PSEB 9th Class Maths Solutions Chapter 1 Number Systems Ex 1.6

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

Question 1.
Find:
(i) \(64^{\frac{1}{2}}\)
Answer:
\(64^{\frac{1}{2}}\) = \(\left(2^{6}\right)^{\frac{1}{2}}\)
= \(2^{6 \times \frac{1}{2}}\)
= 2³ = 8

(ii) \(32^{\frac{1}{5}}\)
Answer:
\(32^{\frac{1}{5}}\) = \(\left(2^{5}\right)^{\frac{1}{5}}\)
= \(2^{5 \times \frac{1}{5}}\)
= 2¹ = 2

(iii) \(125^{\frac{1}{3}}\)
Answer:
\(125^{\frac{1}{3}}\) = \(\left(5^{3}\right)^{\frac{1}{3}}\)
= \(5^{3} \times \frac{1}{3}\)
= 5¹
= 5

Question 2.
Find:
(i) \(9^{\frac{3}{2}}\)
Answer:
\(9^{\frac{3}{2}}\) = \(\left(3^{2}\right)^{\frac{3}{2}}\)
= \(3^{2 \times \frac{3}{2}}\)
= 3³
= 27

(ii) \(32^{\frac{2}{5}}\)
Answer:
\(32^{\frac{2}{5}}\) = \(\left(2^{5}\right)^{\frac{2}{5}}\)
= \(2^{5 \times \frac{2}{5}}\)
= 2² = 4

(iii) \(16^{\frac{3}{4}}\)
Answer:
\(16^{\frac{3}{4}}\) = \(\left(2^{4}\right)^{\frac{3}{4}}\)
= \(2^{4 \times \frac{3}{4}}\)
= 2³ = 8

(iv) \(125^{\frac{-1}{3}}\)
Answer:
\(125^{\frac{-1}{3}}\) = \(\left(5^{3}\right)^{-\frac{1}{3}}\)
= \(5^{3 \times \frac{-1}{3}}\)
= 5^-1 = \(\frac{1}{5}\)

Question 3.
Simplify
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
Answer:
\(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\) = \(2^{\frac{2}{3}+\frac{1}{5}}\)
= \(2^{\frac{10+3}{15}}\)
= \(2^{\frac{13}{15}}\)

(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
Answer:
\(\left(\frac{1}{3^{3}}\right)^{7}\) = (3^-3)⁷
= 3^{– 3 × 7}
= 3^{– 21}

(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
Answer:
\(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\) = \(11^{\frac{1}{2}-\frac{1}{4}}\)
= \(11^{\frac{2-1}{4}}\)
= \(11^{\frac{1}{4}}\)

(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
Answer:
\(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\) = \((7 \cdot 8)^{\frac{1}{2}}\)
= \(56^{\frac{1}{2}}\)