Class 8 Powers Exercise 2.1-1

\begin{array}{l} \text {Q1. Express each of the following as a rational number of the form } \\ \frac {p}{q} \text {, where p and q } \\ \text { are integers and } q \neq 0 \\ \text {(i) } 2^{-3} \quad \text {(ii) } (-4)^{-2} \\ \text {(iii) }\frac{1}{3^{-2}} \\ \text {(iv) } (\frac{1}{2})^{-5} \\ \text {(v) } (\frac{2}{3})^{-2} \\ \text {Sol. We know that for any non zero rational number a and } \\ \text {for any positive integer n, }a^{-n}=\frac{1}{a^n} \\ \text {Thus, we have } \\ \text {(i) } 2^{-3} = \frac{1}{2^{3}} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{1}{8} \\ \\ \text {(ii) } (-4)^{-2}=\frac{1}{(-4)^{2}} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{1}{16} \\ \\ \text {(iii) } \frac{1}{3^{-2}} =3^{2} \quad \quad [\because \frac {1}{a^{-n}} = a^{n} ] \\ =9 \\ \\ \text {(iv) } (\frac{1}{2})^{-5}=\frac{1}{2^{-5}} \quad \quad [\because (\frac{a}{b})^{n} = \frac {a^{n}}{b^{n}} ] \\ =2^{5} \quad \quad [\because \frac {1}{a^{-n}} = a^{n} ] \\ =32 \\ \\ \text {(v) } (\frac{2}{3})^{-2}=\frac{1}{(\frac{2}{3})^{2}} \quad \quad [\because \frac {1}{a^{-n}} = a^{n} ] \\ = \frac{1}{\frac{2^{2}}{3^{2}}} \quad \quad [\because (\frac{a}{b})^{n} = \frac {a^{n}}{b^{n}} ] \\=\frac{1}{\frac{4}{9}}=\frac{9}{4} \\ \\\text {Q2. Find the values of each of the following: } \\ \text {(i) } 3^{-1}+4^{-1} \quad \quad \quad \quad \text {(ii) } (3^{0}+4^{-1}) \times 2^{2} \\ \text {(iii) } (3^{-1}+4^{-1}+5^{-1})^{0} \quad \quad \quad \text {(iv) } [(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1} \\ \\ \text {Sol. We know that for any non zero rational number a and } \\ \text {for any positive integer n, }a^{-n}=\frac{1}{a^n} \\ \text {Thus, we have } \\ \\ \text {(i) } 3^{-1}+4^{-1} = \frac{1}{3}+\frac{1}{4} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{4+3}{12} \\ =\frac{7}{12} \\ \\\text { (ii) } (3^{0}+4^{-1}) \times 2^{2} \\ =(1+\frac{1}{4}) \times 4 \quad \quad [\because a^{-n} = \frac {1}{a^{n}} \text { and } a^{0} = 1] \\ =\frac{5}{4} \times 4 = 5 \\ \\\text { (iii) } (3^{-1}+4^{-1}+5^{-1})^{0}=1 \quad \quad [\because a^{0} = 1] \\ \\\text { (iv) } [(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1} \\ \\=[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1} \\=[\frac{1}{(\frac{1}{3})}-\frac{1}{(\frac{1}{4})}]^{-1} \\=(3-4)^{-1} \\ =(-1)^{-1}=-1 \\ \\\text {Q3. Find the values of each of the following: } \\ \text {(i) } \quad (\frac{1}{2})^{-1}+(\frac{1}{3})^{-1}+(\frac{1}{4})^{-1} \\ \text {(ii) } \quad (\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}\\ \text {(iii) } \quad (2^{-1} \times 4^{-4}) \div 2^{-2} \\ \text {(iv) } \quad (5^{-1} \times 2^{-1}) \div 6^{-1} \\ \\ \text {Sol. (i) } \quad (\frac{1}{2})^{-1}+(\frac{1}{3})^{-1}+(\frac{1}{4})^{-1} \\ =2+3+4 \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ =12 \\ \\\text {(ii) } \quad (\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2} \\ =\frac{1}{(\frac{1}{2})^{2}}+\frac{1}{(\frac{1}{3})^{2}}+\frac{1}{(\frac{1}{4})^{2}} \\ \quad \quad \quad \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{1}{\frac{1}{4}}+\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{16}} \\ =4+9+16=29 \\ \\\text {Or we can also solve like below } \\(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2} \\ = 2^{2} + 3^{2} + 4^{2} \quad \quad [\because (\frac{a}{b})^{-n} = (\frac{b}{a})^{n}] \\ =4+9+16=29 \\ \\\text {(iii) } \quad ((2^{-1} \times 4^{-4}) \div 2^{-2} \\ =[\frac{1}{2} \times \frac{1}{4} ] \div \frac{1}{2^{2}} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{1}{8} \times 4 \\ =\frac{1}{2} \\ \\\text {(iv) } \quad (5^{-1} \times 2^{-1}) \div 6^{-1} \\ =(\frac{1}{5} \times \frac{1}{2}) \div \frac{1}{6} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =\frac{1}{10} \times 6 \\ =\frac{3}{5} \\ \\ \text {Q4. Simplify : } \\ \text {(i) } \quad (4^{-1} \times 3^{-1})^{2} \\ \text {(ii) } \quad (5^{-1} \div 6^{-1})^{3} \\ \text {(iii) } \quad (2^{-1}+3^{-1})^{-1} \\ \text {(iv) } \quad (3^{-1}+4^{-1})^{-1} \times 5^{-1} \\ \\ \text {Sol. (i) } \quad (4^{-1} \times 3^{-1})^{2} \\ =(\frac{1}{4} \times \frac{1}{3})^{2} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =(\frac{1}{12})^{2} \\ =(\frac{1^{2}}{12^{2}}) \quad \quad [\because (\frac{a}{b})^{n} = \frac{a^{n}}{b^{n}}] \\ =\frac{1}{144} \\ \\\text {(ii) }\quad (5^{-1} \div 6^{-1})^{3} \\ =(\frac{1}{5} \div \frac{1}{6})^{3} \quad \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =(\frac{1}{5} \times 6)^{3} \\ =(\frac{6}{5})^{3} \\ =\frac{216}{125} \quad \quad [\because (\frac{a}{b})^{n} = \frac{a^{n}}{b^{n}}] \\ \\\text {(iii) } \quad (2^{-1}+3^{-1})^{-1} \\ =(\frac{1}{2}+\frac{1}{3})^{-1} \quad \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =(\frac{5}{6})^{-1} \\ =(\frac{1}{\frac{5}{6}})=\frac{6}{5} \\ \\\text {(iv) } \quad (3^{-1}+4^{-1})^{-1} \times 5^{-1} \\ =(\frac{1}{3} \times \frac{1}{4})^{-1} \times \frac{1}{5} \quad \quad [\because a^{-n} = \frac {1}{a^{n}}] \\ =(\frac{1}{12})^{-1} \times \frac{1}{5} \\ = 12 \times \frac{1}{5} \\ \\ \end{array}