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} = 7^{4}]
= {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} ×3^{2n + 3} = 3^{11} ÷ 3^{5}
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 3^{n} = 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 27^{1/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 343^{2/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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