Class 8 Powers Exercise 2.2-1

\begin{array}{l} \text {Q1. Write each of the following in exponential form: } \\ (i) \quad (\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \\ (ii) \quad (\frac{2}{5})^{-2} \times(\frac{2}{5})^{-2} \times(\frac{2}{5})^{-2} \\ \\\text {Sol. } (i) \quad (\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \times(\frac{3}{2})^{-1} \\ =(\frac{3}{2})^{(-1-1-1-1)} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\ =(\frac{3}{2})^{-4} \\ \\(ii) \quad (\frac{2}{5})^{-2} \times(\frac{2}{5})^{-2} \times(\frac{2}{5})^{-2} \\ =(\frac{2}{5})^{(-2-2-2)} \quad \quad [\because a^{m} \times a^{n} = a^{m+n}] \\ =(\frac{2}{5})^{-6} \\ \\\text {Q2. Evaluate: } \\ (i) \quad 5^{-2} \\ (ii) \quad (-3)^{-2} \\ (iii) \quad (\frac {1}{3})^{-4} \\ (iv) \quad (\frac {-1}{2})^{-1} \\ \\\text {Sol. } (i) \quad 5^{-2} = \frac {1}{5^{2}} \quad \quad [\because a^{-n} = \frac {1}{a^n}] \\ =\frac {1}{25} \\ \\(ii) \quad (-3)^{-2}= \frac {1}{-3^{2}} \quad \quad [\because a^{-n} = \frac {1}{a^n}] \\ =\frac {1}{9} \\ \\(iii) \quad (\frac{1}{3})^{-4} =3^{4} \quad \quad [\because (\frac{a}{b})^{-n} = \frac{b}{a})^{n}] \\ =81 \\ \\(iv) \quad (\frac {-1}{2})^{-1} \\ =(-2)^{1} \quad \quad [\because (\frac{a}{b})^{-n} = \frac{b}{a})^{n}] \\ =-2 \\ \\ \text {Q3. Express each of the following as a rational number in the form p/q: } \\ (i) \quad 6^{-1} \\ (ii) \quad (-7)^{-1} \\ (iii)\quad (\frac{1}{4})^{-1} \\ (iv) \quad (-4)^{-1} \times(\frac{-3}{2})^{-1} \\ (v) \quad (\frac{3}{5})^{-1} \times(\frac{5}{2})^{-1} \\ \\ \text {Sol. We know that } a^{-n} = \frac{1}{a^{n}} \text { using this rule, we have } \\ \\ (i) \quad 6^{-1}=\frac{1}{6} \\ \\(ii) \quad -7^{-1}= \frac{-1}{7} \\ \\(iii) \quad (\frac{1}{4})^{-1} \\ =\frac{1}{\frac{1}{4}} = 4 \\ \\(iv) \quad (-4)^{-1} \times (\frac{-3}{2})^{-1} \\ =\frac{1}{-4} \times \frac{1}{\frac{-3}{2}} \\ =\frac{1}{-4} \times \frac{2}{-3} \\ =\frac{1}{6} \\ \\(v) \quad (\frac{3}{5})^{-1} \times (\frac{5}{2})^{-1} \\ =\frac{1}{\frac{3}{5}} \times \frac{1}{\frac{5}{2}} \\ =\frac{5}{3} \times \frac{2}{5} \\ =\frac{2}{3} \\ \end{array}\begin{array}{l} \text {Q4. Simplify: } \\ (i) \quad (4^{-1} \times 3^{-1})^{2} \\ (ii) \quad (5^{-1} \div 6^{-1})^{3} \\ (iii) \quad (2^{-1}+3^{-1})^{-1} \\ (iv) \quad (3^{-1} \times 4^{-1})^{-1} \times 5^{-1} \\ (v) \quad (4^{-1}-5^{-1}) \div 3^{-1} \\ \\ \text {Sol. We know that } a^{-n} = \frac{1}{a^{n}} \text { using this rule, we have } \\ \\(i) \quad (4^{-1} \times 3^{-1})^{2} \\ = (\frac{1}{4} \times \frac{1}{3})^{2} \\ =(\frac{1}{12})^{2} = (\frac{1}{144}) \\ \\ (ii) \quad (5^{-1} \div 6^{-1})^{3} \\ =(\frac{1}{5} \div \frac{1}{6})^{3} \\ =(\frac{1}{5} \times \frac{6}{1})^{3} \\ =(\frac{6}{5})^{3} \\ =(\frac{216}{125}) \\ \\(iii) \quad {2^{-1}+3^{-1}}^{-1} \\ =(\frac{1}{2}+\frac{1}{3})^{-1} \\ =(\frac{5}{6})^{-1} \\ =(\frac{6}{5}) \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\(iv) \quad (3^{-1} \times 4^{-1})^{-1} \times 5^{-1} \\ =(\frac{1}{3} \times \frac{1}{4})^{-1} \times \frac{1}{5} \\ =(\frac{1}{12})^{-1} \times \frac{1}{5} \\ =12 \times \frac{1}{5} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ =\frac{12}{5} \\ \\(v) \quad (4^{-1}-5^{-1}) \div 3^{-1} \\ =(\frac{1}{4}-\frac{1}{5}) \div \frac{1}{3} \\ =(\frac{5-4}{20}) \times 3 \\ =\frac{1}{20} \times 3=\frac{3}{20} \\ \\ \text {Q5. Express each of the following rational numbers with a negative exponent: } \\ (i) \quad (\frac {1}{4})^{3} \\ (ii) \quad 3^{5} \\ (iii) \quad (\frac {3}{5})^{4} \\ (iv) \quad [(\frac {3}{2})^{4}]^{-3} \\ (v) \quad [(\frac {7}{3})^{4}]^{-3} \\ \\ \text {Sol. } (i) \quad (\frac{1}{4})^{3} \\ =(\frac{4}{1})^{-3} \quad \quad [\because \frac{1}{a^{n}} = a^{-n} ] \\ =(4)^{-3} \\ \\(ii) \quad 3^{5} =(\frac{1}{3})^{-5} \quad \quad [\because \frac{1}{a^{n}} = a^{-n} ] \\ \\(iii) \quad (\frac {3}{5})^{4} =(\frac{5}{3})^{-4} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\(iv) \quad [(\frac {3}{2})^{4}]^{-3} \\ =(\frac{3}{2})^{[(4) \times (-3)]} \quad \quad [\because {a^{m}}^{n} = a^{mn} ] \\ =(\frac{3}{2})^{-12} \\ \\(v) \quad [(\frac {7}{3})^{4}]^{-3} \\ =(\frac{7}{3})^{[(4) \times (-3)]} \quad \quad [\because {a^{m}}^{n} = a^{mn} ] \\ =(\frac{7}{3})^{-12} \\ \\ \text {Q6. Express each of the following rational numbers with a positive exponent: } \\ (i) \quad (\frac {3}{4})^{-2} \\ (ii) \quad (\frac {5}{4})^{-3} \\ (iii) \quad 4^{3} \times 4^{-9} \\ (iv) \quad [(\frac {4}{3})^{-3}]^{-4} \\ (v) \quad [(\frac {3}{2})^{4}]^{-2} \\ \\ \text {Sol. } (i) \quad (\frac {3}{4})^{-2} \\ =(\frac{4}{3})^{2} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\(ii) \quad (\frac {5}{4})^{-3} \\ =(\frac{4}{5})^{3} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\(iii) \quad 4^{3} \times 4^{-9} \\ =4^{(3-9)} \quad \quad [\because a^{m} \times a^{n} = a^{(m+n)} ] \\ =4^{-6} \\ =(\frac{1}{4})^{6} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\(iv) \quad [(\frac {4}{3})^{-3}]^{-4} \\ =(\frac{4}{3})^{[(-3) \times (-4)]} \quad \quad [\because {a^{m}}^{n} = a^{mn} ] \\ =(\frac{4}{3})^{12} \\ \\(v) \quad [(\frac {3}{2})^{4}]^{-2} \\ =(\frac{3}{2})^{[4 \times (-2)]} \quad \quad [\because {a^{m}}^{n} = a^{mn} ] \\ =(\frac{3}{2})^{-8} \\ =(\frac{2}{3})^{8} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ \\ \end{array}