a9 b 15
9a12 b15
9a12 b15
=
4b14 a9
4b14 a9
12
9
15
14
9a
b
9a3 b1
9
=
= a3 b
4
4
4
2b
a3 b
a
b
apply
(a3 b 5 )3
12 b14
2
apply
9
4a
3
!
3
22 (a 6 )2 (b7 )2
9
=
3
5
5
3 3 b5 ( 2)
2a
3
5
2
3
2a 6 b7
3
apply
1
and (ab)n = an bn
an
=
an
= an
am
1
an
m
4
Solution:
E =
2a
=
2
=
=
=
=
=
3
1
a3
2a
2
b
4
1
( 2a
2 b)4
1
2 1
1 a3 ( 2)4 (a
2
1
3
a
16a 8 b4
2 a8
a3 16b4
2a8
2a8
=
a3 16b4
16a3 b4
8
3
a5
1 2a
/
=
8 2b
/4
8b4
c copyright Hidegkuti, Powell, 2011
apply a
2 )4 b4
n
=
1
an
apply (ab)n = an bn
apply (an )m = anm
apply a
apply
n
=
an
= an
am
1
an
m
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
28.
page 9
2
3p3 q 5
1
(2q 0 p3 )
Solution:
E =
2
3p3 q 5
(2q 0 p3 )
=
3p3 q 5
2
2 1p3
=
3p3 q 5
2
2p3
2
= ( 3)2 p3
1
apply q 0 = 1 and
1
a
= an
n
1
apply (ab)n = an bn
2
q5
2p3
apply (an )m = anm
= 9p3 2 q 5 2 2p3
apply an am = an+m
= 18p6 q 10 p3
= 18p6+3 q 10 = 18p9 q 10
29.
2a
2 b3
22 (a
2
1 b) 3
Solution:
E =
2
2 b3
2a
apply
1 b) 3
22 (a
3
=
22 a 1 b
2a 2 b3
=
4 a 1
b
2
3
2a b
=
4a 1( 3) b
2a 2 b3
=
2a3 b 3
a 2 b3
=
2a3
=
2a3+2 b
=
2a5 b
!2
3
3
3
!2
3 3
= 4a5 2 b
= 4a10 b
4a10
1
c copyright Hidegkuti, Powell, 2011
m
2
apply (ab)n = an bn
2
=
an
= an
am
3 3 2
= ( 2)2 a5
= 4a10
=
n
!2
6 2
62
b
a
apply (an )m = anm
apply
b
n
apply (ab)n = an bn
2
( 2)
a
b
b
6 2
12
apply (an )m = anm
1
apply a n = n
a
1
b12
1
4a10
=
b12
b12
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
x3 y 0 x
y 3
30.
page 10
2
5
Solution:
x3 y 0 x
y 3
E =
x3+( 5)
y 3
=
1x
=
( 1)
x3 y 7 x
y 3
5
y 0 = 1 and an am = an+m
2
2
1x 2
y 3
=
apply
2
x 2(
( 1)2 y
a
b
n
=
an
bn
2
2
(y
=
!
2
apply (ab)n = an bn
3) 2
(y
=
31.
5
x
3)
2
2
apply (an )m = anm and a
2
2)
3( 2)
=
n
=
1
an
x4
x4
=
1y 6
y6
0
Solution: Any non-zero quantity raised to the power zero is 1: So the answer is 1.
32.
x
x
1
+y
2
y
1
2
Solution: This problem is very di¤erent because there are addition and subtraction involved. Because of
that, we can not simply move the expressions with negative exponents. Instead, this will be a problem
involving complex fractions.
E =
=
=
=
=
x 1+y 1
x 2 y 2
1
1
+ 1
1
x
y
1
1
2
x
y2
1 1
+
x y
1
1
2
x
y2
1 y
+
x y
1 y2
x2 y 2
y+x
xy
y 2 x2
x2 y 2
c copyright Hidegkuti, Powell, 2011
bring fractions to the common denominator
1
y
1
y2
x
x
x2
x2
=
x
y
+
xy xy
y2
x2
x2 y 2 x2 y 2
to divide is to multiply by the reciprocal
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
y+x
x2 y 2
xy
y 2 x2
y+x
xy
2
1
y
x2
xy (x + y)
y 2 x2
xy (x + y)
(y x) (y + x)
xy
y x
=
=
=
=
=
33.
2a
2 3 0
b a
2
2a2 ( 2a
aba
page 11
cancel out xy
factor y 2
x2 via the di¤erence of
squares theorem, cancel out x + y
3
2b 2
2 b) 2 ab0
Solution:
E =
=
=
=
=
=
=
=
=
=
=
2a
2 3 0
b a
2
2a2 ( 2a
2a
2 3
b
2
aba
a1+(
2 3
b
2
2a3 (
( 2)
2
a
1a
2a
2
2
2) b1+( 2)
2 b) 2
1b 1
2 3
b (
3
apply (xy)n = xn y n
2
1)
3
(a
2) 2 b 2
a
1
a 2( 2) b3 ( 1) 3 a 1( 3) b
2a3 ( 2) 2 a 2( 2) b 2
( 2) 2 a4 b3 ( 1) 3 a3 b3
2a3 ( 2) 2 a4 b 2
b3 ( 1) 3 b3
2b 2
( 1) 3 b3+3
2b 2
( 1) 3 b6
2b 2
b 6 b2
( 1)3 2
b8
b6+2
=
=
1 2
2
c copyright Hidegkuti, Powell, 2011
3
2 b) 2
2a3 ( 2)
( 2)
a0 = b0 = 1 and xn xm = xn+m
2 b) 2 ab0
2a2+1 ( 2a
2a
3
2b 2
3
b
1
3
apply (xn )m = xnm
1( 3)
cancel out a4 and a3 and ( 2)
2
apply xn xm = xn+m
apply x
n
=
1
xn
apply xn xm = xn+m
b8
2
Last revised: August 5, 2014
Integer Exponents
Lecture Notes
a2 b
34.
5
1a
3
b7 ( ab2 )
!
2
Solution:
a2 b
E =
a2 b
=
b7 (
35.
2
2
=
=
=
=
=
=
a
=
1b5
( 1) 3 b1
3
1
2
= b4(
2)
=b
3 b2( 3)
a
!
!
!
3
2
apply (xn )m = xnm
2
=
2
=
a2 b5 a
b7 ( 1) 3 a
cancel out a
apply a
!
2
=
1 ( 1) b5
b1
2
=
1b5
b1
2
apply
2
= b4
8
=
n
=
3
1
an
xn
= xn
xm
m
apply (xn )m = xnm
1
b8
2
y 3 x0
2yx0 y
2x 2 0
apply a0 = 1 and an am = an+m
yx5 (y 2 x) 3 (2x 1 yx3 ) 1
x
2
2
y3
apply (ab)n = an bn
yx5 (y 2 x) 3 (2x 1+3 y) 1
x
2
2
y3
apply (ab)n = an bn and an am = an+m
2 ) 3 x 3 (2x2 y) 1
x
2
2
y3
apply (an )m = anm
3) (y 2 ) 3 (2) 1 (x2 ) 1 y 1
x
yx2 y
2
1 a 3 b5
( 1) 3 b1 a 3
1 yx3 ) 1
(2x
yx5+(
apply xn xm = xn+m
3b 6
2x 2 0
2yx0 y
yx5 (y
2
5
2
1 ( 1)3 b5
b1
2
5
1( 5) a 5
b5
x
apply (xy)n = xn y n
3 (b2 ) 3
=
3
2
a
a2+( 5) b5
( 1) 3 b7+( 6) a
yx5 (y 2 x)
Solution:
=
3
1)
5
1
=
y 3 x0
E =
3
!
a2 b
b7 ( 1)
=
x
5
1a
b7 ( ab2 )
=
page 12
2( 2) y 3
2( 3) 2 1 x2(
x4 y 3
2 1 y 1+6+( 1) x2+(
x4 y 3
x4 y 3 6
=
2 1
2 1
1
4
4
2x
2 x
= 3
y3
y
x4 y 3
1) y 1
2 1 yx2 y 6 x
x4 y 3
=
2)
2 1 y 6 x0
=
2y 1
apply an am = an+m
an
= an
am
1
= n
a
x0 = 1 and
apply a
n
m
For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture
Notes. E-mail questions or comments to mhidegkuti@ccc.edu.
c copyright Hidegkuti, Powell, 2011
Last revised: August 5, 2014