The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 3 Worksheet 1 Solutions of Exponents and Radicals offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 3 WS 1 Solutions
Question 1.
Express each of the following in exponential form:
(i) \(\sqrt[5]{35}\)
Solution:
\(\sqrt[5]{35}=(35)^{1 / 5}\)
(ii) \(\sqrt[11]{27^2}\)
Solution:
\(\sqrt[11]{27^2}=(27)^{2 / 11}\)
(iii) \(\sqrt[7]{\frac{11}{3}}\)
Solution:
\(\sqrt[7]{\frac{11}{3}}=\left(\frac{11}{3}\right)^{1 / 7}\)
(iv) \(\sqrt[3]{\left(\frac{2}{5}\right)^{-3}}\)
Solution:
\(\sqrt[3]{\left(\frac{2}{5}\right)^{-3}}=\left(\frac{2}{5}\right)^{-3 / 3}=\left(\frac{2}{5}\right)^{-1}\)
(v) \(\sqrt[13]{(111)^3}\)
Solution:
\((111)^{3 / 13}\)
(vi) \(\sqrt[7]{(29)^2}\)
Solution:
\(\sqrt[7]{(29)^2}=(29)^{2 / 7}\)
(vii) \(\sqrt[3]{(2)^{-6}}\)
Solution:
\(\sqrt[3]{(2)^{-6}}=(2)^{-6 / 3}\) = (2)-2
(viii) \(\sqrt[7]{\left(\frac{15}{341}\right)^{-3}}\)
Solution:
\(\sqrt[7]{\left(\frac{15}{341}\right)^{-3}}=\left(\frac{15}{341}\right)^{-3 / 7}\)
Question 2.
Express each of the following as radicals:
(i) \((21)^{1 / 8}\)
Solution:
\((21)^{1 / 8}=\sqrt[8]{21}\)
(ii) \((25)^{3 / 4}\)
Solution:
\((25)^{3 / 4}=\sqrt[4]{(25)^3}\)
(iii) \(\left(\frac{2}{9}\right)^{1 / 9}\)
Solution:
\(\left(\frac{2}{9}\right)^{1 / 9}=\sqrt[9]{\frac{2}{9}}\)
(iv) \((100)^{-5 / 6}\)
Solution:
\((100)^{-5 / 6}=\sqrt[6]{(100)^{-5}}\)
(v) \(\left(\frac{8}{9}\right)^{3 / 4}\)
Solution:
\(\left(\frac{8}{9}\right)^{3 / 4}=\sqrt[4]{\left(\frac{8}{9}\right)^3}\)
(vi) \(\left(\frac{17}{231}\right)^{-5 / 6}\)
Solution:
\(\left(\frac{17}{231}\right)^{-5 / 6}=\sqrt[6]{\left(\frac{17}{231}\right)^{-5}}\)
(vii) \(\left(\frac{15}{21}\right)^{2 / 5}\)
Solution:
\(\left(\frac{15}{21}\right)^{2 / 5}=\sqrt[5]{\left(\frac{14}{21}\right)^2}\)
Question 3.
Express each of the following as radicals:
(i) \(x^{-1 / 2}\)
Solution:
\(x^{-1 / 2}=\frac{1}{x^{1 / 2}}\)
(ii) \(x^{-2 / 5}\)
Solution:
\(x^{-2 / 5}=\frac{1}{x^{2 / 5}}\)
(iii) \(\frac{7}{x^{-5 / 6}}\)
Solution:
\(\frac{7}{x^{-5 / 6}}=7 \cdot x^{5 / 6}\)
(iv) (x-3)4
Solution:
(x-3)4 = \(\left(\frac{1}{x^3}\right)^4\)
DAV Class 8 Maths Chapter 3 Worksheet 1 Notes
- am/n = \(\sqrt[n]{a^m}\)
where \(\sqrt[n]{a^m}\) is called the radical and n is its index. Index is always positive. - a-m/n = \(\frac{1}{a^{m / n}}=\frac{1}{\left(a^m\right)^{1 / n}}\)
Laws of Exponents:
- am × an = am+n
- am ÷ an = am-n
- (am)n = amn
- am × bn = (ab)m
- a0 = 1
Example 1.
Simplify the following:
(i) (- 3)5 × (- 3)– 8
(ii) 36 ÷ 3– 9
Solution:
(i) (- 3)5 × (- 3)– 8 = (- 3)5 – 8
[Using am × an = am + n]
= (- 3)– 3
= \(\frac{1}{(-3)^3}\)
= \(-\frac{1}{27}\)
(ii) (3)6 ÷ (3)– 9 = 36 – (- 9)
[am ÷ an = am – n]
= 39 + 6
= 315
Example 2.
Simplify the following:
(i) \(\left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-4}\right\}+\left(\frac{1}{4}\right)^{-3}\)
(ii) \(\left(\frac{6}{7}\right)^{-5} \times\left(\frac{7}{6}\right)^{-3}\)
Solution:
\(\left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-4}\right\}+\left(\frac{1}{4}\right)^{-3}\)
= {(3)3 – 24} ÷ (4)3
= (27 – 16) ÷ 64
= 11 ÷ 64
[∵ a– m = \(\frac{1}{a^m}\)]
= \(\frac{11}{64}\)
(ii) \(\left(\frac{6}{7}\right)^{-5} \times\left(\frac{7}{6}\right)^{-3}\)
= \(\left(\frac{7}{6}\right)^5 \times\left(\frac{6}{7}\right)^3\)
= \(\frac{7^5}{6^5} \times \frac{6^3}{7^3}\)
[∵ \(\frac{a^m}{a^n}\) = am – n]
= 75 – 3 × 63 – 5
= 52 × 6– 2
= \(\frac{5^2}{6^2}=\frac{25}{36}\)
Example 3.
Find ‘k’ so that: (- 5)k + 1 × (- 5)5 = (- 5)7.
Solution:
(- 5)k + 1 × (- 5)5 = (- 5)7
= (- 5)k + 1 + 5 = (- 5)7
[∵ am × an = am + n]
= (- 5)k + 6 = (- 5)7
Equating the powers of equal bases,
k + 6 = 7
∴ k = 7 – 6 = 1.
Example 4:
Simplify the following and write in exponential form:
(i) (25 ÷ 28)5 × 2– 5
(ii) (- 4)– 3 × (5)– 3 × (- 5)– 3
(iii) \(\frac{1}{27}\) × (3)– 3
(iv) (3)4 × \(\left(\frac{5}{3}\right)^4\)
Solution:
(i) (25 ÷ 28)5 × 2– 5
= (25 – 8)5 × \(\frac{1}{2^5}\)
= \(\frac{2^{-15}}{2^5}\)
= 2– 15 – 5 = 2– 20
(ii) (- 4)– 3 × (5)– 3 × (- 5)– 3 = (- 4 × 5 × – 5)– 3
[∵ am × bm = (ab)m]
= (100)– 3
= (102)– 3
= 10– 6
= \(\frac{1}{10^6}\)
(iii) \(\frac{1}{27}\) × (3)– 3
= \(\frac{1}{27} \times \frac{1}{3^3}\)
[∵ a– n = \(\frac{1}{a^n}\)]
= \(\frac{1}{3^3} \times \frac{1}{3^3}\)
= \(\frac{1}{3^{3+3}}=\frac{1}{3^6}\)
(iv) (- 3)4 × \(\left(\frac{5}{3}\right)^4\)
= [- 3 \(\frac{5}{3}\)]4
[∵ am × bm =(ab)m]
= (- 5)4
= 54
Example 5.
Find the values of \(\left(\frac{9}{16}\right)^{3 / 2} \times\left(\frac{9}{16}\right)^{1 / 2}\) and \(\left(\frac{9}{16}\right)^{4 / 2}\). Are they equal?
Solution:
Example 6.
Verify that: 642/3 × 272/3 = (64 × 27)2/3.
Solution:
Consider 642/3 × 272/3 = \(4^{3 \times \frac{2}{3}} \times 3^{3 \times \frac{2}{3}}\)
= 42 × 32
= 16 × 9 = 144
Now take up 642/3 × 272/3 = (1728)2/3
= (123)2/3
= \(12^{3 \times \frac{2}{3}}\)
= 122 = 144
Hence, 642/3 × 272/3 = (64 × 27)2/3