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.
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