Free Exponents and Powers 01 Practice Test - 8th Grade
Question 1
If m and n are integers, a is a non-zero integer and am × an =ax, then the value of x is _____.
SOLUTION
Solution : B
We have am×an=am+n, where a, m and n are integers, a≠0.
Given am×an=ax
⟹am+n=ax
⟹x=m+n
Question 2
Find the multiplicative inverse of 10−5.
0.00001
- 100000
100000
SOLUTION
Solution : D
10−5
= 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 10−5.
Question 3
In standard form 21600000 is written as___________.
2.16×107
216×107
2.16×105
216×1000000
SOLUTION
Solution : A
21600000 = 2.16 x 10000000
Therefore, in standard form 21600000 is written as :
2.16×107
Question 4
The value of 1000∘ is 1000.
True
False
SOLUTION
Solution : B
Note that a0=1 for any a≠0
Thus, the value of 1000∘ is 1.
Question 5
Express 4.95×105 in usual form.
49500
4950
495000
4950000
SOLUTION
Solution : C
4.95×105=4.95×100000=495000
Question 6
(4×103+3×1+1×10−1+6×100)=
SOLUTION
Solution :4×103+3×1+1×10−1+6×100=(4×1000)+(3×1)+110+(6×1)=4000+3+0.1+6=4009.1
Question 7
Find the value of ‘m‘ if (−2)2×(−5)3=50m
-10
10
SOLUTION
Solution : A
(−2)2×(−5)3=50m
⇒4×(−125)=50m
⇒−500=50m
⇒m=−10
Question 8
If 4x = 64, then
22x=64
2x = 6
x = 3
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 a≠0,1,−1)
Since x=3, 22x = 64 , and 2x =6
Question 9
(9×10−2)+(9×103) in usual form is given by
9000.09
SOLUTION
Solution : A
The usual forms of the expressions 9×10−2 and 9×103 are given by 0.09 and 9000.
Thus, 9000 + 0.09 = 9000.09.
Question 10
Evaluate the exponential expression x5 × x4, for x = 2.
32
128
256
512
SOLUTION
Solution : D
x5 × x4, for x = 2
Substituting 2 in place of x
= 25× 24
= 29 (∵am×an=am+n) = 512