Download Lesson 4.6 Applying the Exponent Laws Exercises (pages 241–243

Lesson 4.6
Applying the Exponent Laws
A
3. Use the product of powers law:
When the bases are the same, add the exponents.
a) x3  x 4  x3  4
 x7
b) a 2  a 5  a 2   5
 a 3
1
 3
a
Write with a positive exponent.
c) b 3  b5  b 3  5
 b2
d) m2  m 3  m 2   3
 m1
1

m
Write with a positive exponent.
4. a) 0.52  0.53
Use the product of powers law:
When the bases are the same, add the exponents.
0.52  0.53  0.52  3
 0.55
b) 0.52  0.53
Use the product of powers law:
When the bases are the same, add the exponents.
0.52  0.53  0.52  ( 3)
 0.51 Write with a positive exponent.
=
1
0.5
Exercises (pages 241–243)
c)
0.52
0.53
Use the quotient of powers law:
When the bases are the same, subtract the exponents.
0.52
 0.52  3
0.53
 0.51 Write with a positive exponent.

d)
1
0.5
0.52
0.53
Use the quotient of powers law:
When the bases are the same, subtract the exponents.
0.52
 0.52  ( 3)
3
0.5
 0.55
5. Use the quotient of powers law:
When the bases are the same, subtract the exponents.
x4
 x4  2
a)
x2
 x2
b)
x2
 x2  5
5
x
 x 3 Write with a positive exponent.
=
c)
1
x3
n 6  n5  n 6  5
 n1
n
d)
a2
 a2  6
a6
 a 4 Write with a positive exponent.
=
1
a4
6. Use the power of a power law: multiply the exponents
a)
n 
2 3
 n(2)(3)
 n6
b)
z 
2 3
 z (2)( 3)
 z 6
1
z6
=
c)
n 
4 3
Write with a positive exponent.
 n( 4)( 3)
 n12
d)
c 
2 2
 c ( 2)(2)
 c 4
Write with a positive exponent.
1
c4
=
7. Use the power of a power law: multiply the exponents
4
(3)(4)
 3 3 
 3
a)      
5
 5  
12
3
 
5
 3 3 
b)   
 5  
4
 3
 
5
(3)( 4)
3
 
5
12
Write with a positive exponent.
12
5
= 
3
c)
 3 3 
  
 5  
4
 3
 
5
( 3)( 4)
12
3
 
5
 3 3 
d)    
 5  
4
 3
  
 5
( 3)( 4)
12
 3
  
 5
8. a) Use the power of a quotient law.
2
a2
a

 
b2
b
b) Use the power of a quotient law.
n2 

 n2 
  
m3
m
3
3
Use the power of a power law.
n(2)(3)
m3
n6
 3
m

 c2 
 2
d 
4
 d2 
  2
c 
d 

c 
4
Writing with a positive exponent
2 4
2 4
d (2)(4)
c (2)(4)
d8
 8
c

c)
Using the power of a quotient law
Using the power of a power law
d) Use the power of a quotient law.
 2b 
 2b 
  
2
 5c 
 5c 
2
2
22 b 2
52 c 2
4b 2

25c 2

Use the power of a product law.
e)
 ab 
2
 a 2b 2
Using the power of a product law
 n m   n   m
2
3
2 3
3
Using the power of a product law
 n(2)(3) m3
6
=nm
f)
c d 
3
2 4
Using the power of a power law
3
  c3    d 2 
4
4
Using the power of a product law
 c (3)( 4) d (2)( 4)
Using the power of a power law
= c 12 d 8
Write with positive exponents.
=
g)
1
c d8
12
 xy    x    y 
1 3
1 3
3
Using the power of a product law
 x3 y ( 1)(3)
Using the power of a power law
= x3 y 3
Write with a positive exponent.
3
=
h)
B
9. a)
x
y3
x 3  x 4  x 3  4
Using the product of powers law
 x1
x
b)
a 4  a 1  a 4  ( 1)
a

c)
5
Write with a positive exponent.
1
a5
b4  b3  b2  b4  ( 3)  2
b
d)
Using the product of powers law
Using the product of powers law
3
m8  m2  m6  m8  ( 2)  ( 6)
m
1
0
Using the product of powers law
e)
x 5
 x 5  2
x2
 x 7

f)
g)
1
x7
Using the quotient of powers law
b 8
 b 8  ( 3)
b 3
 b 5
Using the quotient of powers law
Write with a positive exponent.
1
b5
t 4
 t 4  ( 4)
t 4
 t0
=1
3
10. a)
Write with a positive exponent.
s5
 s 5  ( 5)
s 5
 s10
=
h)
Using the quotient of powers law
1
Using the quotient of powers law
3
1.5 2  1.5 2  1.5 2
 1.5

1
2
Using the product of powers law
4
2
 1.52
 2.25
3
 3 4
b)  
4
5
3
 3 4  3 4
   
4
4
8
 3 4
 
4
3
 
4
9

16
2

5
4
Using the product of powers law
1
c)
5
1
 0.6  3   0.6  3   0.6  3
  0.6 
6
3
  0.6 
2

5
3
Using the product of powers law
 0.36
4
 4 3
d)  
5
4
 
5

4
3
4
 4 3
 
5
4
 
5
1
e)
0.6
0.6
1
2
3
2
1
 0.6 2
 0.6


3
2
 4
  
 3
Using the product of powers law
0
Using the quotient of powers law
2
2
 0.61
Write with a positive exponent.
1
0.6
1

6
10
10

6
5

3

f)
2
 3 3
2  1
  
 
 8     3 3  3


1

8
 3 3 
 
 8
3
 3 3
  
 8
1
 3
  
 8
3

8
Using the quotient of powers law.
g)
5
5
4
0.49 2
2

0.49
4
0.49
5


3
2
 0.49 2
 0.49
8
2
Write with a positive exponent.
1


Using the quotient of powers law
3
0.49 2
1

0.49

3
1
0.73
1

0.343
1

343
1000
1000

343

5
h)
0.027 3
0.027
4
3
5
 0.027 3

4
3
Using the quotient of powers law
1
 0.027 3
 3 0.027
 0.3, or
11. a)
x
1
y 2 
3
3
10
  x 1    y 2 
3
3
= x ( 1)( 3)  y ( 2)( 3)
= x3  y 6
 x3 y 6
Using the power of a product law
Using the power of a power law
b)
c)
 2a

2 2 2
b
 4m n 
2
3 3
  21    a 2    b 2 
2
2
2
Using the power of a product law
= 2(1)( 2)  a ( 2)( 2)  b(2)( 2)
Using the power of a power law
= 22  a 4  b 4
Write with positive exponents.
4

a
2  b4

a4
4b 4
2
  41    m 2    n3 
3
3
3
= 4(1)( 3)  m(2)( 3)  n(3)( 3)
3
6
=4 m n
1
 3
4  m6  n9

9
Using the power of a product law
Using the power of a power law
Write with positive exponents.
1
64m6 n9
d)
 3 2 3 
 m n 
2

4
4
 31 
4
4
     m 2    n 3 
2 
3
= 
2
Using the power of a product law
(1)( 4)
 m( 2)( 4)  n( 3)( 4)
Using the power of a power law
4
3
=    m8  n12
2
4
2
    m8  n12
3
16

 m8  n12
81
16m8 n12

81
Write with a positive exponent.
12. The volume of a cone with base radius r and height h is given by the formula:
1
V  πr 2 h
3
The cone has equal height and radius.
So, substitute: r = h
1
V  πh 2 h
3
1
 πh 2 h1
Use the product of powers law.
3
1
 πh 2  1
3
1 3
 πh
3
Substitute V = 1234, then solve for h.
1
1234  πh3
Multiply each side by 3.
3
1

3 1234   3  πh3 
3

3
3702  πh
Divide each side by π.
3702 πh3

π
π
3702
 h3
π
To solve for h, take the cube root of each side
by raising each side to the one-third power.
1
3
 3702 
3 3

  h 
 π 
1
1
Use the power of a power law.
 3702  3
Use a calculator.

 h
 π 
h  10.5623...
The height of the cone is approximately 10.6 cm.
4
13. The volume, V, of a sphere with radius r is given by the formula: V  πr 3
3
Substitute V = 375, then solve for r.
4
375  πr 3
Multiply each side by 3.
3
4

3  375   3  πr 3 
3

3
1125  4πr
Divide each side by 4π.
1125 4πr 3

4π
4π
1125
 r3
4π
To solve for r , take the cube root of each side
by raising each side to the one-third power.
1
3
 1125 
3 3

  r 
 4π 
1
Use the power of a power law.
1
 1125  3

 r
 4π 
r  4.4735...
Use a calculator.
The surface area, SA, of a sphere with radius r is given by the formula:
Substitute: r  4.4735...
SA  4πr 2
 4π(4.4735...) 2
 251.4808...
The surface area of the sphere is approximately 251 square feet.
a b 
a b
2 1 2
14. a)
3
3

a 2( 2)  b( 1)( 2)
a ( 3)(3)  b(1)(3)
Using the power of a power law
=
a 4  b 2
a 9  b3
Use the quotient of powers law.
 a 4  ( 9)  b 2  3
 a 5  b 1

5
a
b
Write with a positive exponent.
b)
  c 3 d 1 


 c2 d 


2
 c ( 3)( 1)  d (1)( 1) 


c2d


 c3  d 1 
= 2

 cd 
Using the power of a power law inside the large brackets
2
Use the quotient of powers law.
  c3  2  d 1  1 
  c1  d 2 
2
2
2
Use the power of a power law.
 c1( 2)  d ( 2)( 2)
 c 2 d 4
Write with a positive exponent.
4
=
d
c2
15. a) Simplify first.
 a3b2  a 2b3   a3  b2  a 2  b3
 a 3  a 2  b 2  b3
=a
32
b
Use the product of powers law.
23
 a 5b5
Substitute: a = –2, b = 1
a 5b5  (2)5 1
5
 (32)(1)
 32
b) Simplify first.
 a 1b2  a 2b3   a1  b2  a2  b3
 a 1  a 2  b 2  b 3
=a
1  ( 2)
3 5
a b
1
ab
Substitute: a = –2, b = 1
1
1

3 5
ab
(2)3 (1)5
1

(8)(1)
1

8
=
c) Simplify first.
3 5
b
Use the product of powers law.
2  ( 3)
Write with positive exponents.
a 4b5
 a 4  1  b5  3
ab3
= a 5  b 2
Using the quotient of powers law
Write with a positive exponent.
2
b
a5
Substitute: a = –2, b = 1
b2
12

a 5  2 5
=

1
32
d) Simplify first.
 a 7 b7 
 9 10 
a b 
5
  a 7  ( 9)  b7  10 
=  a 2  b 3 
 a2 
= 3
b 
5
 b3 
= 2
a 
b3(5)
= 2(5)
a
b15
 10
a
5
5
5
Using the quotient of powers law inside the brackets
Write the expression inside the brackets with a positive exponent.
Write with a positive exponent.
Use the power of a power law.
Substitute: a = –2, b = 1
b15
115

10
10
a
 2 

2
1
1024
4
2
16. a) m 3  m 3  m 3
=m
6
3
 m2

4
3
Using the product of powers law
b) x

3
2
x

1
4

x
=x
x

3  1
  
2  4
Using the quotient of powers law
6  1
   
4  4

5
4
Write with a positive exponent.
1
5
x4
c)
3
9a 4b 4
2
3a b
1
4
3
9 a 4 b 4


 1
3 a2
b4
Use the quotient of powers law.
3
=  3  a 4  2  b 4

1
4
2
=  3  a 6  b 4
1
  3  a 6  b 2

Write with a positive exponent.
1
2
3b
a6
d)
1
1
 64c 6  3 
3

   64  1  1  c 6 
1

 9  12 


a9
b 2
ab 


Simplify inside the brackets first.
1
1

3
=  64  a 9  b 2  c 6 


1
3
 (64)  a
1
9 
 3
1
3
b
 1  1 
  
 2  3 
Use the power of a power law.
c
1
6 
 3
1
6
 (64)  a 3  b  c 2
1
  4  a 3  b 6  c 2
1
4b 6 c 2
=
a3
Write with a positive exponent.
17. a) In the second line, the exponents were multiplied instead of added.
A correct solution is:
1
 1 
 x2 y 3   x 2 y 1   x2  x 2  y 3  y 1 Use the product of powers law.


x
2
1
2
4
1
2
 x2

 y 3  ( 1)
 y 3  ( 1)
5
2
 x y 4
Write with a positive exponent.
5
2
=
x
y4
b) In the first line, –5 was multiplied by –2 instead of being raised to the power –2.
A correct solution is:
 5a 2 


 12 
 b 
2
 5

b



2
 a 2( 2)
1
   2 
2
 5
2
 a 4
b 1
Using the power of a power law
Write with positive exponents.
b1
 5
2
 a4
b
25 a 4
18. I record the volume of water in the measuring cylinder, in millilitres. Then I carefully place the
marble in the cylinder. The water level rises; this is the total volume of the water and the marble. I
record this volume. I subtract the volume of the water alone from the total volume of the water and
the marble to determine the volume of the marble in millilitres. Since
1 mL = 1 cm3, I can write the volume of the marble in cubic centimetres.
4
The volume, V, of a sphere with radius r is given by the formula: V  πr 3
3
I substitute the volume of the marble in cubic centimetres for V.
I then multiply both sides of the equation by 3, then divide both sides by 4π . To solve for r,
1
I raise each side to the one-third power. I then use the power of a power law to write  r 3  3
as r. Once I have determined r, I multiply the radius by 2 to get the diameter of the marble.
19. a) There are two errors in the first line: the quotient of powers law was used before the power of a
power law, and when using the quotient of powers law, the powers outside the brackets were
subtracted. There is an error in the second line: when using the power of a power law, the product
of 5 and –6 should have been –30.
A correct solution is:
m
m
3
2
 n2 
n
4

3 2

m( 3)( 4)  n 2( 4)
m 2(2)  n 3(2)
Using the power of a power law

m12  n 8
m 4  n 6
Use the quotient of powers law.
= m12  4  n 8  ( 6)
 m8 n 2

Write with a positive exponent.
m8
n2
b) There is an error in the first line: when using the power of a power law, the exponents
added when they should have been multiplied.
A correct solution is:
1
 12  23  2
r  s 


1
  

 


 
 1 1 
  
   
    1
   1
  r 4  s 2   r  2  2   s  2  2   r  4   s  2 


1
1
1
3
 r4  s
r
1 1

4 4
s
1
 r4  s
3 2

4 4
1

5
4
1
2
5
s4
1

1
2
1
Using the power of a power law
Use the product of powers law.
3 1
 
4 2

 r2  s
r
3
4
2
4
r s


1
were
Write with a positive exponent.
20. Width, in metres, of a piece of An paper: 2

2n  1
4

2n  1
Length, in metres, of a piece of An paper: 2 4
a) i) To determine the dimensions of a piece of A3 paper, substitute: n = 3
Width: 2

2n  1
4
2
2


2(3)  1
4

7
4
Write with a positive exponent.
1
7
24
This is the width in metres. Since 1 m = 1000 mm, to write the width in millimetres, multiply
by 1000.
1
1000
 1000 
7
7
24
24
1000
So, the width of the paper is 7 mm.
24
1000
Use a calculator to evaluate:
= 297.3017…
7
4
2
The width of a piece of A3 paper is about 297 mm.
Length: 2

2n  1
4
2
2


2(3)  1
4

5
4
Write with a positive exponent.
1
5
24
This is the length in metres. Since 1 m = 1000 mm, to write the length in millimetres, multiply
by 1000.
1
1000
 1000 
5
5
24
24
1000
mm.
So, the length of the paper is
5
4
2
1000
Use a calculator to evaluate:
= 420.4482…
5
4
2
The length of a piece of A3 paper is about 420 mm.
1000
1000
A piece of A3 paper has dimensions 5 mm by 7 mm, or 297 mm by 420 mm.
24
24
ii) To determine the dimensions of a piece of A4 paper, substitute: n = 4
Width: 2

2n  1
4
2
2

2(4)  1
4

9
4
Write with a positive exponent.
1

9
24
This is the width in metres. To write the width in millimetres, multiply by 1000.
1
1000
 1000 
9
9
24
24
1000
So, the width of the paper is 9 mm.
24
1000
Use a calculator to evaluate:
= 210.2241…
9
4
2
The width of a piece of A4 paper is about 210 mm.
Length: 2

2n  1
4
2
2

2(4)  1
4

7
4
Write with a positive exponent.
1

7
24
From part i,
1000
7
4
= 297.3017…
2
The length of a piece of A4 paper is about 297 mm.
1000
1000
A piece of A4 paper has dimensions 7 mm by 9 mm, or 210 mm by 297 mm.
24
24
iii) To determine the dimensions of a piece of A5 paper, substitute: n = 5
Width: 2

2n  1
4
2
2


2(5)  1
4

11
4
Write with a positive exponent.
1
11
24
This is the width in metres. To write the width in millimetres, multiply by 1000.
1
11
4
 1000 
1000
11
24
2
So, the width of the paper is
Use a calculator to evaluate:
1000
11
4
2
1000
11
4
mm.
= 148.6508…
2
The width of a piece of A5 paper is about 149 mm.
Length: 2

2n  1
4
2
2


2(5)  1
4

9
4
Write with a positive exponent.
1
9
24
1000
From part ii, 9 = 210.2241…
24
The length of a piece of A5 paper is about 210 mm.
1000
1000
A piece of A5 paper has dimensions 9 mm by 11 mm, or 149 mm by 210 mm.
24
24
b) i) The paper is folded along a line perpendicular to its length.
So, one dimension of a folded piece of A3 paper, in millimetres,
half the original length:
1000 1
1000
  5
5
2
24
2 4  21
1000
 5
1
24
1000
 5 4

24 4
1000
 9
24
is one-
The other dimension is the original width:
1000
2
7
4
So, a folded piece of A3 paper has dimensions
mm
1000
2
7
4
mm by
1000
2
9
4
mm.
ii) The paper is folded along a line perpendicular to its length.
So, one dimension of a folded piece of A4 paper, in millimetres,
is one-half the original length:
1000 1
1000
  7
7
2
24
2 4  21
1000
 7
1
24
1000
 7 4

24 4
1000
 11
24
1000
The other dimension is the original width: 9 mm
24
1000
1000
So, a folded piece of A4 paper has dimensions 9 mm by 11 mm.
24
24
iii) The paper is folded along a line perpendicular to its length.
So, one dimension of a folded piece of A5 paper, in millimetres,
is one-half the original length:
1000 1
1000
  9
9
2
24
2 4  21
1000
 9
1
24
1000
 9 4

24 4
1000
 13
24
1000
The other dimension is the original width: 11 mm
24
1000
1000
So, a folded piece of A5 paper has dimensions 11 mm by 13 mm.
24
24
c) I noticed that a piece of A4 paper has the same dimensions as a folded piece of A3 paper, and a
piece of A5 paper has the same dimensions as a folded piece of A4 paper.
C
21. a)
4
1
 a  3 4  b1 4   

 a 3b   c5 
c5 1






 2   4 3 
2 4 
4 1
3 1 





c
b
 c  a b 

 a

 a12  b 4   c 5 
 
   4 3 
8
 c
 a b 
 a12b 4 c8  a 4b 3c 5
12  4 4  3
=a
b
c
Using the power of a power law
Rewrite without using fractions.
Use the product of powers law.
85
 a16b 7 c3
Write with a positive exponent.
16 3
=
b)
 2a b c 
 4a bc 
1 4 3 2
2
4 2

a c
b7
22  a ( 1) 2  b 4 2   c ( 3) 2 
42  a 2 2  b 2  c ( 4) 2
Using the power of a power law

22  a 2  b 8  c 6
16  a 4  b 2  c 8
Rewrite 16 as a power of 2.

22  a 2  b 8  c 6
24  a 4  b 2  c 8
Use the quotient of powers law.
 22  4  a 2  4  b 8  2  c 6   8
 26  a 2  b 10  c14
14
c
2 a 2 b10
c14

64a 2b10

6
Write with positive exponents.
22. a)
2
2
1
 12 23 
 2
 2
3
2
x
y

x

y




4
Using the power of a power law
2
Substitute: x = a 2 , y  a 3
 x  y3
4
 a 2
 2 3
  a3 
 
Use the power of a power law.
2 4

3
 a 2  a 3
8
 a 2  a 9
a
a
2 
8
9

18 8

9 9

10
a 9
1
 10
a9
Use the product of powers law.
Write with a positive exponent.
b)

x  y

3
4
1

2
 3 
  x4 
  1 
  y2 


3
3

x
y
Rewriting as a fraction
3
(3)
4
Using the power of a power law
 1
  3
 2
9

2
x4
y

Substitute: x = a 2 , y  a 3
3
2
9
2 4

a 
 
a 
 
2
3

a
 2 

Use the power of a power law.
3
2
9
4
2  3
  
 2
a3

9
a 2
 1
a
a
a
a


9
  1
2

9 2

2 2

7
2
Use the quotient of powers law.
Write with a positive exponent.
1
7
a2
23. a) The product of powers law is: xm  xn  x m  n
3
So, for the product to be x 2 , I need to find two exponents whose sum is
For example:
1
1
x 2  x1  x 2
3
 x2

2
2
1
5
1
x4  x4  x4
6
 x4
3
 x2

5
4
11
1
11
x 8  x8  x 8

12
 x8
3
 x2
1
8
3
.
2
b) The quotient of powers law is: xm  xn  xm  n
3
So, for the quotient to be x 2 , I need to find two exponents whose difference is
For example:
1
x2  x 2  x
2
1
2
5
5
1
5

x 2  x1  x 2
3
 x2
 x2
7
7
2
7

x 2  x2  x 2
2
2
 x2
3
4
2
3
 x2
 x2
c) The power of a power law is:  x m   x mn
n
3
So, for the result to be x 2 , I need to find two exponents whose product is
For example:
x
1
3 2

x
1
3 
2
3
 x2
9
  
 13  2
  
 3  2 
x

x
 
 
1
x
x
9
1
 x9  6  x
1
9 
6
9
9
6
 x6
3
2
 x2
3
3
.
2
3
.
2
1
1
1
3  1 2
AC AD
 2 2
 1 2
24. In

, substitute AB =   , AC =   , AE =   , then solve for AD.
2 3
AB AE
3
3
1
3  1 2
 
2 3
2
 
3
1
2
1

 3  1 2
 
1  2 3
  
1
 3    2 2
 3
  
1
2
1
1

 1 2
Multiply both sides by   .
3
AD
1
 1 2
 
 3



1 
  1  2  AD
  
1
  3    1 2

  3

 







1
3  1 2  1 2
   
2 3  3
1
 AD
Use the product of powers law.
 2 2
 
3
1
3  1 2
 
2 3

1
2
 AD
1
2
2
 
3
31
 
2 3
 AD
1
2
2
 
3
1
2  AD
2
 
3
 1  2 
  
 2  3 

1
2
1
2
1
 AD
 1  3  2
    AD
 2  2 
Use a calculator to evaluate:
1
 1  3  2
AD    
 2  2 
 0.6123...
Use a negative exponent to move the denominator to the numerator.
Write with a positive exponent.
1
 1  3  2
The length of AD is    cm, or approximately 0.6 cm.
 2  2 