# Free Exponents and Powers 01 Practice Test - 7th grade

3×1000+2×10+410+31000 is the expanded form of the number ____.

A.

3200.430

B.

3002.403

C.

32.430

D.

3020.403

#### SOLUTION

Solution : D

3×1000+2×10+410+31000

=3000+20+0.4+0.003

=3020.403

In power notation, 8116 can be expressed as ___.

A.

(3424)

B.

(34)3

C.

(32)4

D.

(32)3

#### SOLUTION

Solution : A and C

We can write:
81=3×3×3×3
16=2×2×2×2

So, 8116 =
(3424)

And we know that, (ambm)=(ab)m
8116 = (3424) =(32)4

Third power of 14× fourth power of 4 equals to ______.

A.

42

B.

43

C.

4

D.

40

#### SOLUTION

Solution : C

Third power of 14=(14)3
Fourth power of  4 is equal to 44.

(14)3×44
= 14×14×14×4×4×4×4
= 4×4×4×44×4×4
= 4443
= 4

If a = 2 and b = 3, find the value of (ab)a.

A.

16

B.

32

C.

64

D.

128

#### SOLUTION

Solution : C

On substituting a = 2 and b = 3, we get

(23)2=23×2        [(xm)n=x(m×n)]

=26

=64

The value of 43×23 is

#### SOLUTION

Solution :

Notice that here the two terms 43 and 23 have different bases, but the same exponents.

43×23=(4×4×4)×(2×2×2)
=(4×2)×(4×2)×(4×2)=83=512

Express 5126000 in standard form.

A.

0.5126×107

B.

5.126×106

C.

51.26×105

D.

5126×103

#### SOLUTION

Solution : B

In standard form, any number is expressed in powers of 10. The decimal number must be in between 1 and 10.

5.126×1000000=5.126×106

Express 256×243 in exponential form.

A.

28×35

B.

27×34

C.

26×33

D.

22×32

#### SOLUTION

Solution : A

We can write:
256=(2×2×2×2×2×2×2×2)=28

243=(3×3×3×3×3)=35

256×243=28×35

Find the value of (2  5)7÷(2  5)6.

A.

(2  5)2

B.

25

C.

2  5

D.

45

#### SOLUTION

Solution : C

abac=abc      [a is a non-zero integer]

Using this law, we get,

(2  5)7÷(2  5)6

(2  5)76=(2  5)1=2  5

23×33
___

#### SOLUTION

Solution :

23×33=(2×3)3=63=216

The value of (3020)×52 is:

A.

1

B.

25

C.

0

D.

9

#### SOLUTION

Solution : C

For any number a 0, a0 = 1.

30=20=1

(3020)×52=(11)×25                             =0