exponents and powers grade 7 - edugain · pdf fileexponents and powers ... rule 3: when you...

Grade 7Exponents and Powers

For more such worksheets visit www.edugain.com

Answer the questions

(1)

Find value of x.

(2) Anthony plants a jasmine on his 8th birthday. If the plant has one jasmine to start with, andnumber of jasmines doubles every week, then how many jasmines will be there af ter x weeks?

(3) Find number f or f ollowing expanded f orms

A) 3×104 + 0×106 + 9×100 + 0×101 + 6×100

B) 2×105 + 0×103 + 4×103 + 2×101 + 0×103 + 0×105 + 3×105 + 7×105

(4)

(5) Simplif y f ollowing and write answer in exponential f orm

A) 173

177 × 175 × 174

B) 139 × 132 × 132 × 133

133 × 138 × 135

C) 53

56 × 57 × 54 × 56 × 55

D) 74 × 76 × 73 × 74

76 × 77 × 75 × 76 × 76

(6) = ?

(7) (-4)2 × (-3)x = -48Find value of x.

(8) If x=7 and y=1, f ind the value of (xy)y

(9) What is the remainder when 250 is divided by 10?

ID : ww-7-Exponents-and-Powers [1]

(C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited

http://www.edugain.com

Choose correct answer(s) from given choice

(10) If x=1 and y=5, f ind the value of

a. 26

5 b.

26

6

c. 28

5 d.

25

5

(11) (-7)3 = ?

a. 343 b. -357

c. -343 d. -336

(12)

Find value of x.

a. 2 b. 3

c. -3 d. -2

Fill in the blanks


A)252 × 1253 = 5

B)495 × 495 ÷ 75 = 7

(14) Find value of f ollowing

A)

=

B)

=

C)

=

D)

=


A) 1258

1259 × 55 × 1252 = 5



B) 492

76 × 496 = 7

© 2016 Edugain (www.edugain.com). All Rights Reserved

Many more such worksheets can begenerated at www.edugain.com



http://www.edugain.com

Answers

(1) -1

Step 1

Rules of the exponent:Rule 1: When you multiply two numbers with the same base, you add the exponents.Rule 2: When you divide two numbers with the same base, you subtract the exponents.Rule 3: When you have an exponent expression that is raised to a power, you can multiplythe exponent and power.

Rule 4: (x)-n can be written as 1

(x)n

Rule 5: If (y)x = (y)10 then we can write x = 10.

Step 2

Now

[{( 1

5 )-1}1]x =

1

5

⇒ {( 1

5 )-1}1x =

1

(5)1

⇒ ( 1

5 )-1x =

1

(5)1

⇒ ( 1

5 )-1x = (

1

5 )1

⇒ -1x = 1⇒ x = -1

Step 3

Theref ore the value of x is -1 .



(2) 2x

Step 1

If you look at the question caref ully, you will notice that Anthony plants a jasmine on his 8th

birthday. If the plant has one jasmine to start with, and number of jasmines doubles everyweek.

Step 2

Number of jasmine at starting = 1 Number of jasmine after 1 week = 21

Number of jasmine after 2 week = 22

Number of jasmine after 3 week = 23 --------------------------------- --------------------------------- Number of jasmine after x week = 2x

Step 3

Theref ore the total number of jasmines af ter x weeks will be 2x.

(3) A) 30015

Step 1

3×104 + 0×106 + 9×100 + 0×101 + 6×100 can be solved by the f ollowing steps:

3×104 + 0×106 + 9×100 + 0×101 + 6×100

= 30000 + 0 + 9 + 0 + 6= 30015

Step 2

Now the number f or expended f orm 3×104 + 0×106 + 9×100 + 0×101 + 6×100 is30015.

B) 1204020

Step 1

2×105 + 0×103 + 4×103 + 2×101 + 0×103 + 0×105 + 3×105 + 7×105 can be solvedby the f ollowing steps:

2×105 + 0×103 + 4×103 + 2×101 + 0×103 + 0×105 + 3×105 + 7×105

= 200000 + 0 + 4000 + 20 + 0 + 0 + 300000 + 700000= 1204020

Step 2

Now the number f or expended f orm 2×105 + 0×103 + 4×103 + 2×101 + 0×103 +

0×105 + 3×105 + 7×105 is 1204020 .



(4) 64

1

Step 1

If you look at the question caref ully, you will notice that

you have to f ind out the value of ( 4

2 )3 × (

4

2 )3 .

Step 2

By solving each term separately

( 4

2 )3 =

64

8

and

( 4

2 )3 =

64

8

Step 3

Now ( 4

2 )3 × (

4

2 )3 =

64

8 ×

64

8

= 64

1

Step 4

Theref ore the value of ( 4

2 )3 × (

4

2 )3 is

64

1 .



(5) A) 175

Step 1

Rules of the exponent:Rule 1: When you multiply two numbers with the same base, you add theexponents.Rule 2: When you divide two numbers with the same base, you subtract theexponents.Rule 3: When you have an exponent expression that is raised to a power, youcan multiply the exponent and power.

Step 2

173

177 × 175 × 174 can be simplif ied as:

173

177 × 175 × 174

= 17 (3)

17 (7) × 17 (5 + 4)

= 173

177 × 179

= 173 ÷ 177 × 179

= 17 (3 - 7 + 9)

= 175

Step 3

Theref ore the value of 173

177 × 175 × 174 in exponential f orm is 175 .



B) 1310

Step 1


Step 2

139 × 132 × 132 × 133


139 × 132 × 132 × 133

133 × 138 × 135

= 13(9 + 2 + 2 + 3)

13(3 + 8) × 13(5)

= 1316

1311 × 135

= 1316 ÷ 1311 × 135

= 13(16 - 11 + 5)

= 1310

Step 3

Theref ore the value of 139 × 132 × 132 × 133

133 × 138 × 135 in exponential f orm is

1310 .



C) 519

Step 1


Step 2

53

56 × 57 × 54 × 56 × 55 can be simplif ied as:

53

56 × 57 × 54 × 56 × 55

= 5(3)

5(6) × 5(7 + 4 + 6 + 5)

= 53

56 × 522

= 53 ÷ 56 × 522

= 5(3 - 6 + 22)

= 519

Step 3


56 × 57 × 54 × 56 × 55 in exponential f orm is 519 .



D) 721

Step 1


Step 2

74 × 76 × 73 × 74

76 × 77 × 75 × 76 × 76 can be simplif ied as:

74 × 76 × 73 × 74

76 × 77 × 75 × 76 × 76

= 7 (4 + 6 + 3 + 4)

7 (6 + 7) × 7 (5 + 6 + 6)

= 717

713 × 717

= 717 ÷ 713 × 717

= 7 (17 - 13 + 17)

= 721

Step 3

Theref ore the value of 74 × 76 × 73 × 74

76 × 77 × 75 × 76 × 76 in exponential f orm is

721.



(6)

1

81

Step 1



(x)n

Step 2

Now [ { (- 1

3 )1}2]2 can be simplif ied as:

= { (- 1

3 )1} 2 × 2

={ (- 1

3 )1}4

= (- 1

3 ) 1 × 4

=(- 1

3 )4

= 1

81

Step 3

Theref ore the value of [ { (- 1

3 )1}2]2 is

1

81 .



(7) 1

Step 1

If you look at the question caref ully, you will notice that you have to f ind out the value of x

f rom (-4)2 × (-3)x = -48

Step 2

Now (-4)2 × (-3)x = -48

⇒ 16 × (-3)x = -48

⇒ (-3)x = -48

16

⇒ (-3)x = -3

⇒ (-3)x = (-3)1

⇒ x = 1

Step 3

Theref ore the value of x is 1.

(8) 7

Step 1

If you look at the question caref ully, you will notice that x = 7 and y = 1,

you have to f ind out the value of (xy)y.

Step 2

Now put the value of x and y in (xy)y

(xy)y = (7 × 1)1

= (7)1

= 7

Step 3

Theref ore the value of (xy)y at x = 7 and y = 1 is 7 .



(9) 4

Step 1

Let us try to observe the pattern of remainders when we divide the number by 10 on smallerpower ranges.Number Dividend Divisor Remainder

21 2 10 2

22 4 10 4

23 8 10 8

24 16 10 6

25 32 10 2

26 64 10 4

27 128 10 8

28 256 10 6 .......

If we see closely, we notice that when we divide 2n by 10, the remainder is repeated af ter

every 4 th power of 2.

Step 2

Hence, if we divide 50 (the power of 2 in question) by 4, the remainder will tell us whatpower to use to simplif y the problem.

Step 3

When we divide 50 by 4, we get the remainder of 2.

Theref ore we should use 22 to simplif y the problem.

Step 4

This implies that when 250 is divided by 10, we will get the same remainder as we get when

22 is divided by 10. Which is 4.

Step 5

Thus, the remainder when 250 is divided by 10 is 4 .



(10)a.

26

5

Step 1

If you look at the question caref ully, you will notice that you have to f ind out the value of (

x

y +

y

x )x at x = 1 and y = 5.

Step 2

Now put x = 1 and y = 5 in ( x

y +

y

x )x

( x

y +

y

x )x = (

1

5 +

5

1 )1

= ( 1 + 25

5 )1

= ( 26

5 )1

= 26

5

Step 3

Theref ore the value of ( x

y +

y

x )x at x = 1 and y = 5 is

26

5 .

(11) c. -343

Step 1

(-7)3 can be written as:

(-7)3 = (-7) × (-7) × (-7) = -343

Step 2

Now the value of (-7)3 is -343.



(12) d. -2

Step 1



(x)n

Rule 5: If (y)x = (y)10 then we can write x = 10.

Step 2

Now

{( 1

3 )x}-1 =

1

9

⇒ ( 1

3 )x × -1 =

1

9

⇒ ( 1

3 )(x)(-1) = (

1

32 )

⇒ ( 1

3 )(x)(-1) = (

1

3 )2

Comparing both side ⇒ (x)(-1) = 2 ⇒ x = -2

Step 3

Theref ore the value of x is -2 .



(13) A) 13

Step 1


Step 2

First of all you have to simplif y each number with the same base.

Now 252 × 1253 can be simplif ied as:

252 × 1253

= (5 × 5)2 × (5 × 5 × 5)3

= (52)2 × (53)3

= 54 × 59

= 5(4 + 9)

= 513

Step 3

Theref ore the value of 252 × 1253 in exponential f orm is 513.



B) 15

Step 1


Step 2

First of all you have to simplif y each number with the same base.

Now 495 × 495 ÷ 75 can be simplif ied as:

495 × 495 ÷ 75

= (7 × 7)5 × (7 × 7)5 ÷ 75

= (72)5 × (72)5 ÷ 75

= 710 × 710 ÷ 75

= 7 (10 + 10 - 5)

= 715

Step 3

Theref ore the value of 495 × 495 ÷ 75 in exponential f orm is 715 .



(14) A) 36 1

Step 1


you have to f ind out the value of ( -3

2 )2 ÷ (

-1

2 )4 .

Step 2


( -3

2 )2 =

9

4

and

( -1

2 )4 =

1

16

Step 3

Now ( -3

2 )2 ÷ (

-1

2 )4 =

9

4 ÷

1

16

= 9

4 ×

16

1

= 36

1

Step 4

Theref ore the value of ( -3

2 )2 ÷ (

-1

2 )4 is

36

1 .



B) 243 1600

Step 1


you have to f ind out the value of ( 3

4 )3 ÷ (

-5

3 )2 .

Step 2


( 3

4 )3 =

27

64

and

( -5

3 )2 =

25

9

Step 3

Now ( 3

4 )3 ÷ (

-5

3 )2 =

27

64 ÷

25

9

= 27

64 ×

9

25

= 243

1600

Step 4

Theref ore the value of ( 3

4 )3 ÷ (

-5

3 )2 is

243

1600 .



C) 144 625

Step 1



4 )2 ÷ (

-5

4 )4 .

Step 2


( -3

4 )2 =

9

16

and

( -5

4 )4 =

625

256

Step 3

Now ( -3

4 )2 ÷ (

-5

4 )4 =

9

16 ÷

625

256

= 9

16 ×

256

625

= 144

625

Step 4


4 )2 ÷ (

-5

4 )4 is

144

625 .



D) 25 81

Step 1



3 )4 ÷ (

4

5 )2 .

Step 2


( -2

3 )4 =

16

81

and

( 4

5 )2 =

16

25

Step 3

Now ( -2

3 )4 ÷ (

4

5 )2 =

16

81 ÷

16

25

= 16

81 ×

25

16

= 25

81

Step 4


3 )4 ÷ (

4

5 )2 is

25

81 .



(15) A) 8

Step 1


Step 2

1258


1258

1259 × 55 × 1252

= (5 × 5 × 5)8

(5 × 5 × 5)9 × (5)5 × (5 × 5 × 5)2

= (53)8

(53)9 × (5)5 × (53)2

= 524

527 × 55 × 56

= 5(24)

5(27) × 5(5 + 6)

= 524

527 × 511

= 524 ÷ 527 × 511

= 5(24 - 27 + 11)

= 58

Step 3


1259 × 55 × 1252 in exponential f orm is 58.



B) 10

Step 1


Step 2

492

76 × 496 can be simplif ied as:

492

76 × 496

= (7 × 7)2

(7)6 × (7 × 7)6

= (72)2

(7)6 × (72)6

= 74

76 × 712

= 7 (4)

7 (6) × 7 (12)

= 74

76 × 712

= 74 ÷ 76 × 712

= 7 (4 - 6 + 12)

= 710

Step 3


76 × 496 in exponential f orm is 710 .



exponents and powers grade 7 - edugain · pdf fileexponents and powers ... rule 3: when you...

Documents