# Free Exponents and Powers 01 Practice Test - 8th Grade

If m and n are integers, a is a non-zero integer and am × an =ax, then the value of x is _____.

A. mn
B. m+n
C. mn
D. mn

#### SOLUTION

Solution : B

We have am×an=am+n,​​ where a, m and n are integers, a0.
Given am×an=ax
am+n=ax
x=m+n

Find the multiplicative inverse of 105.

A. - 50
B.

0.00001

C.

- 100000

D.

100000

#### SOLUTION

Solution : D

105
= 1105
= 1100000

When a number is multiplied with its multiplicative inverse, the result is 1. To make the given number, which is 1100000, equal to 1, we need to multiply it with 100000, which is the reciprocal of 1100000.

Therefore, 100000, or 105 is the multiplicative inverse of 105.

In standard form  21600000 is written as___________.

A.

2.16×107

B.

216×107

C.

2.16×105

D.

216×1000000

#### SOLUTION

Solution : A

21600000 = 2.16 x 10000000
Therefore, in standard form 21600000 is written as :
2.16×107

The value of 1000 is 1000.

A.

True

B.

False

#### SOLUTION

Solution : B

Note that a0=1 for any a0
Thus, the value of 1000 is 1.

Express  4.95×105 in usual form.

A.

49500

B.

4950

C.

495000

D.

4950000

#### SOLUTION

Solution : C

4.95×105=4.95×100000=495000

(4×103+3×1+1×101+6×100)=

___

#### SOLUTION

Solution :

4×103+3×1+1×101+6×100=(4×1000)+(3×1)+110+(6×1)=4000+3+0.1+6=4009.1

Find the value of ‘m‘ if (2)2×(5)3=50m

A.

-10

B.

10

C. 50
D. -50

#### SOLUTION

Solution : A

(2)2×(5)3=50m
4×(125)=50m
500=50m
m=10

If 4x = 64, then

A.

22x=64

B.

2x = 6

C.

x = 3

D.

10x=10000

#### SOLUTION

Solution : A, B, and C

4x=64
4x=43            (as 64 is 43 )
x=3   (am = an if and only if m=n for a0,1,1)
Since x=3,  22x = 64 ,  and 2x =6

(9×102)+(9×103) in usual form is given by

A.

9000.09

B. 900.9
C. 9000.9
D. 900.09

#### SOLUTION

Solution : A

The usual forms of the expressions 9×102 and 9×103 are given by 0.09 and 9000.
Thus, 9000 + 0.09 = 9000.09.

Evaluate the exponential expression x5 × x4, for x = 2.

A.

32

B.

128

C.

256

D.

512

#### SOLUTION

Solution : D

x5 × x4, for x = 2
Substituting 2 in place of x
= 25× 24
= 29   (am×an=am+n) = 512