# DAV Class 8 Maths Chapter 3 Worksheet 1 Solutions

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:

1. am × an = am+n
2. am ÷ an = am-n
3. (am)n = amn
4. am × bn = (ab)m
5. 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