# Free Exponents and Powers 03 Practice Test - 7th grade

Express 343 as a power of 7.

A.

72

B.

73

C.

74

D.

70

#### SOLUTION

Solution : B

343 can be written as

343=7×7×7
so, 343=73

Express 32000 as a product of powers of prime factors.

A.

28×52

B.

28×53

C.

27×54

D.

28×54

#### SOLUTION

Solution : B

32000 can be written as

32000=(2×2×2×2×2)×1000=25×103
(as 32=2×2×2×2×2)

= 25×(2×2×2×5×5×5)
= 25×23×53
= 28×53

If a = 2 and b = 3. Find the value of ab×ba.

A.

56

B.

72

C.

64

D.

8

#### SOLUTION

Solution : B

Putting a=2 and b=3
We get:
23=2×2×2=8 and 32=9
Hence, 8×9=72

Find the value of 33×32.

A.

35

B.

343

C.

31

D.

243

#### SOLUTION

Solution : A and D

Here the two terms have  33 and 32 have different exponents, but the same base.Hence by adding the powers  we get,

33×32=33+2 = 35=243

Express the given number in standard form:
21700000

A.

2.1×107

B.

2.17×107

C.

1.17×107

D.

1.1701×107

#### SOLUTION

Solution : B

21700000 can be written as

2.17×10000000=2.17×107

The value of 33×23 is:

A.

216

B.

215

C.

201

D.

219

#### SOLUTION

Solution : A

Expanding the terms:

(3×2)3=63=216

(102)2__.

#### SOLUTION

Solution :

From the law of exponent, we have,

(am)n=(a)mn
Therefore,
(102)2=104=10000

In 104, 10 is the base and 4 is the __.

#### SOLUTION

Solution :

Here, 4 is called the exponent.

(25)2=(25)2

A.

True

B.

False

#### SOLUTION

Solution : B

2252 = 425

2252 = 425

Hence, the given statement is false.

am×bm=(ab)m

A.

True

B.

False

#### SOLUTION

Solution : A

From the law of exponents, we know that
am×bm=(ab)m
Thus, the given statement is true.