# Free Exponents and Powers 02 Practice Test - 8th Grade

Find the value of 73.

A.

343

B.

1343

C.

121

D.

149

#### SOLUTION

Solution : B

We know that am=1am for any non-zero integer a and integer m.
73=173
=17×7×7=1343

Find the multiplicative inverse of 107.

A.

102

B.

102

C.

107

D.

106

#### SOLUTION

Solution : C

Multiplicative inverse of a non-zero x is 1x; so that when these two are multiplied, the product value is 1. Therefore, the multiplicative inverse of 107should be a number which when multiplied with 107 gives 1.
Let the multiplicative inverse of 107 be x.
Then, 107×x=1.
x=1107=107
(an=1an1an=11an=an)
Thus, the multiplicative inverse of 107 is 107.

Simplify and write in exponential form: 43×42×162

A.

43

B.

44

C.

45

D.

46

#### SOLUTION

Solution : C

Given expression is

43×42×162=43×42×((4)2)2=43×42×44   ((am)n=amn)=432+4   (am×an=am+n)=45

Simplify  (15)3(13)4(14)3.

A.

1213

B.

1312

C.

1116

D.

1611

#### SOLUTION

Solution : C

an=1an for a0
(15)3=1(15)3=(51)3=53=125
Similarly, (13)4=1(13)4=(31)4=34=81 and
(14)3=1(14)3=(41)3=43=64

(15)3(13)4(14)3=1258164=4464=1116

The standard exponential form of 1,540,000,000 is 1.54×109.

A.

True

B.

False

#### SOLUTION

Solution : A

1,540,000,000=1.54×1000000000=1.54×109

Identify the base  in the given example:  53
Base: ___

#### SOLUTION

Solution :

The number written at the bottom is the base whereas the number at the superscript is known as the exponent. In the given example, 5 is the base and 3 is the exponent.

If 3676.48 can be written in expanded form as:
3×10a+b×102+7×10c+6×100+4×10d+e×102 .
Find the value of a+bcde.

__

#### SOLUTION

Solution :

3676.48 can be written in expanded form as:
3×103+6×102+7×101+6×100+4×101+8×102
a=3,b=6,c=1,d=1,e=8a+bcde=3+61+18=1

If (2)m+1×25=42 the value of m is

___

#### SOLUTION

Solution :

Given, (2)m+1×25=42
(2)m+1×25=((2)2)2
(2)m+1+5=(2)4
Since the bases on  both the sides  are equal, therefore the exponents must also be equal.  This means, m+1+5=4, or m=10.

Express in usual form: 5.06x104

__

#### SOLUTION

Solution :

5.06×104=5.0610000=0.000506

The distance between Earth and Mars is 2.2253x108 kms. The distance between Mars and Saturn is 1.19666×109 kms. A spaceship starts from Earth and travels to Saturn via Mars. Find the total distance covered by the spaceship in metres.

A.

1.41919x1012 m

B. 14191900000 m
C. 1419190000000 m
D. 1.41919x109 m

#### SOLUTION

Solution : A and C

The total distance travelled by the spaceship between Earth and Saturn = Distance between (Earth & Mars + Mars & Saturn)
2.2253x108  kms + 1.19666 x 109 kms
=
2.2253x108  kms + 11.9666 x 108 kms
=14.1919 x108 kms
= 1.41919x109×103 m
= 1.41919x 1012m=1419190000000 m