Q1. Commpute the following:
(i) $1^8 \times 3^0 \times 5^3 \times2^2$
(ii) $(4^7)^2 \times (4^{-3})^4$
(iii) $(2^{-9} \div 2^{-11})^3$
(iv) $(\dfrac{2}{3})^{-4} \times (\dfrac{27}{8})^{-2}$
(v) $(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}$
(vi) $(12)^{-2} \times (3)^3$
(vii) $(-5)^4 \times (5)^6 \div (-5)^9$
(viii) $(-\dfrac{1}{3}^4) \div (-\dfrac{1}{3}^{-8}) \times -\dfrac{1}{3}^5$
(ix) $9^0 \times 4^{-1} \div 2^{-4}$
(x) $(625)^{-\frac{3}{4}}$
(xi) $(\dfrac{27}{64})^{-\frac{2}{3}}$
(xii) $(\dfrac{1}{32})^{-\frac{2}{5}}$
(xiii) $(27)^{\frac{2}{3}} \div (\dfrac{81}{16})^{-\frac{1}{4}}$
(xiv) $(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 \div (64)^{\frac{2}{3}}$
(xv) $(243)^{\frac{2}{5}} \div (32)^{-\frac{2}{5}}$
(xvi) $(125)^{-\frac{2}{3}} \div (8)^{\frac{2}{3}}$
Show Answer
(i) $1^8 \times 3^0 \times 5^3 \times2^2$
$\Rightarrow 1 \times 1 \times 125 \times 4$
$\Rightarrow 1 \times 500 = 500$ Ans
(ii) $(4^7)^2 \times (4^{-3})^4$
$\Rightarrow 4^{7 \times 2} \times 4^{-3 \times 4}$
$\Rightarrow 4^{14} \times 4^{-12}$
$\Rightarrow 4^{14 - 12}$
$\Rightarrow 4^2 = 16$ Ans
(iii) $(2^{-9} \div 2^{-11})^3$
$\Rightarrow (2^{-9 \times 3}) \div (2^{-11 \times 3})$
$\Rightarrow (2^{-27}) \div (2^{-33})$
$\Rightarrow \dfrac{1}{2^{27}} \div \dfrac{1}{2^{33}}$
$\Rightarrow \dfrac{\dfrac{1}{2^{27}}}{\dfrac{1}{2^{33}}}$
$\Rightarrow \dfrac{2^{33}}{2^{27}}$
$\Rightarrow 2^{33 - 27}$
$\Rightarrow 2^6 = 64$ Ans
The previous solution can also be done as following:
(iii) $(2^{-9} \div 2^{-11})^3$
$\Rightarrow (2^{-9 \times 3}) \div (2^{-11 \times 3})$
$\Rightarrow (2^{-27}) \div (2^{-33})$
$\Rightarrow 2^{-27} \div \dfrac{1}{2^{33}}$
$\Rightarrow 2^{-27} \times 2^{33}$
$\Rightarrow 2^{-27 + 33}$
$\Rightarrow 2^6 = 64$ Ans
(iv) $(\dfrac{2}{3})^{-4} \times (\dfrac{27}{8})^{-2}$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times (\dfrac{3^3}{2^3})^{}-2$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times \dfrac{3^{3 \times -2}}{2^{3 \times -2}}$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times \dfrac{3^{-6}}{2^{-6}}$
$\Rightarrow \dfrac{\dfrac{1}{16}}{\dfrac{1}{81}} \times \dfrac{\dfrac{1}{729}}{\dfrac{1}{64}}$
$\Rightarrow \dfrac{81}{16} \times \dfrac{64}{729} = \dfrac{4}{9}$ Ans
(v) $(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}$
$\Rightarrow 1 \div (\dfrac{2^3}{5^3}) \times \dfrac{16}{25}$
$\Rightarrow 1 \div \dfrac{8}{125} \times \dfrac{16}{25}$
$\Rightarrow 1 \times \dfrac{125}{8} \times \dfrac{16}{25}$
$\Rightarrow 1 \times 5 \times 2 = 16$ Ans
(vi) $(12)^{-2} \times (3)^3$
$\Rightarrow \dfrac{1}{12^2} \times 27$
$\Rightarrow \dfrac{1}{144} \times 27$
$\Rightarrow \dfrac{27}{144} = \dfrac{9}{48} = \dfrac{3}{16}$ Ans
(vii) $(-5)^4 \times (5)^6 \div (-5)^9$
$\Rightarrow (-5)^4 \times (5)^6 \times \dfrac{1}{(-5)^9}$
$\Rightarrow (-5)^4 \times (5)^6 \times (-5)^{-9}$
$\Rightarrow (-5)^4 \times (-5)^{6 - 9}$
$\Rightarrow (-5)^4 \times (-5)^{-3}$
$\Rightarrow (-5)^{4 - 3} = -5$ Ans
(viii) $(-\dfrac{1}{3}^4) \div (-\dfrac{1}{3}^{-8}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^4 \times \dfrac{1}{-\dfrac{1}{3}^{8}}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^4 \times -\dfrac{1}{3}^{-8}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^{4 - 8}) \times -\dfrac{1}{3}^5$
$\Rightarrow -\dfrac{1}{3}^{-4} \times -\dfrac{1}{3}^5$
$\Rightarrow -\dfrac{1}{3}^{-4 + 5} = (-\dfrac{1}{3})$ Ans
(ix) $9^0 \times 4^{-1} \div 2^{-4}$
$\Rightarrow 1 \times (\dfrac{1}{4} \div \dfrac{1}{2^4})$
$\Rightarrow 1 \times (\dfrac{1}{4} \div \dfrac{1}{16})$
$\Rightarrow 1 \times (\dfrac{1}{4} \times 16)$
$\Rightarrow 1 \times (4) = 4$ Ans
(x) $(625)^{-\frac{3}{4}}$
$\Rightarrow (625)^{-\frac{3}{4}} = (5^4)^{-\frac{3}{4}}$
$\Rightarrow (5)^{-\frac{3}{4} \times 4} = (5)^{-3}$
$\therefore (5)^{-3} = \dfrac{1}{5^3} = \dfrac{1}{125}$ Ans
(xi) $(\dfrac{27}{64})^{-\frac{2}{3}}$
$\Rightarrow (\dfrac{3^3}{4^3})^{-\frac{2}{3}} = [(\dfrac{3}{4})^3]^{-\frac{2}{3}}$
$\Rightarrow [\dfrac{3}{4}]^{-\frac{2}{3} \times 3} = [\dfrac{3}{4}]^{-2}$
$\Rightarrow \dfrac{3^{-2}}{4^{-2}} = \dfrac{\dfrac{1}{3^2}}{\dfrac{1}{4^2}}$
$\therefore \dfrac{\dfrac{1}{9}}{\dfrac{1}{16}} = \dfrac{16}{9} = 1\dfrac{7}{9}$ Ans
(xii) $(\dfrac{1}{32})^{-\frac{2}{5}}$
$\Rightarrow (\dfrac{1}{(2)^5})^{-\frac{2}{5}} = [(2)^{-5}]^{-\frac{2}{5}}$
$\Rightarrow (2)^{-\frac{2}{5} \times -5} = 2^2 = 4$ Ans
(xiii) $(125)^{-\frac{2}{3}} \div (8)^{\frac{2}{3}}$
$\Rightarrow [5^3]^{-\frac{2}{3}} \div [2^3]^{\frac{2}{3}}$
$\Rightarrow 5^{-\frac{2}{3} \times 3} \div 2^{\frac{2}{3} \times 3}$
$\Rightarrow 5^{-2} \div 2^2$
$\Rightarrow \dfrac{1}{25} \div 4$
$\Rightarrow \dfrac{1}{25} \times \dfrac{1}{4} = \dfrac{1}{100}$ Ans
(xiv) $(243)^{\frac{2}{5}} \div (32)^{-\frac{2}{5}}$
$\Rightarrow [3^5]^{\frac{2}{5}} \div [2^5]^{-\frac{2}{5}}$
$\Rightarrow [3]^{\frac{2}{5} \times 5} \div [2]^{-\frac{2}{5} \times 5}$
$\Rightarrow 3^2 \div 2^{-2}$
$\Rightarrow 3^2 \div \dfrac{1}{2^2}$
$\Rightarrow 9 \div \dfrac{1}{4}$
$\Rightarrow 9 \times 4 = 36$ Ans
(xv) $(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 \div (64)^{\frac{2}{3}}$
$\Rightarrow 3^4 - 1 \times (-2)^5 \div [(4)^3]^{\frac{2}{3}}$
$\Rightarrow 81 - 1 \times - 32 \div [4]^{\frac{2}{3} \times 3}$
$\Rightarrow 81 - 1 \times - 32 \div 4^2$
$\Rightarrow 81 - 1 \times - 32 \times \dfrac{1}{16}$
$\Rightarrow 81 - 1 \times - 2$
$\therefore 81 (- 1 \times - 2) = 81 + 2 = 83$ Ans
(xvi) $(27)^{\frac{2}{3}} \div (\dfrac{81}{16})^{-\frac{1}{4}}$
$\Rightarrow [(3)^3]^{\frac{2}{3}} \div [(\dfrac{3^4}{2^4})]^{-\frac{1}{4}}$
$\Rightarrow [3]^{\frac{2}{3} \times 3} \div [\dfrac{3}{2}]^{-\frac{1}{4} \times 4}$
$\Rightarrow 3^2 \div [\dfrac{3}{2}]^{-1}$
$\Rightarrow 9 \div \dfrac{1}{\dfrac{3}{2}}$
$\Rightarrow 9 \div \dfrac{2}{3}$
$\therefore 9 \times \dfrac{3}{2} = \dfrac{27}{2} = 13\dfrac{1}{2}$ Ans
$\Rightarrow 1 \times 1 \times 125 \times 4$
$\Rightarrow 1 \times 500 = 500$ Ans
(ii) $(4^7)^2 \times (4^{-3})^4$
$\Rightarrow 4^{7 \times 2} \times 4^{-3 \times 4}$
$\Rightarrow 4^{14} \times 4^{-12}$
$\Rightarrow 4^{14 - 12}$
$\Rightarrow 4^2 = 16$ Ans
(iii) $(2^{-9} \div 2^{-11})^3$
$\Rightarrow (2^{-9 \times 3}) \div (2^{-11 \times 3})$
$\Rightarrow (2^{-27}) \div (2^{-33})$
$\Rightarrow \dfrac{1}{2^{27}} \div \dfrac{1}{2^{33}}$
$\Rightarrow \dfrac{\dfrac{1}{2^{27}}}{\dfrac{1}{2^{33}}}$
$\Rightarrow \dfrac{2^{33}}{2^{27}}$
$\Rightarrow 2^{33 - 27}$
$\Rightarrow 2^6 = 64$ Ans
The previous solution can also be done as following:
(iii) $(2^{-9} \div 2^{-11})^3$
$\Rightarrow (2^{-9 \times 3}) \div (2^{-11 \times 3})$
$\Rightarrow (2^{-27}) \div (2^{-33})$
$\Rightarrow 2^{-27} \div \dfrac{1}{2^{33}}$
$\Rightarrow 2^{-27} \times 2^{33}$
$\Rightarrow 2^{-27 + 33}$
$\Rightarrow 2^6 = 64$ Ans
(iv) $(\dfrac{2}{3})^{-4} \times (\dfrac{27}{8})^{-2}$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times (\dfrac{3^3}{2^3})^{}-2$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times \dfrac{3^{3 \times -2}}{2^{3 \times -2}}$
$\Rightarrow \dfrac{2^{-4}}{3^{-4}} \times \dfrac{3^{-6}}{2^{-6}}$
$\Rightarrow \dfrac{\dfrac{1}{16}}{\dfrac{1}{81}} \times \dfrac{\dfrac{1}{729}}{\dfrac{1}{64}}$
$\Rightarrow \dfrac{81}{16} \times \dfrac{64}{729} = \dfrac{4}{9}$ Ans
(v) $(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}$
$\Rightarrow 1 \div (\dfrac{2^3}{5^3}) \times \dfrac{16}{25}$
$\Rightarrow 1 \div \dfrac{8}{125} \times \dfrac{16}{25}$
$\Rightarrow 1 \times \dfrac{125}{8} \times \dfrac{16}{25}$
$\Rightarrow 1 \times 5 \times 2 = 16$ Ans
(vi) $(12)^{-2} \times (3)^3$
$\Rightarrow \dfrac{1}{12^2} \times 27$
$\Rightarrow \dfrac{1}{144} \times 27$
$\Rightarrow \dfrac{27}{144} = \dfrac{9}{48} = \dfrac{3}{16}$ Ans
(vii) $(-5)^4 \times (5)^6 \div (-5)^9$
$\Rightarrow (-5)^4 \times (5)^6 \times \dfrac{1}{(-5)^9}$
$\Rightarrow (-5)^4 \times (5)^6 \times (-5)^{-9}$
$\Rightarrow (-5)^4 \times (-5)^{6 - 9}$
$\Rightarrow (-5)^4 \times (-5)^{-3}$
$\Rightarrow (-5)^{4 - 3} = -5$ Ans
(viii) $(-\dfrac{1}{3}^4) \div (-\dfrac{1}{3}^{-8}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^4 \times \dfrac{1}{-\dfrac{1}{3}^{8}}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^4 \times -\dfrac{1}{3}^{-8}) \times -\dfrac{1}{3}^5$
$\Rightarrow (-\dfrac{1}{3}^{4 - 8}) \times -\dfrac{1}{3}^5$
$\Rightarrow -\dfrac{1}{3}^{-4} \times -\dfrac{1}{3}^5$
$\Rightarrow -\dfrac{1}{3}^{-4 + 5} = (-\dfrac{1}{3})$ Ans
(ix) $9^0 \times 4^{-1} \div 2^{-4}$
$\Rightarrow 1 \times (\dfrac{1}{4} \div \dfrac{1}{2^4})$
$\Rightarrow 1 \times (\dfrac{1}{4} \div \dfrac{1}{16})$
$\Rightarrow 1 \times (\dfrac{1}{4} \times 16)$
$\Rightarrow 1 \times (4) = 4$ Ans
(x) $(625)^{-\frac{3}{4}}$
$\Rightarrow (625)^{-\frac{3}{4}} = (5^4)^{-\frac{3}{4}}$
$\Rightarrow (5)^{-\frac{3}{4} \times 4} = (5)^{-3}$
$\therefore (5)^{-3} = \dfrac{1}{5^3} = \dfrac{1}{125}$ Ans
(xi) $(\dfrac{27}{64})^{-\frac{2}{3}}$
$\Rightarrow (\dfrac{3^3}{4^3})^{-\frac{2}{3}} = [(\dfrac{3}{4})^3]^{-\frac{2}{3}}$
$\Rightarrow [\dfrac{3}{4}]^{-\frac{2}{3} \times 3} = [\dfrac{3}{4}]^{-2}$
$\Rightarrow \dfrac{3^{-2}}{4^{-2}} = \dfrac{\dfrac{1}{3^2}}{\dfrac{1}{4^2}}$
$\therefore \dfrac{\dfrac{1}{9}}{\dfrac{1}{16}} = \dfrac{16}{9} = 1\dfrac{7}{9}$ Ans
(xii) $(\dfrac{1}{32})^{-\frac{2}{5}}$
$\Rightarrow (\dfrac{1}{(2)^5})^{-\frac{2}{5}} = [(2)^{-5}]^{-\frac{2}{5}}$
$\Rightarrow (2)^{-\frac{2}{5} \times -5} = 2^2 = 4$ Ans
(xiii) $(125)^{-\frac{2}{3}} \div (8)^{\frac{2}{3}}$
$\Rightarrow [5^3]^{-\frac{2}{3}} \div [2^3]^{\frac{2}{3}}$
$\Rightarrow 5^{-\frac{2}{3} \times 3} \div 2^{\frac{2}{3} \times 3}$
$\Rightarrow 5^{-2} \div 2^2$
$\Rightarrow \dfrac{1}{25} \div 4$
$\Rightarrow \dfrac{1}{25} \times \dfrac{1}{4} = \dfrac{1}{100}$ Ans
(xiv) $(243)^{\frac{2}{5}} \div (32)^{-\frac{2}{5}}$
$\Rightarrow [3^5]^{\frac{2}{5}} \div [2^5]^{-\frac{2}{5}}$
$\Rightarrow [3]^{\frac{2}{5} \times 5} \div [2]^{-\frac{2}{5} \times 5}$
$\Rightarrow 3^2 \div 2^{-2}$
$\Rightarrow 3^2 \div \dfrac{1}{2^2}$
$\Rightarrow 9 \div \dfrac{1}{4}$
$\Rightarrow 9 \times 4 = 36$ Ans
(xv) $(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 \div (64)^{\frac{2}{3}}$
$\Rightarrow 3^4 - 1 \times (-2)^5 \div [(4)^3]^{\frac{2}{3}}$
$\Rightarrow 81 - 1 \times - 32 \div [4]^{\frac{2}{3} \times 3}$
$\Rightarrow 81 - 1 \times - 32 \div 4^2$
$\Rightarrow 81 - 1 \times - 32 \times \dfrac{1}{16}$
$\Rightarrow 81 - 1 \times - 2$
$\therefore 81 (- 1 \times - 2) = 81 + 2 = 83$ Ans
(xvi) $(27)^{\frac{2}{3}} \div (\dfrac{81}{16})^{-\frac{1}{4}}$
$\Rightarrow [(3)^3]^{\frac{2}{3}} \div [(\dfrac{3^4}{2^4})]^{-\frac{1}{4}}$
$\Rightarrow [3]^{\frac{2}{3} \times 3} \div [\dfrac{3}{2}]^{-\frac{1}{4} \times 4}$
$\Rightarrow 3^2 \div [\dfrac{3}{2}]^{-1}$
$\Rightarrow 9 \div \dfrac{1}{\dfrac{3}{2}}$
$\Rightarrow 9 \div \dfrac{2}{3}$
$\therefore 9 \times \dfrac{3}{2} = \dfrac{27}{2} = 13\dfrac{1}{2}$ Ans
Q2. Simplify the following:
(i) $8^{\frac{4}{3}} + 25^{\frac{3}{2}} - [\dfrac{1}{27}]^{-\frac{2}{3}}$
(ii) $[64^{-2}]^{-3} \div [\lbrace(-8)^{2}\rbrace^3]^2$
(iii) $(2^{-3} - 2^{-4}) (2^{-3} + 2^{-4})$
Show Answer
(i) $8^{\frac{4}{3}} + 25^{\frac{3}{2}} - [\dfrac{1}{27}]^{-\frac{2}{3}}$
$\Rightarrow [2^3]^{\frac{4}{3}} + [5^2]^{\frac{3}{2}} - [\dfrac{1}{3^3}]^{-\frac{2}{3}}$
$\Rightarrow [2^3]^{\frac{4}{3}} + [5^2]^{\frac{3}{2}} - [3^{-3}]^{-\frac{2}{3}}$
$\Rightarrow 2^{\frac{4}{3} \times 3} + 5^{\frac{3}{2} \times 2} - 3^{-\frac{2}{3} \times (-3)}$
$\Rightarrow 2^4 + 5^3 - 3^2$
$\Rightarrow 16 + 125 - 9$
$\therefore 141 - 9 = 132$ Ans
(ii) $[64^{-2}]^{-3} \div [\lbrace(-8)^{2}\rbrace^3]^2$
$\Rightarrow [(8^2)^{-2}]^{-3} \div [\lbrace(-2^3)^{2}\rbrace^3]^2$
$\Rightarrow [(8)^{-2 \times 2}]^{-3} \div [\lbrace(-2)^{2 \times 3}\rbrace^3]^2$
$\Rightarrow [(8)^{-4}]^{-3} \div [\lbrace(-2)^{6}\rbrace^3]^2$
$\Rightarrow [8]^{-3 \times -4} \div [(-2)^{3 \times 6}]^2$
$\Rightarrow 8^{12} \div [(-2)^{18}]^2$
$\Rightarrow (2^3)^{12} \div (-2)^{2 \times 18}$
$\Rightarrow 2^{12 \times 3} \div (-2)^{36}$
$\Rightarrow 2^{36} \div (-2)^{36}$
$\Rightarrow 2^{36} \div 2^{36}$ (an even power changes a negative no. into positive)
$\Rightarrow 2^{36} \times \dfrac{1}{2^{36}}$
$\Rightarrow 2^{36} \times 2^{-36}$
$\therefore 2^{36 + (-36)} = 2^{36 - 36} = 2^0 = 1$ Ans
(iii) $(2^{-3} - 2^{-4}) (2^{-3} + 2^{-4})$
$\Rightarrow 2^{-3}(2^{-3} + 2^{-4}) - 2^{-4}(2^{-3} + 2^{-4})$
$\Rightarrow (2^{-3 - 3} + 2^{-4 - 3}) - (2^{-3 - 4} + 2^{-4 - 4})$
$\Rightarrow (2^{-6} + 2^{-7}) - (2^{-7} + 2^{-8})$
$\Rightarrow 2^{-6} + 2^{-7} - 2^{-7} - 2^{-8}$
$\Rightarrow 2^{-6} - 2^{-8}$
$\Rightarrow \dfrac{1}{2^6} - \dfrac{1}{2^8}$
$\therefore \dfrac{1}{64} - \dfrac{1}{256} = \dfrac{4 - 1}{256} = \dfrac{3}{256}$ Ans
$\Rightarrow [2^3]^{\frac{4}{3}} + [5^2]^{\frac{3}{2}} - [\dfrac{1}{3^3}]^{-\frac{2}{3}}$
$\Rightarrow [2^3]^{\frac{4}{3}} + [5^2]^{\frac{3}{2}} - [3^{-3}]^{-\frac{2}{3}}$
$\Rightarrow 2^{\frac{4}{3} \times 3} + 5^{\frac{3}{2} \times 2} - 3^{-\frac{2}{3} \times (-3)}$
$\Rightarrow 2^4 + 5^3 - 3^2$
$\Rightarrow 16 + 125 - 9$
$\therefore 141 - 9 = 132$ Ans
(ii) $[64^{-2}]^{-3} \div [\lbrace(-8)^{2}\rbrace^3]^2$
$\Rightarrow [(8^2)^{-2}]^{-3} \div [\lbrace(-2^3)^{2}\rbrace^3]^2$
$\Rightarrow [(8)^{-2 \times 2}]^{-3} \div [\lbrace(-2)^{2 \times 3}\rbrace^3]^2$
$\Rightarrow [(8)^{-4}]^{-3} \div [\lbrace(-2)^{6}\rbrace^3]^2$
$\Rightarrow [8]^{-3 \times -4} \div [(-2)^{3 \times 6}]^2$
$\Rightarrow 8^{12} \div [(-2)^{18}]^2$
$\Rightarrow (2^3)^{12} \div (-2)^{2 \times 18}$
$\Rightarrow 2^{12 \times 3} \div (-2)^{36}$
$\Rightarrow 2^{36} \div (-2)^{36}$
$\Rightarrow 2^{36} \div 2^{36}$ (an even power changes a negative no. into positive)
$\Rightarrow 2^{36} \times \dfrac{1}{2^{36}}$
$\Rightarrow 2^{36} \times 2^{-36}$
$\therefore 2^{36 + (-36)} = 2^{36 - 36} = 2^0 = 1$ Ans
(iii) $(2^{-3} - 2^{-4}) (2^{-3} + 2^{-4})$
$\Rightarrow 2^{-3}(2^{-3} + 2^{-4}) - 2^{-4}(2^{-3} + 2^{-4})$
$\Rightarrow (2^{-3 - 3} + 2^{-4 - 3}) - (2^{-3 - 4} + 2^{-4 - 4})$
$\Rightarrow (2^{-6} + 2^{-7}) - (2^{-7} + 2^{-8})$
$\Rightarrow 2^{-6} + 2^{-7} - 2^{-7} - 2^{-8}$
$\Rightarrow 2^{-6} - 2^{-8}$
$\Rightarrow \dfrac{1}{2^6} - \dfrac{1}{2^8}$
$\therefore \dfrac{1}{64} - \dfrac{1}{256} = \dfrac{4 - 1}{256} = \dfrac{3}{256}$ Ans
Q3. Evaluate the following:
(i) $(-5)^0$
(ii) $8^0 + 4^0 + 2^0$
(iii) $(8 + 4 + 2)^0$
(iv) $4x^0$
(v) $(4x)^0$
(vi) $[(10^3)^0]^5$
(vii) $(7x^0)^2$
(viii) $9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}}$
Show Answer
(i) $(-5)^0 = 1$ Ans
(ii) $8^0 + 4^0 + 2^0 = 1 + 1 + 1 = 3$ Ans
(iii) $(8 + 4 + 2)^0 = 1$ Ans
(iv) $4x^0 = 4$ Ans
(v) $(4x)^0 = 1$ Ans
(vi) $[(10^3)^0]^5 = [1]^5 = 1$ Ans
(vii) $(7x^0)^2 = 7^2 = 49$ Ans
(viii) $9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}}$
$\Rightarrow 1 + (3^2)^{-1} - (3^2)^{-2} + (3^2)^{\frac{1}{2}} - (3^2)^{-\frac{1}{2}}$
$\Rightarrow 1 + 3^{-1 \times 2} - 3^{-2 \times 2} + 3^{\frac{1}{2} \times 2} - 3^{-\frac{1}{2} \times 2}$
$\Rightarrow 1 + 3^{-2} - 3^{-4} + 3^1 - 3^{-1}$
$\Rightarrow 1 + \dfrac{1}{3^2} - \dfrac{1}{3^4} + 3 - \dfrac{1}{3}$
$\Rightarrow 1 + \dfrac{1}{9} - \dfrac{1}{81} + 3 - \dfrac{1}{3}$
$\Rightarrow (1 + 3) + \dfrac{1}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow 4 + \dfrac{1}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow (\dfrac{36 + 1}{9}) - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow \dfrac{37}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow (\dfrac{333 - 1}{81}) - \dfrac{1}{3}$
$\Rightarrow \dfrac{332}{81} - \dfrac{1}{3}$
$\therefore \dfrac{332 - 27}{81} = \dfrac{305}{81} = 3\dfrac{62}{81}$ Ans
(ii) $8^0 + 4^0 + 2^0 = 1 + 1 + 1 = 3$ Ans
(iii) $(8 + 4 + 2)^0 = 1$ Ans
(iv) $4x^0 = 4$ Ans
(v) $(4x)^0 = 1$ Ans
(vi) $[(10^3)^0]^5 = [1]^5 = 1$ Ans
(vii) $(7x^0)^2 = 7^2 = 49$ Ans
(viii) $9^0 + 9^{-1} - 9^{-2} + 9^{\frac{1}{2}} - 9^{-\frac{1}{2}}$
$\Rightarrow 1 + (3^2)^{-1} - (3^2)^{-2} + (3^2)^{\frac{1}{2}} - (3^2)^{-\frac{1}{2}}$
$\Rightarrow 1 + 3^{-1 \times 2} - 3^{-2 \times 2} + 3^{\frac{1}{2} \times 2} - 3^{-\frac{1}{2} \times 2}$
$\Rightarrow 1 + 3^{-2} - 3^{-4} + 3^1 - 3^{-1}$
$\Rightarrow 1 + \dfrac{1}{3^2} - \dfrac{1}{3^4} + 3 - \dfrac{1}{3}$
$\Rightarrow 1 + \dfrac{1}{9} - \dfrac{1}{81} + 3 - \dfrac{1}{3}$
$\Rightarrow (1 + 3) + \dfrac{1}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow 4 + \dfrac{1}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow (\dfrac{36 + 1}{9}) - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow \dfrac{37}{9} - \dfrac{1}{81} - \dfrac{1}{3}$
$\Rightarrow (\dfrac{333 - 1}{81}) - \dfrac{1}{3}$
$\Rightarrow \dfrac{332}{81} - \dfrac{1}{3}$
$\therefore \dfrac{332 - 27}{81} = \dfrac{305}{81} = 3\dfrac{62}{81}$ Ans
Q4. Simplify the following:
(i) $\dfrac{a^5b^2}{a^2b^{-3}}$
(ii) $15y^8 \div 3y^3$
(iii) $x^{10}y^6 \div x^3y^{-2}$
(iv) $5z^{16} \div 15z^{-11}$
(v) $(36x^2)^{\frac{1}{2}}$
(vi) $(125x^{-3})^{\frac{1}{3}}$
(vii) $(2x^2y^{-3})^{-2}$
(viii) $(27x^{-3}y^6)^{\frac{2}{3}}$
(ix) $(-2x^{\frac{2}{3}}y^{-\frac{3}{2}})^6$
Show Answer
(i) $\dfrac{a^5b^2}{a^2b^{-3}} = a^{5 - 2}b^{2 - (-3)}$
$\Rightarrow a^3b^{2 + 3} = a^3b^5$ Ans
(ii) $15y^8 \div 3y^3$
$\Rightarrow \dfrac{15y^8}{3y^3} = 5y^{8 - 3} = 5y^5$ Ans
(iii) $x^{10}y^6 \div x^3y^{-2}$
$\Rightarrow \dfrac{x^{10}y^6}{x^3y^{-2}} = x^{10 - 3}y^{6 - (-2)}$
$\Rightarrow = x^7y^{6 + 2} = x^7y^8$ Ans
(iv) $5z^{16} \div 15z^{-11}$
$\Rightarrow \dfrac{5z^{16}}{15z^{-11}} = (\dfrac{1}{3})z^{16 - (-11)}$
$\Rightarrow (\dfrac{1}{3})z^{16 + 11} = (\dfrac{1}{3})z^{27}$ Ans
Explanation:
$\because \dfrac{z^{16}}{z^{-11}} = \dfrac{z^{16}}{\dfrac{1}{z^{11}}}$
$\therefore z^{16} \times z^{11} = z^{16 + 11} = z^{27}$
(v) $(36x^2)^{\frac{1}{2}}$
$\Rightarrow [(6^2)(x^2)]^{\frac{1}{2}} = [(6x)^2]^{\frac{1}{2}}$
$\Rightarrow [6x]^{\frac{1}{2} \times 2} = (6x)^1 = 6x$ Ans
(vi) $(125x^{-3})^{\frac{1}{3}}$
$\Rightarrow (5^3x^{-3})^{\frac{1}{3}} = 5^{\frac{1}{3} \times 3}x^{\frac{1}{3} \times -3}$
$\therefore 5^1x^{-1} = 5x^{-1}$ Ans
(vii) $(2x^2y^{-3})^{-2}$
$\Rightarrow (2)^{-2}(x)^{2 \times -2}(y)^{-3 \times -2}$
$\Rightarrow \dfrac{1}{2^2}(x)^{-4}(y)^6$
$\therefore \dfrac{1}{4}x^{-4}y^6$ Ans
(viii) $(27x^{-3}y^6)^{\frac{2}{3}}$
$\Rightarrow ((3)^3(x)^{-3}(y)^6)^{\frac{2}{3}}$
$\Rightarrow (3)^{\frac{2}{3} \times 3}(x)^{\frac{2}{3} \times -3}(y)^{\frac{2}{3} \times 6}$
$\therefore 3^2x^{-2}y^{2\times2} = 9x^{-2}y^4$ Ans
(ix) $(-2x^{\frac{2}{3}}y^{-\frac{3}{2}})^6$
$\Rightarrow (-2)^6(x)^{\frac{2}{3} \times 6}(y)^{-\frac{3}{2} \times 6}$
$\Rightarrow (64)(x)^{2 \times 2}(y)^{-3 \times 3}$
$\therefore 64x^{4}y^{-9}$ Ans
$\Rightarrow a^3b^{2 + 3} = a^3b^5$ Ans
(ii) $15y^8 \div 3y^3$
$\Rightarrow \dfrac{15y^8}{3y^3} = 5y^{8 - 3} = 5y^5$ Ans
(iii) $x^{10}y^6 \div x^3y^{-2}$
$\Rightarrow \dfrac{x^{10}y^6}{x^3y^{-2}} = x^{10 - 3}y^{6 - (-2)}$
$\Rightarrow = x^7y^{6 + 2} = x^7y^8$ Ans
(iv) $5z^{16} \div 15z^{-11}$
$\Rightarrow \dfrac{5z^{16}}{15z^{-11}} = (\dfrac{1}{3})z^{16 - (-11)}$
$\Rightarrow (\dfrac{1}{3})z^{16 + 11} = (\dfrac{1}{3})z^{27}$ Ans
Explanation:
$\because \dfrac{z^{16}}{z^{-11}} = \dfrac{z^{16}}{\dfrac{1}{z^{11}}}$
$\therefore z^{16} \times z^{11} = z^{16 + 11} = z^{27}$
(v) $(36x^2)^{\frac{1}{2}}$
$\Rightarrow [(6^2)(x^2)]^{\frac{1}{2}} = [(6x)^2]^{\frac{1}{2}}$
$\Rightarrow [6x]^{\frac{1}{2} \times 2} = (6x)^1 = 6x$ Ans
(vi) $(125x^{-3})^{\frac{1}{3}}$
$\Rightarrow (5^3x^{-3})^{\frac{1}{3}} = 5^{\frac{1}{3} \times 3}x^{\frac{1}{3} \times -3}$
$\therefore 5^1x^{-1} = 5x^{-1}$ Ans
(vii) $(2x^2y^{-3})^{-2}$
$\Rightarrow (2)^{-2}(x)^{2 \times -2}(y)^{-3 \times -2}$
$\Rightarrow \dfrac{1}{2^2}(x)^{-4}(y)^6$
$\therefore \dfrac{1}{4}x^{-4}y^6$ Ans
(viii) $(27x^{-3}y^6)^{\frac{2}{3}}$
$\Rightarrow ((3)^3(x)^{-3}(y)^6)^{\frac{2}{3}}$
$\Rightarrow (3)^{\frac{2}{3} \times 3}(x)^{\frac{2}{3} \times -3}(y)^{\frac{2}{3} \times 6}$
$\therefore 3^2x^{-2}y^{2\times2} = 9x^{-2}y^4$ Ans
(ix) $(-2x^{\frac{2}{3}}y^{-\frac{3}{2}})^6$
$\Rightarrow (-2)^6(x)^{\frac{2}{3} \times 6}(y)^{-\frac{3}{2} \times 6}$
$\Rightarrow (64)(x)^{2 \times 2}(y)^{-3 \times 3}$
$\therefore 64x^{4}y^{-9}$ Ans
Q5. Simplify: $(x^{a + b})^{a - b}(x^{b + c})^{b - c}(x^{c + a})^{c - a}$
Show Answer
$\Rightarrow (x^{a + b})^{a - b}(x^{b + c})^{b - c}(x^{c + a})^{c - a}$
Solving for exponent:
$\Rightarrow [(a + b)(a - b)].[(b + c)(b - c)].[(c + a)(c - a)]$
$\Rightarrow [a(a + b) - b(a + b)].[b(b + c) - c(b + c)].[c(c + a) - a(c + a)]$
$\Rightarrow [a^2 + ab - ab - b^2].[b^2 + bc - bc - c^2].[c^2 + ca - ca - a^2]$
$\therefore [a^2 - b^2].[b^2 - c^2].[c^2 - a^2]$
So, we have:
$\Rightarrow (x)^{a^2 - b^2}(x)^{b^2 - c^2}(x)^{c^2 - a^2}$
$\Rightarrow x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}$
$\Rightarrow x^{(a^2 - a^2) + (b^2 - b^2) + (c^2 - c^2)}$
$\therefore x^0 = 1$ Ans
Solving for exponent:
$\Rightarrow [(a + b)(a - b)].[(b + c)(b - c)].[(c + a)(c - a)]$
$\Rightarrow [a(a + b) - b(a + b)].[b(b + c) - c(b + c)].[c(c + a) - a(c + a)]$
$\Rightarrow [a^2 + ab - ab - b^2].[b^2 + bc - bc - c^2].[c^2 + ca - ca - a^2]$
$\therefore [a^2 - b^2].[b^2 - c^2].[c^2 - a^2]$
So, we have:
$\Rightarrow (x)^{a^2 - b^2}(x)^{b^2 - c^2}(x)^{c^2 - a^2}$
$\Rightarrow x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}$
$\Rightarrow x^{(a^2 - a^2) + (b^2 - b^2) + (c^2 - c^2)}$
$\therefore x^0 = 1$ Ans
Q6. Simplify the following:
(i)$\sqrt[5]{x^{20}y^{-10}z^5} \div \dfrac{x^3}{y^3}$
(ii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{2}{4}}$
(iii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{3}{4}}$
Show Answer
(i)$\sqrt[5]{x^{20}y^{-10}z^5} \div \dfrac{x^3}{y^3}$
$\Rightarrow (x^{\frac{20}{5}}y^{-\frac{10}{5}}z^{\frac{5}{5}}) \div \dfrac{x^3}{y^3}$
$\Rightarrow (x^{4}y^{-2}z) \div \dfrac{x^3}{y^3}$
$\Rightarrow x^{4}y^{-2}z \times \dfrac{y^3}{x^3}$
$\Rightarrow \dfrac{x^{4}y^{-2 + 3}z}{x^3} = \dfrac{x^{4}yz}{x^3}$
$\Rightarrow x^{4-3}yz = xyz$ Ans
(ii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{2}{4}}$
$\Rightarrow\lbrace\dfrac{(4)^4a^{16}}{(3)^4b^4}\rbrace^{-\frac{2}{4}}$
$\Rightarrow \lbrace\dfrac{(4)^{-\frac{2}{4} \times 4}(a)^{-\frac{2}{4} \times 16}}{(3)^{-\frac{2}{4} \times 4}(b)^{-\frac{2}{4} \times 4}}\rbrace$
$\Rightarrow\lbrace\dfrac{(4)^{-2}(a)^{-8}}{(3)^{-2}(b)^{-2}}\rbrace$
$\Rightarrow \dfrac{\dfrac{1}{4^2} \times \dfrac{1}{a^8}}{\dfrac{1}{3^2} \times \dfrac{1}{b^2}}$
$\Rightarrow \dfrac{3^2 \times b^2}{4^2 \times a^8} = \dfrac{9b^2}{16a^8}$
$\therefore \dfrac{9}{16}a^{-8}b^2$ Ans
(iii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{3}{4}}$
$\Rightarrow\lbrace\dfrac{(4)^4a^{16}}{(3)^4b^4}\rbrace^{-\frac{3}{4}}$
$\Rightarrow \lbrace\dfrac{(4)^{-\frac{3}{4} \times 4}(a)^{-\frac{3}{4} \times 16}}{(3)^{-\frac{3}{4} \times 4}(b)^{-\frac{3}{4} \times 4}}\rbrace$
$\Rightarrow\lbrace\dfrac{(4)^{-3}(a)^{-12}}{(3)^{-3}(b)^{-3}}\rbrace$
$\Rightarrow \dfrac{\dfrac{1}{4^3} \times \dfrac{1}{a^{12}}}{\dfrac{1}{3^3} \times \dfrac{1}{b^3}}$
$\Rightarrow \dfrac{3^3 \times b^3}{4^3 \times a^{12}} = \dfrac{27b^3}{64a^{12}}$
$\therefore \dfrac{9}{16}a^{-12}b^3$ Ans
$\Rightarrow (x^{\frac{20}{5}}y^{-\frac{10}{5}}z^{\frac{5}{5}}) \div \dfrac{x^3}{y^3}$
$\Rightarrow (x^{4}y^{-2}z) \div \dfrac{x^3}{y^3}$
$\Rightarrow x^{4}y^{-2}z \times \dfrac{y^3}{x^3}$
$\Rightarrow \dfrac{x^{4}y^{-2 + 3}z}{x^3} = \dfrac{x^{4}yz}{x^3}$
$\Rightarrow x^{4-3}yz = xyz$ Ans
(ii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{2}{4}}$
$\Rightarrow\lbrace\dfrac{(4)^4a^{16}}{(3)^4b^4}\rbrace^{-\frac{2}{4}}$
$\Rightarrow \lbrace\dfrac{(4)^{-\frac{2}{4} \times 4}(a)^{-\frac{2}{4} \times 16}}{(3)^{-\frac{2}{4} \times 4}(b)^{-\frac{2}{4} \times 4}}\rbrace$
$\Rightarrow\lbrace\dfrac{(4)^{-2}(a)^{-8}}{(3)^{-2}(b)^{-2}}\rbrace$
$\Rightarrow \dfrac{\dfrac{1}{4^2} \times \dfrac{1}{a^8}}{\dfrac{1}{3^2} \times \dfrac{1}{b^2}}$
$\Rightarrow \dfrac{3^2 \times b^2}{4^2 \times a^8} = \dfrac{9b^2}{16a^8}$
$\therefore \dfrac{9}{16}a^{-8}b^2$ Ans
(iii) $(\dfrac{256a^{16}}{81b^4})^{-\frac{3}{4}}$
$\Rightarrow\lbrace\dfrac{(4)^4a^{16}}{(3)^4b^4}\rbrace^{-\frac{3}{4}}$
$\Rightarrow \lbrace\dfrac{(4)^{-\frac{3}{4} \times 4}(a)^{-\frac{3}{4} \times 16}}{(3)^{-\frac{3}{4} \times 4}(b)^{-\frac{3}{4} \times 4}}\rbrace$
$\Rightarrow\lbrace\dfrac{(4)^{-3}(a)^{-12}}{(3)^{-3}(b)^{-3}}\rbrace$
$\Rightarrow \dfrac{\dfrac{1}{4^3} \times \dfrac{1}{a^{12}}}{\dfrac{1}{3^3} \times \dfrac{1}{b^3}}$
$\Rightarrow \dfrac{3^3 \times b^3}{4^3 \times a^{12}} = \dfrac{27b^3}{64a^{12}}$
$\therefore \dfrac{9}{16}a^{-12}b^3$ Ans
Q7. Simplify and express as positive indices:
(i) $(a^{-2}b)^{-2} \times (ab^{-3})$
(ii) $(x^ny^{-m})^4 \times (x^3y^{-2})^{-n}$
(iii) $(\dfrac{125a^{-3}}{y^6})^{-\frac{1}{3}}$
(iv) $(\dfrac{32x^{-5}}{243y^{-5}})^{-\frac{1}{5}}$
(v) $(a^{-2}b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}}$
(vi) $(xy)^{m - n} \times (yz)^{n - l} \times (zx)^{l - m}$
Show Answer
(i) $(a^{-2}b)^{-2} \times (ab^{-3})$
$\Rightarrow (a^{\lbrace-2 + (-2)\rbrace}b^{-2}) \times (a^{-3}b^{-3})$
$\Rightarrow (a^{4}b^{-2}) \times (a^{-3}b^{-3})$
$\Rightarrow (a^{4} \times \dfrac{1}{b^2}) \times (\dfrac{1}{a^3} \times \dfrac{1}{b^3})$
$\Rightarrow (\dfrac{a^{4}}{b^2}) \times (\dfrac{1}{a^3b^3})$
$\Rightarrow \dfrac{a^4}{a^3b^{2 + 3}} = \dfrac{a^{4-3}}{b^5} = \dfrac{a}{b^5}$ Ans
(ii) $(x^ny^{-m})^4 \times (x^3y^{-2})^{-n}$
$\Rightarrow (x^{4 \times n}y^{4 \times (-m)}) \times (x^{3 \times (-n)}y^{(-2) \times (-n)})$
$\Rightarrow (x^{4n}y^{-4m}) \times (x^{-3n}y^{2n})$
$\Rightarrow (x^{4n}y^{-4m}) \times (x^{-3n}y^{2n})$
$\Rightarrow (x^{4n} \times \dfrac{1}{y^{4m}}) \times (\dfrac{1}{x^{3n}} \times y^{2n})$
$\Rightarrow \dfrac{x^{4n}y^{2n}}{x^{3n}y^{4m}} = \dfrac{x^{4n - 3n}y^{2n}}{y^{4m}}$
$\Rightarrow \dfrac{x^ny^{2n}}{y^{4m}} = x^ny^{2n - 4m}$ Ans
(iii) $(\dfrac{125a^{-3}}{y^6})^{-\frac{1}{3}}$
$\Rightarrow (\dfrac{(5)^3a^{-3}}{y^6})^{-\frac{1}{3}}$
$\Rightarrow \dfrac{(5)^{-\frac{1}{3} \times 3}(a)^{-\frac{1}{3} \times (-3)}}{y^{-\frac{1}{3} \times 6}}$
$\Rightarrow \dfrac{5^{-1}a}{y^{-2}}$
$\Rightarrow \dfrac{\dfrac{1}{5} \times a}{\dfrac{1}{y^2}} = \dfrac{\dfrac{a}{5}}{\dfrac{1}{y^2}} = \dfrac{ay^2}{5}$ Ans
(iv) $(\dfrac{32x^{-5}}{243y^{-5}})^{-\frac{1}{5}}$
$\Rightarrow \lbrace\dfrac{(2)^5(x)^{-5}}{(3)^5(y)^{-5}}\rbrace^{-\frac{1}{5}}$
$\Rightarrow \lbrace\dfrac{(2)^{-\frac{1}{5} \times 5}(x)^{-\frac{1}{5} \times (-5)}}{(3)^{-\frac{1}{5} \times 5}(y)^{-\frac{1}{5} \times (-5)}}\rbrace$
$\Rightarrow \dfrac{2^{-1}x}{3^{-1}y} = \dfrac{\dfrac{1}{2} \times x}{\dfrac{1}{3} \times y}$
$\Rightarrow \dfrac{\dfrac{x}{2}}{\dfrac{y}{3}} = \dfrac{3x}{2y}$ Ans
(v) $(a^{-2}b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}}$
$\Rightarrow \lbrace(a)^{\frac{1}{2} \times (-2)}(b)^{\frac{1}{2}}\rbrace \times \lbrace(a)^{\frac{1}{3}}(b)^{\frac{1}{3} \times (-3)}\rbrace$
$\Rightarrow (a^{-1}b^{\frac{1}{2}}) \times (a^{\frac{1}{3}}b^{-1})$
$\Rightarrow (a)^{(-1) + \frac{1}{3}}(b)^{\frac{1}{2} - 1}$
$\Rightarrow (a)^{\frac{(-3) + 1}{3}}(b)^{\frac{1 - 2}{2}}$
$\Rightarrow (a)^{-\frac{2}{3}}(b)^{-\frac{1}{2}}$
$\Rightarrow (a)^{-\frac{2}{3}}(b)^{-\frac{1}{2}}$
$\Rightarrow \dfrac{1}{a^{\frac{2}{3}}} \times \dfrac{1}{b^{\frac{1}{2}}}$
$\Rightarrow \dfrac{1}{a^{\frac{2}{3}}b^{\frac{1}{2}}}$ Ans
(vi) $(xy)^{m - n} \times (yz)^{n - l} \times (zx)^{l - m}$
$\Rightarrow (x^{m - n}y^{m - n}) (y^{n - l}z^{n - l}) (z^{l - m}x^{l - m})$
$\Rightarrow (x^{m - n} \times x^{l - m}) (y^{m - n} \times y^{n - l}) (z^{n - l} \times z^{l - m})$
$\Rightarrow (x^{m - n + l - m}) (y^{m - n + n - l}) (z^{n - l + l - m})$
$\Rightarrow (x^{-n + l}) (y^{m - l}) (z^{n - m})$
$\Rightarrow (x^{-n})(x^l)(y^m)(y^{-l})(z^n)(z^{-m})$
$\Rightarrow (\dfrac{1}{x^n})(x^l)(y^m)(\dfrac{1}{y^l})(z^n)(\dfrac{1}{z^m})$
$\Rightarrow \dfrac{x^ly^mz^n}{x^ny^lz^m}$
$\Rightarrow x^{l - n}y^{m - l}z^{n - m}$ Ans
$\Rightarrow (a^{\lbrace-2 + (-2)\rbrace}b^{-2}) \times (a^{-3}b^{-3})$
$\Rightarrow (a^{4}b^{-2}) \times (a^{-3}b^{-3})$
$\Rightarrow (a^{4} \times \dfrac{1}{b^2}) \times (\dfrac{1}{a^3} \times \dfrac{1}{b^3})$
$\Rightarrow (\dfrac{a^{4}}{b^2}) \times (\dfrac{1}{a^3b^3})$
$\Rightarrow \dfrac{a^4}{a^3b^{2 + 3}} = \dfrac{a^{4-3}}{b^5} = \dfrac{a}{b^5}$ Ans
(ii) $(x^ny^{-m})^4 \times (x^3y^{-2})^{-n}$
$\Rightarrow (x^{4 \times n}y^{4 \times (-m)}) \times (x^{3 \times (-n)}y^{(-2) \times (-n)})$
$\Rightarrow (x^{4n}y^{-4m}) \times (x^{-3n}y^{2n})$
$\Rightarrow (x^{4n}y^{-4m}) \times (x^{-3n}y^{2n})$
$\Rightarrow (x^{4n} \times \dfrac{1}{y^{4m}}) \times (\dfrac{1}{x^{3n}} \times y^{2n})$
$\Rightarrow \dfrac{x^{4n}y^{2n}}{x^{3n}y^{4m}} = \dfrac{x^{4n - 3n}y^{2n}}{y^{4m}}$
$\Rightarrow \dfrac{x^ny^{2n}}{y^{4m}} = x^ny^{2n - 4m}$ Ans
(iii) $(\dfrac{125a^{-3}}{y^6})^{-\frac{1}{3}}$
$\Rightarrow (\dfrac{(5)^3a^{-3}}{y^6})^{-\frac{1}{3}}$
$\Rightarrow \dfrac{(5)^{-\frac{1}{3} \times 3}(a)^{-\frac{1}{3} \times (-3)}}{y^{-\frac{1}{3} \times 6}}$
$\Rightarrow \dfrac{5^{-1}a}{y^{-2}}$
$\Rightarrow \dfrac{\dfrac{1}{5} \times a}{\dfrac{1}{y^2}} = \dfrac{\dfrac{a}{5}}{\dfrac{1}{y^2}} = \dfrac{ay^2}{5}$ Ans
(iv) $(\dfrac{32x^{-5}}{243y^{-5}})^{-\frac{1}{5}}$
$\Rightarrow \lbrace\dfrac{(2)^5(x)^{-5}}{(3)^5(y)^{-5}}\rbrace^{-\frac{1}{5}}$
$\Rightarrow \lbrace\dfrac{(2)^{-\frac{1}{5} \times 5}(x)^{-\frac{1}{5} \times (-5)}}{(3)^{-\frac{1}{5} \times 5}(y)^{-\frac{1}{5} \times (-5)}}\rbrace$
$\Rightarrow \dfrac{2^{-1}x}{3^{-1}y} = \dfrac{\dfrac{1}{2} \times x}{\dfrac{1}{3} \times y}$
$\Rightarrow \dfrac{\dfrac{x}{2}}{\dfrac{y}{3}} = \dfrac{3x}{2y}$ Ans
(v) $(a^{-2}b)^{\frac{1}{2}} \times (ab^{-3})^{\frac{1}{3}}$
$\Rightarrow \lbrace(a)^{\frac{1}{2} \times (-2)}(b)^{\frac{1}{2}}\rbrace \times \lbrace(a)^{\frac{1}{3}}(b)^{\frac{1}{3} \times (-3)}\rbrace$
$\Rightarrow (a^{-1}b^{\frac{1}{2}}) \times (a^{\frac{1}{3}}b^{-1})$
$\Rightarrow (a)^{(-1) + \frac{1}{3}}(b)^{\frac{1}{2} - 1}$
$\Rightarrow (a)^{\frac{(-3) + 1}{3}}(b)^{\frac{1 - 2}{2}}$
$\Rightarrow (a)^{-\frac{2}{3}}(b)^{-\frac{1}{2}}$
$\Rightarrow (a)^{-\frac{2}{3}}(b)^{-\frac{1}{2}}$
$\Rightarrow \dfrac{1}{a^{\frac{2}{3}}} \times \dfrac{1}{b^{\frac{1}{2}}}$
$\Rightarrow \dfrac{1}{a^{\frac{2}{3}}b^{\frac{1}{2}}}$ Ans
(vi) $(xy)^{m - n} \times (yz)^{n - l} \times (zx)^{l - m}$
$\Rightarrow (x^{m - n}y^{m - n}) (y^{n - l}z^{n - l}) (z^{l - m}x^{l - m})$
$\Rightarrow (x^{m - n} \times x^{l - m}) (y^{m - n} \times y^{n - l}) (z^{n - l} \times z^{l - m})$
$\Rightarrow (x^{m - n + l - m}) (y^{m - n + n - l}) (z^{n - l + l - m})$
$\Rightarrow (x^{-n + l}) (y^{m - l}) (z^{n - m})$
$\Rightarrow (x^{-n})(x^l)(y^m)(y^{-l})(z^n)(z^{-m})$
$\Rightarrow (\dfrac{1}{x^n})(x^l)(y^m)(\dfrac{1}{y^l})(z^n)(\dfrac{1}{z^m})$
$\Rightarrow \dfrac{x^ly^mz^n}{x^ny^lz^m}$
$\Rightarrow x^{l - n}y^{m - l}z^{n - m}$ Ans
Q8. Show that: $(\dfrac{x^a}{x^{-b}})^{a-b}.(\dfrac{x^b}{x^{-c}})^{b-c}.(\dfrac{x^c}{x^{-a}})^{c-a} = 1$
Show Answer
$\Rightarrow \dfrac{x^{a(a-b)}}{x^{-b(a-b)}}.\dfrac{x^{b(b-c)}}{x^{-c(b-c)}}.\dfrac{x^{c(c-a)}}{x^{-a(c-a)}}$
$\Rightarrow \dfrac{x^{a^2-ab}}{x^{-ab+b^2}}.\dfrac{x^{b^2-bc}}{x^{-bc+c^2}}.\dfrac{x^{c^2-ac}}{x^{-ac+a^2}}$
$\Rightarrow \dfrac{x^{a^2-ab+b^2-bc+c^2-ac}}{x^{-ab+b^2+(-bc+c^2)+(-ac+a^2)}}$
$\Rightarrow \dfrac{x^{a^2-ab+b^2-bc+c^2-ac}}{x^{-ab+b^2-bc+c^2-ac+a^2}}$
$\therefore \dfrac{x^{a^2+b^2+c^2-ab-bc-ac}}{x^{a^2+b^2+c^2-ab-bc-ac}} = 1$ Ans
$\Rightarrow \dfrac{x^{a^2-ab}}{x^{-ab+b^2}}.\dfrac{x^{b^2-bc}}{x^{-bc+c^2}}.\dfrac{x^{c^2-ac}}{x^{-ac+a^2}}$
$\Rightarrow \dfrac{x^{a^2-ab+b^2-bc+c^2-ac}}{x^{-ab+b^2+(-bc+c^2)+(-ac+a^2)}}$
$\Rightarrow \dfrac{x^{a^2-ab+b^2-bc+c^2-ac}}{x^{-ab+b^2-bc+c^2-ac+a^2}}$
$\therefore \dfrac{x^{a^2+b^2+c^2-ab-bc-ac}}{x^{a^2+b^2+c^2-ab-bc-ac}} = 1$ Ans
Q9. Evaluate: $\dfrac{x^{5+n}.(x^2)^{3n+1}}{x^{7n-2}}$
Show Answer
$\Rightarrow \dfrac{x^5.x^n.x^{2(3n+1)}}{x^{7n}.x^{-2}}$
$\Rightarrow \dfrac{x^5.x^n.x^{6n+2}}{x^{7n}.x^{-2}}$
$\Rightarrow \dfrac{x^5.x^n.x^{6n}.x^2}{x^{7n}.x^{-2}}$
$\Rightarrow \dfrac{x^{5+2}.x^{6n+n}}{x^{7n}.\dfrac{1}{x^2}}$
$\Rightarrow \dfrac{x^7.x^{7n}}{x^{7n}.\dfrac{1}{x^2}}$
$\Rightarrow \dfrac{x^{7}}{\dfrac{1}{x^2}}$
$\Rightarrow x^7.x^2 = x^{7+2} = x^9$ Ans
$\Rightarrow \dfrac{x^5.x^n.x^{6n+2}}{x^{7n}.x^{-2}}$
$\Rightarrow \dfrac{x^5.x^n.x^{6n}.x^2}{x^{7n}.x^{-2}}$
$\Rightarrow \dfrac{x^{5+2}.x^{6n+n}}{x^{7n}.\dfrac{1}{x^2}}$
$\Rightarrow \dfrac{x^7.x^{7n}}{x^{7n}.\dfrac{1}{x^2}}$
$\Rightarrow \dfrac{x^{7}}{\dfrac{1}{x^2}}$
$\Rightarrow x^7.x^2 = x^{7+2} = x^9$ Ans
Q10. Evaluate: $\dfrac{a^{2n+1}.a^{(2n+1)(2n-1)}}{a^{n(4n-1)}.(a^2)^{2n+3}}$
Show Answer
$\Rightarrow \dfrac{(a^{2n}.a)(a^{2n(2n-1)+1(2n-1)})}{a^{4n^2-n}.a^{4n+6}}$
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-2n+2n-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-2n+2n-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^{-1}}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^{-1}}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.\dfrac{1}{a}}{a^{4n^2}.\dfrac{1}{a^n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^n}{a^{4n^2}.a.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n+n}}{a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{3n}}{a^{4n}.a^6}$
$\Rightarrow \dfrac{1}{a^{4n-3n}.a^6}$
$\Rightarrow \dfrac{1}{a^n.a^6} = \dfrac{1}{a^{n+6}}$ Ans
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-2n+2n-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-2n+2n-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{(a^{2n}.a)(a^{4n^2-1})}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^{-1}}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^{-1}}{a^{4n^2}.a^{-n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.\dfrac{1}{a}}{a^{4n^2}.\dfrac{1}{a^n}.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n}.a.a^{4n^2}.a^n}{a^{4n^2}.a.a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{2n+n}}{a^{4n}.a^6}$
$\Rightarrow \dfrac{a^{3n}}{a^{4n}.a^6}$
$\Rightarrow \dfrac{1}{a^{4n-3n}.a^6}$
$\Rightarrow \dfrac{1}{a^n.a^6} = \dfrac{1}{a^{n+6}}$ Ans
Q11. Prove that: $(m+n)^{-1}(m^{-1}+n^{-1}) = (mn)^{-1}$
Show Answer
$\Rightarrow (m+n)^{-1}(m^{-1}+n^{-1})$
$\Rightarrow (\dfrac{1}{m+n})(\dfrac{1}{m}+\dfrac{1}{n})$
$\Rightarrow (\dfrac{1}{m+n})(\dfrac{n+m}{mn})$
$\because \dfrac{1}{m+n} \times \dfrac{n+m}{mn} = \dfrac{1}{mn}$
$\therefore \dfrac{1}{mn} = (mn)^{-1}$ Hence Proved
$\Rightarrow (\dfrac{1}{m+n})(\dfrac{1}{m}+\dfrac{1}{n})$
$\Rightarrow (\dfrac{1}{m+n})(\dfrac{n+m}{mn})$
$\because \dfrac{1}{m+n} \times \dfrac{n+m}{mn} = \dfrac{1}{mn}$
$\therefore \dfrac{1}{mn} = (mn)^{-1}$ Hence Proved
Q12. Prove that:
(i) $(\dfrac{x^a}{x^b})^{\frac{1}{ab}}.(\dfrac{x^b}{x^c})^{\frac{1}{bc}}.(\dfrac{x^c}{x^a})^{\frac{1}{ca}} = 1$
(ii) $\dfrac{1}{1+x^{a-b}} + \dfrac{1}{1+x^{b-a}} = 1$
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(i) $(\dfrac{x^a}{x^b})^{\frac{1}{ab}}.(\dfrac{x^b}{x^c})^{\frac{1}{bc}}.(\dfrac{x^c}{x^a})^{\frac{1}{ca}}$
$\Rightarrow (\dfrac{x^{\frac{1}{ab} \times a}}{x^{\frac{1}{ab} \times b}}).(\dfrac{x^{\frac{1}{bc} \times b}}{x^{\frac{1}{bc} \times c}}).(\dfrac{x^{\frac{1}{ca} \times c}}{x^{\frac{1}{ca} \times a}})$
$\Rightarrow (\dfrac{x^{\frac{1}{b}}}{x^{\frac{1}{a}}}).(\dfrac{x^{\frac{1}{c}} }{x^{\frac{1}{b}}}).(\dfrac{x^{\frac{1}{a}}}{x^{\frac{1}{c}}})$
$\Rightarrow \dfrac{x^{\frac{1}{b} + \frac{1}{c} + \frac{1}{a}}}{x^{\frac{1}{b} + \frac{1}{c} + \frac{1}{a}}} = 1$ Hence Proved
(ii) $\dfrac{1}{1+x^{a-b}} + \dfrac{1}{1+x^{b-a}}$
$\because \text{LCD} = (1 + x^{a-b})(1 + x^{b-a})$
$\therefore \dfrac{(1 + x^{b-a}) + (1 + x^{a-b})}{(1 + x^{a-b})(1 + x^{b-a})}$
Now solving for the numerator:
$\Rightarrow (1 + x^{b-a}) + (1 + x^{a-b})$
$\Rightarrow 1 + x^{b-a} + 1 + x^{a-b}$
$\Rightarrow 2 + x^{b-a} + x^{a-b}$ Required Numerator
Now solving for the denominator:
$\Rightarrow (1 + x^{a-b})(1 + x^{b-a})$
$\Rightarrow 1(1 + x^{b-a}) + x^{a-b}(1 + x^{b-a})$
$\Rightarrow (1 + x^{b-a}) + (x^{a-b} + x^{a-b+b-a})$
$\Rightarrow (1 + x^{b-a}) + (x^{a-b} + x^0)$
$\Rightarrow 1 + x^{b-a} + x^{a-b} + 1$
$\Rightarrow 2 + x^{b-a} + x^{a-b}$ Required Denominator
As such, we have:
$\Rightarrow \dfrac{2 + x^{b-a} + x^{a-b}}{2 + x^{b-a} + x^{a-b}} = 1$ Hence Proved
$\Rightarrow (\dfrac{x^{\frac{1}{ab} \times a}}{x^{\frac{1}{ab} \times b}}).(\dfrac{x^{\frac{1}{bc} \times b}}{x^{\frac{1}{bc} \times c}}).(\dfrac{x^{\frac{1}{ca} \times c}}{x^{\frac{1}{ca} \times a}})$
$\Rightarrow (\dfrac{x^{\frac{1}{b}}}{x^{\frac{1}{a}}}).(\dfrac{x^{\frac{1}{c}} }{x^{\frac{1}{b}}}).(\dfrac{x^{\frac{1}{a}}}{x^{\frac{1}{c}}})$
$\Rightarrow \dfrac{x^{\frac{1}{b} + \frac{1}{c} + \frac{1}{a}}}{x^{\frac{1}{b} + \frac{1}{c} + \frac{1}{a}}} = 1$ Hence Proved
(ii) $\dfrac{1}{1+x^{a-b}} + \dfrac{1}{1+x^{b-a}}$
$\because \text{LCD} = (1 + x^{a-b})(1 + x^{b-a})$
$\therefore \dfrac{(1 + x^{b-a}) + (1 + x^{a-b})}{(1 + x^{a-b})(1 + x^{b-a})}$
Now solving for the numerator:
$\Rightarrow (1 + x^{b-a}) + (1 + x^{a-b})$
$\Rightarrow 1 + x^{b-a} + 1 + x^{a-b}$
$\Rightarrow 2 + x^{b-a} + x^{a-b}$ Required Numerator
Now solving for the denominator:
$\Rightarrow (1 + x^{a-b})(1 + x^{b-a})$
$\Rightarrow 1(1 + x^{b-a}) + x^{a-b}(1 + x^{b-a})$
$\Rightarrow (1 + x^{b-a}) + (x^{a-b} + x^{a-b+b-a})$
$\Rightarrow (1 + x^{b-a}) + (x^{a-b} + x^0)$
$\Rightarrow 1 + x^{b-a} + x^{a-b} + 1$
$\Rightarrow 2 + x^{b-a} + x^{a-b}$ Required Denominator
As such, we have:
$\Rightarrow \dfrac{2 + x^{b-a} + x^{a-b}}{2 + x^{b-a} + x^{a-b}} = 1$ Hence Proved