Solved Examples on Exponents
Here are some solved examples on exponents using the laws of exponents.
1. Evaluate the exponent:
(i) 5
-3
(ii) (
1/
3)
-4
(iii) (
5/
2)
-3
(iv) (-2)
-5
(v) (
-3/
4)
-4
We have:
(i) 5
-3 = 1/5
3 = 1/125
(ii) (1/3)
-4 = (3/1)
4 = 3
4 = 81
(iii) (5/2)
-3 = (2/5)
3 = 2
3/5
3 = 8/125
(iv) (-2)
-5 = 1/(-2)
-5 = 1/-2
5 = 1/-32 = -1/32
(v) (-3/4)
-4 = (4/-3)
4 = (-4/3)
4 = (-4)
4/3
4 = 4
4/3
4 = 256/81
2. Evaluate: (-2/7)-4 × (-5/7)2
Solution:
(
-2/
7)
-4 × (
-5/
7)
2
= (7/-2)
4 × (-5/7)
2
= (-7/2)
4 × (-5/7)
2 [Since, (7/-2) = (-7/2)]
= (-7)
4/2
4 × (-5)
2/7
2
= {7
4 × (-5)
2}/{2
4 × 7
2 }
[Since, (-7)4 = 74]
= {7
2 × (-5)
2 }/2
4
= [49 × (-5) × (-5)]/16
= 1225/16
3. Evaluate: (-1/4)-3 × (-1/4)-2
Solution:
(-1/4)
-3 × (-1/4)
-2
= (4/-1)
3 × (4/-1)
2
= (-4)
3 × (-4)
2
= (-4)
(3 + 2)
= (-4)
5
= -4
5
= -1024.
4. Evaluate: {[(-3)/2]2}-3
Solution:
{[(-3)/2]
2}
-3
= (-3/2)
2 × (-3)
= (-3/2)
-6
= (2/-3)
6
= (-2/3)
6
= (-2)
6/3
6
= 2
6/3
6
= 64/729
5. Simplify:
(i) (2
-1 × 5
-1)
-1 ÷ 4
-1
(ii) (4
-1 + 8
-1) ÷ (2/3)
-1
Solution:
(i) (2
-1 × 5
-1)
-1 ÷ 4
-1
= (1/2 × 1/5)
-1 ÷ (4/1)
-1
= (1/10)
-1 ÷ (1/4)
=
10/
1 ÷
1/
4
= (10 ÷
1/
4)
= (10 × 4)
= 40.
(ii) (4-1 + 8-1) ÷ (2/3)-1
= (1/4 + 1/8) ÷ (3/2)
= (2 + 1)/8 ÷ 3/2
= (3/8 ÷ 3/2)
= (3/8 ÷ 2/3)
= 1/4
6. Simplify: (1/2)-2 + (1/3)-2 + (1/4)-2
Solution:
(1/2)
-2 + (1/3)
-2 + (1/4)
-2
= (2/1)
2 + (3/1)
2 + (4/1)
2
= (2
2 + 3
2 + 4
2)
= (4 + 9 + 16)
= 29.
7. By what number should (1/2)-1 be multiplied so that the product is (-5/4)-1?
Solution:
Let the required number be x. Then,
x × (1/2)
-1 = (-5/4)
-1
⇒ x × (2/1) = (4/-5)
⇒ 2x = -4/5
⇒ x = (1/2 × -4/5) = -2/5
Hence, the required number is -2/5.
8. By what number should (-3/2)-3 be divided so that the quotient is (9/4)-2?
Solution:
Let the required number be x. Then,
(-3/2)
-3/x = (9/4)
-2
⇒ (-2/3)
3 = (4/9)
2 × x
⇒ (-2)
3/3
3 = 4
2/9
2 × x
⇒ -8/27 = 16/81 × x
⇒ x = {-8/27 × 81/16}
⇒ x = -3/2
Hence, the required number is -3/2
9. If a = (2/5)2 ÷ (9/5)0 find the value of a-3.
Solution:
a
-3 = [(2/5)
2 ÷ (9/5)
0]
-3
= [(2/5)
2 ÷ 1]
-3
= [(2/5)
2]
-3
= (2/5)
-6
= (5/2)
6
10. Find the value of n, when 3-7 ×32n + 3 = 311 ÷ 35
Solution:
3
2n + 3 = 3
11 ÷ 3
5/3
-7
⇒ 3
2n + 3 = 3
11 - 5/3
-7
⇒ 3
2n + 3 = 3
6/3
-7
⇒ 3
2n + 3 = 3
6 - (-7)
⇒ 3
2n + 3 = 3
6 + 7
⇒ 3
2n + 3 = 3
13
Since the bases are same and equating the powers, we get 2n + 3 = 13
2n = 13 – 3
2n = 10
n = 10/2
Therefore, n = 5
11. Find the value of n, when (5/3)2n + 1 (5/3)5 = (5/3)n + 2
Solution:
(5/3)
2n + 1 + 5 = (5/3)
n + 2
= (5/3)
2n + 6 = (5/3)
n + 2
Since the bases are same and equating the powers, we get 2n + 6 = n + 2
2n – n = 2 – 6
=> n = -4
12. Find the value of n, when 3n = 243
Solution:
3
n = 3
5
Since, the bases are same, so omitting the bases, and equating the powers we get, n = 5.
13. Find the value of n, when 271/n = 3
Solution:
(27) = 3
n
⇒ (3)
3 = 3
n
Since, the bases are same and equating the powers, we get
⇒ n = 3
14. Find the value of n, when 3432/n = 49
Solution:
[(7)
3]
2/n = (7)
2
⇒ (7)
6/n = (7)
2
⇒ 6/n = 2
Since, the bases are same and equating the powers, we get n =
6/
2 = 3.
● Exponents
Exponents
Laws of Exponents
Rational Exponent
Integral Exponents of a Rational Numbers
Solved Examples on Exponents
Practice Test on Exponents
● Exponents - Worksheets
Worksheet on Exponents
8th Grade Math Practice
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