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Chapter 6

Indices — Exercise 6

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 6

Question 1(i)

Evaluate :

(125)13(125)^{\dfrac{1}{3}}

Answer

Given,

(125)13(125)^{\dfrac{1}{3}}

Simplifying the expression :

⇒(125)13⇒[(5)3]13⇒53×13⇒5.\Rightarrow (125)^{\dfrac{1}{3}} \\[1em] \Rightarrow [(5)^3]^{\dfrac{1}{3}} \\[1em] \Rightarrow 5^{3 \times \dfrac{1}{3}} \\[1em] \Rightarrow 5.

Hence, (125)13(125)^{\dfrac{1}{3}} = 5.

Question 1(ii)

Evaluate :

(8)23(8)^{\dfrac{2}{3}}

Answer

Given,

(8)23(8)^{\dfrac{2}{3}}

Simplifying the expression :

⇒(8)23⇒(23)23⇒23×23⇒22⇒4.\Rightarrow (8)^{\dfrac{2}{3}} \\[1em] \Rightarrow (2^3)^{\dfrac{2}{3}} \\[1em] \Rightarrow 2^{3 \times \dfrac{2}{3}} \\[1em] \Rightarrow 2^2 \\[1em] \Rightarrow 4.

Hence, (8)23=4(8)^{\dfrac{2}{3}} = 4.

Question 1(iii)

Evaluate :

(15)−2\Big(\dfrac{1}{5}\Big)^{-2}

Answer

Given,

(15)−2\Big(\dfrac{1}{5}\Big)^{-2}

Simplifying the expression :

⇒(15)−2⇒52⇒25.\Rightarrow \Big(\dfrac{1}{5}\Big)^{-2} \\[1em] \Rightarrow 5^2 \\[1em] \Rightarrow 25.

Hence, (15)−2=25\Big(\dfrac{1}{5}\Big)^{-2} = 25.

Question 1(iv)

Evaluate :

(16)−34(16)^{\dfrac{-3}{4}}

Answer

Given,

(16)−34(16)^{\dfrac{-3}{4}}

Simplifying the expression :

⇒(16)−34⇒(24)−34⇒24×−34⇒(2)−3⇒(12)3⇒18.\Rightarrow (16)^{\dfrac{-3}{4}} \\[1em] \Rightarrow (2^4)^{\dfrac{-3}{4}} \\[1em] \Rightarrow 2^{4 \times -\dfrac{3}{4}} \\[1em] \Rightarrow (2)^{-3} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3} \\[1em] \Rightarrow \dfrac{1}{8}.

Hence, (16)−34=18(16)^{\dfrac{-3}{4}} = \dfrac{1}{8}.

Question 1(v)

Evaluate :

(32)−45(32)^{\dfrac{-4}{5}}

Answer

Given,

(32)−45(32)^{\dfrac{-4}{5}}

Simplifying the expression :

⇒(32)−45⇒(25)−45⇒25×−45⇒(2)−4⇒(12)4⇒116.\Rightarrow (32)^{\dfrac{-4}{5}} \\[1em] \Rightarrow (2^5)^{\dfrac{-4}{5}} \\[1em] \Rightarrow 2^{5 \times -\dfrac{4}{5}} \\[1em] \Rightarrow (2)^{-4} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{4} \\[1em] \Rightarrow \dfrac{1}{16}.

Hence, (32)−45=116(32)^{\dfrac{-4}{5}} = \dfrac{1}{16}.

Question 1(vi)

Evaluate :

(8125)−13\Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}}

Answer

Given,

(8125)−13\Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}}

Simplifying the expression :

⇒(8125)−13⇒[(25)3]−13⇒(25)3×−13⇒(25)−1⇒52.\Rightarrow \Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\dfrac{2}{5}\Big)^3\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{3 \times -\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{-1} \\[1em] \Rightarrow \dfrac{5}{2}.

Hence, (8125)−13=52\Big(\dfrac{8}{125}\Big)^{-\dfrac{1}{3}} = \dfrac{5}{2}.

Question 1(vii)

Evaluate :

(−27)23(-27)^{\dfrac{2}{3}}

Answer

Given,

(−27)23(-27)^{\dfrac{2}{3}}

Simplifying the expression :

⇒(−27)23⇒[(−3)3]23⇒(−3)3×23⇒(−3)2⇒9.\Rightarrow (-27)^{\dfrac{2}{3}} \\[1em] \Rightarrow [(-3)^3]^{\dfrac{2}{3}} \\[1em] \Rightarrow (-3)^{3 \times \dfrac{2}{3}} \\[1em] \Rightarrow (-3)^2 \\[1em] \Rightarrow 9.

Hence, (−27)23=9(-27)^{\dfrac{2}{3}} = 9.

Question 1(viii)

Evaluate :

(0.001)−13(0.001)^{-\dfrac{1}{3}}

Answer

Given,

(0.001)−13(0.001)^{-\dfrac{1}{3}}

Simplifying the expression :

⇒(0.001)−13⇒(11000)−13⇒[(110)3]−13⇒(110)3×−13⇒(110)−1⇒10.\Rightarrow (0.001)^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{1}{1000}\Big)^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{10}\Big)^3\Big]^{-\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{1}{10}\Big)^{3 \times -\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{1}{10}\Big)^{-1} \\[1em] \Rightarrow 10.

Hence, (0.001)−13=10(0.001)^{-\dfrac{1}{3}} = 10.

Question 1(ix)

Evaluate :

(0.027)−23(0.027)^{\dfrac{-2}{3}}

Answer

Given,

(0.027)−23(0.027)^{\dfrac{-2}{3}}

Simplifying the expression :

⇒(0.027)−23⇒(271000)−23⇒[(310)3]−23⇒(310)3×−23⇒(310)−2⇒(103)2⇒1009.\Rightarrow (0.027)^{\dfrac{-2}{3}} \\[1em] \Rightarrow \Big(\dfrac{27}{1000}\Big)^{-\dfrac{2}{3}} \\[1em] \Rightarrow \Big[\Big(\dfrac{3}{10}\Big)^3\Big]^{-\dfrac{2}{3}} \\[1em] \Rightarrow \Big(\dfrac{3}{10}\Big)^{3 \times -\dfrac{2}{3}} \\[1em] \Rightarrow \Big(\dfrac{3}{10}\Big)^{-2} \\[1em] \Rightarrow \Big(\dfrac{10}{3}\Big)^{2} \\[1em] \Rightarrow \dfrac{100}{9}.

Hence, (0.027)−23=1009(0.027)^{\dfrac{-2}{3}} = \dfrac{100}{9}.

Question 2(i)

Evaluate the following :

(14)−2−3×(8)23×50+(916)−12\Big(\dfrac{1}{4}\Big)^{-2} - 3 \times (8)^{\dfrac{2}{3}} \times 5^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}}

Answer

Given,

(14)−2−3×(8)23×50+(916)−12\Big(\dfrac{1}{4}\Big)^{-2} - 3 \times (8)^{\dfrac{2}{3}} \times 5^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}}

Simplifying the expression :

⇒(14)−2−3×(8)23×50+(916)−12⇒(4)2−3×[(2)3]23×1+(169)12⇒16−3×(2)2×1+[(43)2]12⇒16−12+43⇒4+43⇒12+43⇒163=513.\Rightarrow \Big(\dfrac{1}{4}\Big)^{-2} - 3 \times (8)^{\dfrac{2}{3}} \times 5^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}} \\[1em] \Rightarrow (4)^2 - 3 \times [(2)^3]^{\dfrac{2}{3}} \times 1 + \Big(\dfrac{16}{9}\Big)^{\dfrac{1}{2}} \\[1em] \Rightarrow 16 - 3 \times (2)^2 \times 1 + \Big[\Big(\dfrac{4}{3}\Big)^2\Big]^{\dfrac{1}{2}} \\[1em] \Rightarrow 16 - 12 + \dfrac{4}{3} \\[1em] \Rightarrow 4 + \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{12 + 4}{3} \\[1em] \Rightarrow \dfrac{16}{3} = 5\dfrac{1}{3}.

Hence, (14)−2−3×(8)23×50+(916)−12=513\Big(\dfrac{1}{4}\Big)^{-2} - 3 \times (8)^{\dfrac{2}{3}} \times 5^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}} = 5\dfrac{1}{3}.

Question 2(ii)

Evaluate the following :

14+(0.01)−12−(27)23×30\sqrt{\dfrac{1}{4}} + (0.01)^{\dfrac{-1}{2}} - (27)^{\dfrac{2}{3}} \times 3^0

Answer

Given,

14+(0.01)−12−(27)23×30\sqrt{\dfrac{1}{4}} + (0.01)^{\dfrac{-1}{2}} - (27)^{\dfrac{2}{3}} \times 3^0

Simplifying the expression :

⇒14+(0.01)−12−(27)23×30⇒12+(1100)−12−[(3)3]23×1⇒12+(100)12−(3)2⇒12+[(10)2]12−9⇒12+10−9⇒12+1⇒1+22⇒32⇒112.\Rightarrow \sqrt{\dfrac{1}{4}} + (0.01)^{\dfrac{-1}{2}} - (27)^{\dfrac{2}{3}} \times 3^0 \\[1em] \Rightarrow \dfrac{1}{2} + \Big(\dfrac{1}{100}\Big)^{\dfrac{-1}{2}} - [(3)^3]^{\dfrac{2}{3}} \times 1 \\[1em] \Rightarrow \dfrac{1}{2} + (100)^{\dfrac{1}{2}} - (3)^2 \\[1em] \Rightarrow \dfrac{1}{2} + [(10)^2]^{\dfrac{1}{2}} - 9 \\[1em] \Rightarrow \dfrac{1}{2} + 10 - 9 \\[1em] \Rightarrow \dfrac{1}{2} + 1 \\[1em] \Rightarrow \dfrac{1 + 2}{2} \\[1em] \Rightarrow \dfrac{3}{2} \\[1em] \Rightarrow 1\dfrac{1}{2}.

Hence, 14+(0.01)−12−(27)23×30=112\sqrt{\dfrac{1}{4}} + (0.01)^{\dfrac{-1}{2}} - (27)^{\dfrac{2}{3}} \times 3^0 = 1\dfrac{1}{2}.

Question 3(i)

Evaluate the following :

(8116)−34×[(259)−32÷(52)−3]\Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} ÷ {\Big(\dfrac{5}{2}\Big)^{-3}}\Big]

Answer

Given,

(8116)−34×[(259)−32÷(52)−3]\Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} ÷ {\Big(\dfrac{5}{2}\Big)^{-3}}\Big]

Simplifying the expression :

⇒(8116)−34×[(259)−32÷(52)−3]⇒(1681)34×[(925)32÷(25)3]⇒[(23)4]34×[[(35)2]32÷(8125)]⇒(23)3×[(35)3÷(8125)]⇒(827)×[(27125)÷(8125)]⇒(827)×[(27125)×(1258)]⇒(827)×(278)⇒1.\Rightarrow \Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} ÷ {\Big(\dfrac{5}{2}\Big)^{-3}}\Big] \\[1em] \Rightarrow \Big(\dfrac{16}{81}\Big)^{\dfrac{3}{4}} \times \Big[\Big(\dfrac{9}{25}\Big)^{\dfrac{3}{2}} ÷ {\Big(\dfrac{2}{5}\Big)^{3}}\Big] \\[1em] \Rightarrow \Big[\Big(\dfrac{2}{3}\Big)^4\Big]^{\dfrac{3}{4}} \times \Big[\Big[\Big(\dfrac{3}{5}\Big)^2\Big]^{\dfrac{3}{2}} ÷ {\Big(\dfrac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^3 \times \Big[\Big(\dfrac{3}{5}\Big)^3 ÷ {\Big(\dfrac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\dfrac{8}{27}\Big) \times \Big[\Big(\dfrac{27}{125}\Big) ÷ {\Big(\dfrac{8}{125}\Big)}\Big] \\[1em] \Rightarrow \Big(\dfrac{8}{27}\Big) \times \Big[\Big(\dfrac{27}{125}\Big) \times {\Big(\dfrac{125}{8}\Big)}\Big] \\[1em] \Rightarrow \Big(\dfrac{8}{27}\Big) \times \Big(\dfrac{27}{8}\Big) \\[1em] \Rightarrow 1.

Hence, (8116)−34×[(259)−32÷(52)−3]=1\Big(\dfrac{81}{16}\Big)^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} ÷ {\Big(\dfrac{5}{2}\Big)^{-3}}\Big] = 1.

Question 3(ii)

Evaluate the following :

[(64)23×2−2÷70]−12\Big[(64)^{\dfrac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\dfrac{1}{2}}

Answer

Given,

[(64)23×2−2÷70]−12\Big[(64)^{\dfrac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\dfrac{1}{2}}

Simplifying the expression :

⇒[[(4)3]23×(12)2÷1]−12⇒[(4)3×23×(12)2÷1]−12⇒[(4)2×(14)÷1]−12⇒[16×(14)]−12⇒(4)−12⇒(14)12⇒[(12)2]12⇒12.\Rightarrow \Big[[(4)^3]^{\dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^{2} ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[(4)^{3 \times \dfrac{2}{3}} \times \Big(\dfrac{1}{2}\Big)^{2} ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[(4)^2 \times \Big(\dfrac{1}{4}\Big) ÷ 1\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big[16 \times \Big(\dfrac{1}{4}\Big)\Big]^{-\dfrac{1}{2}} \\[1em] \Rightarrow (4)^{-\dfrac{1}{2}} \\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big)^{\dfrac{1}{2}} \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{2}\Big)^2\Big]^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{1}{2}.

Hence, [(64)23×2−2÷70]−12=12\Big[(64)^{\dfrac{2}{3}} \times 2^{-2} ÷ 7^0\Big]^{-\dfrac{1}{2}} = \dfrac{1}{2}.

Question 4(i)

Evaluate the following :

(81)34−(132)−25+(8)13.(12)−1.(2)0(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} . \Big(\dfrac{1}{2}\Big)^{-1} . (2)^0

Answer

Given,

(81)34−(132)−25+(8)13.(12)−1.(2)0(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} . \Big(\dfrac{1}{2}\Big)^{-1} . (2)^0

Simplifying the expression :

⇒(81)34−(132)−25+(8)13×(12)−1×(2)0⇒[(3)4]34−(32)25+[(2)3]13×(2)×1⇒(3)3−[(2)5]25+21×2⇒27−(2)2+4⇒31−4⇒27.\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} \times \Big(\dfrac{1}{2}\Big)^{-1} \times (2)^0 \\[1em] \Rightarrow [(3)^4]^{\dfrac{3}{4}} - (32)^{\dfrac{2}{5}} + [(2)^3]^{\dfrac{1}{3}} \times (2) \times 1 \\[1em] \Rightarrow (3)^3 - [(2)^5]^{\dfrac{2}{5}} + 2^1 \times 2 \\[1em] \Rightarrow 27 - (2)^2 + 4 \\[1em] \Rightarrow 31 - 4 \\[1em] \Rightarrow 27.

Hence, (81)34−(132)−25+(8)13×(12)−1×(2)0=27(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + (8)^{\dfrac{1}{3}} \times \Big(\dfrac{1}{2}\Big)^{-1} \times (2)^0 = 27.

Question 4(ii)

Evaluate the following :

(1681)−34×(499)32÷(343216)23\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}}

Answer

Given,

(1681)−34×(499)32÷(343216)23\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}}

Simplifying the expression :

⇒(1681)−34×(499)32÷(343216)23⇒(8116)34×[(73)2]32÷[(76)3]23⇒[(32)4]34×[(73)2]32÷[(76)3]23⇒(32)3×(73)3÷(76)2⇒(278)×[(34327)÷(4936)]⇒(278)×[(34327)×(3649)]⇒(278)×(73)×4⇒(92)×7⇒632=3112.\Rightarrow \Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} \\[1em] \Rightarrow \Big(\dfrac{81}{16}\Big)^{\dfrac{3}{4}} \times \Big[\Big(\dfrac{7}{3}\Big)^2\Big]^{\dfrac{3}{2}} ÷ \Big[\Big(\dfrac{7}{6}\Big)^3\Big]^{\dfrac{2}{3}} \\[1em] \Rightarrow \Big[\Big(\dfrac{3}{2}\Big)^4\Big]^{\dfrac{3}{4}} \times \Big[\Big(\dfrac{7}{3}\Big)^2\Big]^{\dfrac{3}{2}} ÷ \Big[\Big(\dfrac{7}{6}\Big)^3\Big]^{\dfrac{2}{3}} \\[1em] \Rightarrow \Big(\dfrac{3}{2}\Big)^3 \times \Big(\dfrac{7}{3}\Big)^3 ÷ \Big(\dfrac{7}{6}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{27}{8}\Big) \times \Big[\Big(\dfrac{343}{27}\Big) ÷ \Big(\dfrac{49}{36}\Big)\Big] \\[1em] \Rightarrow \Big(\dfrac{27}{8}\Big) \times \Big[\Big(\dfrac{343}{27}\Big) \times \Big(\dfrac{36}{49}\Big)\Big] \\[1em] \Rightarrow \Big(\dfrac{27}{8}\Big) \times \Big(\dfrac{7}{3}\Big) \times 4 \\[1em] \Rightarrow \Big(\dfrac{9}{2}\Big) \times 7 \\[1em] \Rightarrow \dfrac{63}{2} = 31\dfrac{1}{2}.

Hence, (1681)−34×(499)32÷(343216)23=3112\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} = 31\dfrac{1}{2}.

Question 5(i)

Evaluate the following :

(64125)−23÷1(256625)14+(25643)0\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0

Answer

Given,

(64125)−23÷1(256625)14+(25643)0\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0

Simplifying the expression :

⇒(64125)−23÷1(256625)14+(25643)0⇒(12564)23÷1(4454)14+1⇒(5343)23÷1(45)4×14+1⇒(54)3×23÷1(45)+1⇒(54)2÷54+1⇒2516÷54+1⇒2516×45+1⇒54+1⇒5+44⇒94⇒214.\Rightarrow \Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 \\[1em] \Rightarrow \Big(\dfrac{125}{64}\Big)^{\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{4^4}{5^4}\Big)^{\dfrac{1}{4}}} + 1 \\[1em] \Rightarrow \Big(\dfrac{5^3}{4^3}\Big)^{\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{4}{5}\Big)^{4 \times \dfrac{1}{4}}} + 1 \\[1em] \Rightarrow \Big(\dfrac{5}{4}\Big)^{3 \times \dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{4}{5}\Big)} + 1 \\[1em] \Rightarrow \Big(\dfrac{5}{4}\Big)^2 ÷ \dfrac{5}{4} + 1 \\[1em] \Rightarrow \dfrac{25}{16} ÷ \dfrac{5}{4} + 1 \\[1em] \Rightarrow \dfrac{25}{16} \times \dfrac{4}{5} + 1 \\[1em] \Rightarrow \dfrac{5}{4} + 1 \\[1em] \Rightarrow \dfrac{5 + 4}{4} \\[1em] \Rightarrow \dfrac{9}{4} \\[1em] \Rightarrow 2\dfrac{1}{4}.

Hence, (64125)−23÷1(256625)14+(25643)0=214\Big(\dfrac{64}{125}\Big)^{-\dfrac{2}{3}} ÷ \dfrac{1}{\Big(\dfrac{256}{625}\Big)^{\dfrac{1}{4}}} + \Big(\dfrac{\sqrt{25}}{\sqrt[3]{64}}\Big)^0 = 2\dfrac{1}{4}.

Question 5(ii)

Evaluate the following :

(32)25×(4)−12×(8)132−2÷(64)−13\dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{\dfrac{-1}{3}}}

Answer

Given,

(32)25×(4)−12×(8)132−2÷(64)−13\dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{\dfrac{-1}{3}}}

Simplifying the expression :

⇒[(2)5]25×[(2)2]−12×[(2)3]13(12)2÷(43)−13⇒(2)2×2−1×21(12)2÷4−1⇒(2)2×12×2(14)÷(14)⇒4×12×2(14)×4⇒4.\Rightarrow \dfrac{[(2)^5]^{\dfrac{2}{5}} \times [(2)^2]^{-\dfrac{1}{2}} \times {[(2)^3]^{\dfrac{1}{3}}}}{\Big(\dfrac{1}{2}\Big)^{2} ÷ (4^3)^{-\dfrac{1}{3}}} \\[1em] \Rightarrow \dfrac{(2)^2 \times 2^{-1} \times 2^1}{\Big(\dfrac{1}{2}\Big)^{2} ÷ 4^{-1}} \\[1em] \Rightarrow \dfrac{(2)^2 \times \dfrac{1}{2} \times 2}{\Big(\dfrac{1}{4}\Big) ÷ \Big(\dfrac{1}{4}\Big)} \\[1em] \Rightarrow \dfrac{4 \times \dfrac{1}{2} \times 2}{\Big(\dfrac{1}{4}\Big) \times 4} \\[1em] \Rightarrow 4.

Hence, (32)25×(4)−12×(8)132−2÷(64)−13=4\dfrac{(32)^{\dfrac{2}{5}} \times (4)^{-\dfrac{1}{2}} \times (8)^{\dfrac{1}{3}}}{2^{-2} ÷ (64)^{\dfrac{-1}{3}}} = 4.

Question 6(i)

Evaluate the following :

(27)43+(32)0.8+(0.8)−1+(0.8)0(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0

Answer

Given,

(27)43+(32)0.8+(0.8)−1+(0.8)0(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0

Simplifying the expression :

⇒(27)43+(32)0.8+(0.8)−1+(0.8)0⇒[(3)3]43+[(2)5]0.8+(810)−1+1⇒(3)3×43+[(2)5]0.8+(810)−1+1⇒(3)4+(2)4+(108)1+1⇒81+16+108+1⇒97+1.25+1⇒99.25\Rightarrow (27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0 \\[1em] \Rightarrow [(3)^3]^{\dfrac{4}{3}} + [(2)^5]^{0.8} + \Big(\dfrac{8}{10}\Big)^{-1} + 1 \\[1em] \Rightarrow (3)^{3 \times \dfrac{4}{3}} + [(2)^5]^{0.8} + \Big(\dfrac{8}{10}\Big)^{-1} + 1 \\[1em] \Rightarrow (3)^4 + (2)^4 + \Big(\dfrac{10}{8}\Big)^{1} + 1 \\[1em] \Rightarrow 81 + 16 + \dfrac{10}{8} + 1 \\[1em] \Rightarrow 97 + 1.25 + 1 \\[1em] \Rightarrow 99.25

Hence, (27)43+(32)0.8+(0.8)−1+(0.8)0=99.25(27)^{\dfrac{4}{3}} + (32)^{0.8} + (0.8)^{-1} + (0.8)^0 = 99.25.

Question 6(ii)

Evaluate the following :

[(27)−39−3]13\Big[\dfrac{(27)^{-3}}{9^{-3}}\Big]^{\dfrac{1}{3}}

Answer

Given,

[(27)−39−3]13\Big[\dfrac{(27)^{-3}}{9^{-3}}\Big]^{\dfrac{1}{3}}

Simplifying the expression :

⇒[(27)−39−3]13⇒[(279)−3]13⇒(279)−3×13⇒(279)−1⇒927⇒13.\Rightarrow \Big[\dfrac{(27)^{-3}}{9^{-3}}\Big]^{\dfrac{1}{3}} \\[1em] \Rightarrow \Big[\Big(\dfrac{27}{9}\Big)^{-3}\Big]^{\dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{27}{9}\Big)^{-3 \times \dfrac{1}{3}} \\[1em] \Rightarrow \Big(\dfrac{27}{9}\Big)^{-1} \\[1em] \Rightarrow \dfrac{9}{27} \\[1em] \Rightarrow \dfrac {1}{3}.

Hence, [(27)−39−3]13=13\Big[\dfrac{(27)^{-3}}{9^{-3}}\Big]^{\dfrac{1}{3}} = \dfrac{1}{3}.

Question 7(i)

Evaluate the following :

(32−5)13(32+5)13(\sqrt{32}-\sqrt{5})^{\dfrac{1}{3}} (\sqrt{32}+\sqrt{5})^{\dfrac{1}{3}}

Answer

Given,

(32−5)13(32+5)13(\sqrt{32}-\sqrt{5})^{\dfrac{1}{3}} (\sqrt{32}+\sqrt{5})^{\dfrac{1}{3}}

Simplifying the expression :

⇒[(32−5)(32+5)]13⇒[(32)2−(5)2]13⇒[32−5]13⇒[27]13⇒(33)13⇒3.\Rightarrow [(\sqrt{32}-\sqrt{5})(\sqrt{32}+\sqrt{5})]^{\dfrac{1}{3}} \\[1em] \Rightarrow [(\sqrt{32})^2-(\sqrt{5})^2]^{\dfrac{1}{3}} \\[1em] \Rightarrow [32 - 5]^{\dfrac{1}{3}} \\[1em] \Rightarrow [27]^{\dfrac{1}{3}} \\[1em] \Rightarrow (3^3)^{\dfrac{1}{3}} \\[1em] \Rightarrow 3.

Hence, (32−5)13(32+5)13=3(\sqrt{32}-\sqrt{5})^{\dfrac{1}{3}} \\ (\sqrt{32}+\sqrt{5})^{\dfrac{1}{3}} = 3.

Question 7(ii)

Evaluate the following :

(9)52−3×(4)0−(181)−12(9)^{\dfrac{5}{2}} - 3 \times (4)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}

Answer

Given,

(9)52−3×(4)0−(181)−12(9)^{\dfrac{5}{2}} - 3 \times (4)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}

Simplifying the expression :

⇒(9)52−3(4)0−(181)−12⇒[(3)2]52−3(1)−(81)12⇒35−3−(92)12⇒(3)5−3−9⇒243−3−9⇒231.\Rightarrow (9)^{\dfrac{5}{2}} - 3(4)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} \\[1em] \Rightarrow [(3)^2]^{\dfrac{5}{2}} - 3(1) - (81)^\dfrac{1}{2} \\[1em] \Rightarrow 3^5 - 3 - (9^2)^{\dfrac{1}{2}} \\[1em] \Rightarrow (3)^5 - 3 - 9 \\[1em] \Rightarrow 243 - 3 - 9 \\[1em] \Rightarrow 231.

Hence, (9)52−3.(4)0−(181)−12=231(9)^{\dfrac{5}{2}} - 3.(4)^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} = 231.

Question 8

Simplify :

3n×9n+13n−1×9n−1\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}}

Answer

Given,

3n×9n+13n−1×9n−1\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}}

Simplifying the expression :

⇒3n×9n+13n−1×9n−1⇒3n×(32)n+13n−1×(32)n−1⇒3n×(3)2n+23n−1×(3)2n−2⇒3n+2n+23n−1+2n−2⇒33n+233n−3⇒3(3n+2)−(3n−3)⇒3(3n+2−3n+3)⇒35.\Rightarrow \dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} \\[1em] \Rightarrow \dfrac{3^n \times (3^2)^{n + 1}}{3^{n - 1} \times (3^2)^{n - 1}} \\[1em] \Rightarrow \dfrac{3^n \times (3)^{2n + 2}}{3^{n - 1} \times (3)^{2n - 2}} \\[1em] \Rightarrow \dfrac{3^{n + 2n + 2}}{3^{n - 1 + 2n - 2}} \\[1em] \Rightarrow \dfrac{3^{3n + 2}}{3^{3n - 3}} \\[1em] \Rightarrow 3^{(3n + 2) - (3n - 3)} \\[1em] \Rightarrow 3^{(3n + 2 - 3n + 3)} \\[1em] \Rightarrow 3^5.

Hence, 3n×9n+13n−1×9n−1=35\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 9^{n - 1}} = 3^5.

Question 9

Simplify :

(27)2n3×(8)−n6(18)−n2\dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}}

Answer

Given,

(27)2n3×(8)−n6(18)−n2\dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}}

Simplifying the expression :

⇒[(3)3]2n3×[(2)3]−n6[(3)2×2]−n2⇒[(3)2n]×(2)−n2[(3)2]−n2×(2)−n2⇒(3)2n3−n⇒(3)2n−(−n)⇒(3)2n+n⇒33n.\Rightarrow \dfrac{[(3)^3]^{\dfrac{2n}{3}} \times [(2)^3]^{-\dfrac{n}{6}}}{[(3)^2 \times 2]^{-\dfrac{n}{2}}} \\[1em] \Rightarrow \dfrac{[(3)^{2n}] \times (2)^{-\dfrac{n}{2}}}{[(3)^2]^{-\dfrac{n}{2}} \times (2)^{-\dfrac{n}{2}}} \\[1em] \Rightarrow \dfrac{(3)^{2n}}{3^{-n}} \\[1em] \Rightarrow (3)^{2n - (-n)} \\[1em] \Rightarrow (3)^{2n + n} \\[1em] \Rightarrow 3^{3n}.

Hence, (27)2n3×(8)−n6(18)−n2=33n\dfrac{(27)^{\dfrac{2n}{3}} \times (8)^{-\dfrac{n}{6}}}{(18)^{-\dfrac{n}{2}}} = 3^{3n}.

Question 10

Simplify :

52(n+6)×(25)−7+2n(125)2n\dfrac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}}

Answer

Given,

52(n+6)×(25)−7+2n(125)2n\dfrac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}}

Simplifying the expression :

⇒52(n+6)×(25)−7+2n(125)2n⇒52n+12×[(5)2]−7+2n[(5)3]2n⇒52n+12×52(−7+2n)(5)6n⇒52n+12×5−14+4n(5)6n⇒52n+12+(−14+4n)(5)6n⇒56n−2(5)6n⇒56n−2−6n⇒5−2⇒(15)2⇒125.\Rightarrow \dfrac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}} \\[1em] \Rightarrow \dfrac{5^{2n + 12} \times [(5)^2]^{-7 + 2n}}{[(5)^3]^{2n}} \\[1em] \Rightarrow \dfrac{5^{2n + 12} \times 5^{2(-7 + 2n)}}{(5)^{6n}} \\[1em] \Rightarrow \dfrac{5^{2n + 12} \times 5^{-14 + 4n}}{(5)^{6n}} \\[1em] \Rightarrow \dfrac{5^{2n + 12 + (-14 + 4n)}}{(5)^{6n}} \\[1em] \Rightarrow \dfrac{5^{6n - 2}}{(5)^{6n}} \\[1em] \Rightarrow 5^{6n - 2 - 6n} \\[1em] \Rightarrow 5^{-2} \\[1em] \Rightarrow \Big(\dfrac{1}{5}\Big)^{2} \\[1em] \Rightarrow \dfrac{1}{25}.

Hence, 52(n+6)×(25)−7+2n(125)2n=125\dfrac{5^{2(n + 6)} \times (25)^{-7 + 2n}}{(125)^{2n}} = \dfrac{1}{25}.

Question 11

Simplify :

5n+3−16×5n+112×5n−2×5n+1\dfrac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}}

Answer

Given,

5n+3−16×5n+112×5n−2×5n+1\dfrac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}}

Simplifying the expression :

⇒5n+3−16×5n+112×5n−2×5n+1⇒5n+2+1−16×5n+112×5n−2×5n+1⇒5n+1×52−16×5n+112×5n−2×5n×51⇒5n+1(52−16)5n(12−2×5)⇒5n+1−n(25−16)(12−10)⇒5×9(2)⇒452=22.5\Rightarrow \dfrac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} \\[1em] \Rightarrow \dfrac{5^{n + 2 + 1} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} \\[1em] \Rightarrow \dfrac{5^{n + 1} \times 5^{2} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n} \times 5 ^{1}} \\[1em] \Rightarrow \dfrac{5^{n + 1}(5^2 - 16)}{5^n (12 - 2 \times 5)} \\[1em] \Rightarrow \dfrac{5^{n + 1 - n}(25 - 16)}{(12 - 10)} \\[1em] \Rightarrow \dfrac{5 \times 9}{(2)} \\[1em] \Rightarrow \dfrac{45}{2} = 22.5

Hence, 5n+3−16×5n+112×5n−2×5n+1=22.5\dfrac{5^{n + 3} - 16 \times 5^{n + 1}}{12 \times 5^n - 2 \times 5^{n + 1}} = 22.5.

Question 12

Simplify :

3×(27)n+1+9×3(3n−1)8×33n−5×(27)n\dfrac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n}

Answer

Given,

3×(27)n+1+9×3(3n−1)8×33n−5×(27)n\dfrac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n}

Simplifying the expression :

⇒3×(27)n+1+9×3(3n−1)8×33n−5×(27)n⇒3×[(3)3]n+1+32×3(3n−1)8×33n−5×[(3)3]n⇒3×33n+3+32×33n−18×3n−5×3n⇒(3)3n+3+1+3(3n−1+2)8×33n−5×(3)3n⇒(3)3n+1+3+3(3n+1)33n(8−5)⇒(3)3n+1×33+3(3n+1)33n.31⇒33n+1.(33+1)33n+1⇒(3)3n+1.(27+1)33n+1⇒28.\Rightarrow \dfrac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n} \\[1em] \Rightarrow \dfrac{3 \times [(3)^3]^{n + 1} + 3^2 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times [(3)^3]^n} \\[1em] \Rightarrow \dfrac{3 \times 3^{3n + 3} + 3^2 \times 3^{3n - 1}}{8 \times 3^n - 5 \times 3^n} \\[1em] \Rightarrow \dfrac{(3)^{3n + 3 + 1} + 3^{(3n - 1 + 2)}}{8 \times 3^{3n} - 5 \times (3)^{3n}} \\[1em] \Rightarrow \dfrac{(3)^{3n + 1 + 3} + 3^{(3n + 1)}}{3^{3n}(8- 5)} \\[1em] \Rightarrow \dfrac{(3)^{3n + 1} \times 3^3 + 3^{(3n + 1)}}{3^{3n}.3^1} \\[1em] \Rightarrow \dfrac{3^{3n + 1}.(3^3 + 1)}{3^{3n + 1}} \\[1em] \Rightarrow \dfrac{(3)^{3n + 1}.(27 + 1)}{3^{3n + 1}} \\[1em] \Rightarrow 28.

Hence, 3×(27)n+1+9×3(3n−1)8×33n−5×(27)n=28\dfrac{3 \times (27)^{n + 1} + 9 \times 3^{(3n - 1)}}{8 \times 3^{3n} - 5 \times (27)^n} = 28.

Question 13

Simplify :

5×(25)n+1−25×52n5×5(2n+3)−(25)n+1\dfrac{5 \times (25)^{n + 1} - 25 \times 5^{2n}}{5 \times 5^{(2n + 3)} - (25)^{n + 1}}

Answer

Given,

5×(25)n+1−25×52n5×5(2n+3)−(25)n+1\dfrac{5 \times (25)^{n + 1} - 25 \times 5^{2n}}{5 \times 5^{(2n + 3)} - (25)^{n + 1}}

Simplifying the expression :

⇒5×(25)n+1−52×52n51×52n+3−(25)n+1⇒5×[(5)2]n+1−52n+252n+3+1−[(5)2]n+1⇒51×(5)2n+2−52n+252n+4−(5)2n+2⇒52n+2+1−52n+252n+4−52n+2⇒52n+3−52n+252n+4−52n+2⇒52n+2(5−1)52n+2(52−1)⇒4(25−1)⇒424⇒16.\Rightarrow \dfrac{5 \times (25)^{n + 1} - 5^2 \times 5^{2n}}{5^1 \times 5^{2n + 3} - (25)^{n + 1}} \\[1em] \Rightarrow \dfrac{5 \times [(5)^2]^{n + 1} - 5^{2n + 2}}{5^{2n + 3 + 1} - [(5)^2]^{n + 1}} \\[1em] \Rightarrow \dfrac{5^1 \times (5)^{2n + 2} - 5^{2n + 2}}{5^{2n + 4} - (5)^{2n + 2}} \\[1em] \Rightarrow \dfrac{5^{2n + 2 + 1} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] \Rightarrow \dfrac{5^{2n + 3} - 5^{2n + 2}}{5^{2n + 4} - 5^{2n + 2}} \\[1em] \Rightarrow \dfrac{5^{2n + 2}(5 - 1)}{5^{2n + 2}(5^2 - 1)} \\[1em] \Rightarrow \dfrac{4}{(25 - 1)} \\[1em] \Rightarrow \dfrac{4}{24} \\[1em] \Rightarrow \dfrac{1}{6}.

Hence, 5×(25)n+1−25×52n5×52n+3−(25)n+1=16\dfrac{5 \times (25)^{n + 1} - 25 \times 5^{2n}}{5 \times 5^{2n + 3} - (25)^{n + 1}} = \dfrac{1}{6}.

Question 14

Simplify :

7(2n+3)−(49)n+2[(343)n+1]23\dfrac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\dfrac{2}{3}}}

Answer

Given,

7(2n+3)−(49)n+2[(343)n+1]23\dfrac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\dfrac{2}{3}}}

Simplifying the expression :

⇒7(2n+3)−(49)n+2[(343)n+1]23⇒7(2n+3)−[(7)2]n+2[[(7)3]n+1]23⇒7(2n+3)−(7)2n+473×(n+1)×23⇒7(2n+3)−72n+3+17(n+1)×2⇒7(2n+3)(1−7)(7)2n+2⇒7(2n+3)−(2n+2)×−6⇒72n+3−2n−2×−6⇒71×−6⇒−42.\Rightarrow \dfrac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{(2n + 3)} - [(7)^2]^{n + 2}}{\Big[[(7)^3]^{n + 1}\Big]^{\dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{(2n + 3)} - (7)^{2n + 4}}{7^{3 \times (n + 1) \times \dfrac{2}{3}}} \\[1em] \Rightarrow \dfrac{7^{(2n + 3)} - 7^{2n + 3 + 1}}{7^{{(n + 1)} \times 2}} \\[1em] \Rightarrow \dfrac{7^{(2n + 3)}(1 - 7)}{(7)^{2n + 2}} \\[1em] \Rightarrow 7^{(2n + 3) - (2n + 2)} \times -6 \\[1em] \Rightarrow 7^{2n + 3 - 2n - 2} \times -6 \\[1em] \Rightarrow 7^1 \times -6 \\[1em] \Rightarrow -42.

Hence, 7(2n+3)−(49)n+2[(343)n+1]23=−42\dfrac{7^{(2n + 3)} - (49)^{n + 2}}{\Big[(343)^{n + 1}\Big]^{\dfrac{2}{3}}} = - 42.

Question 15

Simplify :

(x13−x−13)(x23+1+x−23)\Big(x^{\dfrac{1}{3}} - x^{\dfrac{-1}{3}}\Big)(x^{\dfrac{2}{3}} + 1 + x^{\dfrac{-2}{3}}\Big)

Answer

Given,

(x13−x−13)(x23+1+x−23)\Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)(x^{\dfrac{2}{3}} + 1 + x^{-\dfrac{2}{3}}\Big)

Simplifying the expression:

⇒(x13−x−13)(x23+1+x−23)⇒(x13−x−13)[(x13)2+x13⋅x−13+(x−13)2]\Rightarrow \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)(x^{\dfrac{2}{3}} + 1 + x^{-\dfrac{2}{3}}\Big) \\[1em] \Rightarrow \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big) \Big[(x^{\dfrac{1}{3}})^2 + x^{\dfrac{1}{3}} ⋅ x^{-\dfrac{1}{3}} + (x^{-\dfrac{1}{3}})^2\Big]

Simplifying the expression using the identity,

(a − b)(a2 + ab + b2) = a3 − b3

⇒([x13]3−[x−13]3)⇒(x13×3−x−13×3)⇒(x−x−1)⇒x−1x\Rightarrow \Big([x^{\dfrac{1}{3}}]^3 - [x^{-\dfrac{1}{3}}]^3\Big) \\[1em] \Rightarrow \Big(x^{\dfrac{1}{3} \times 3} - x^{-\dfrac{1}{3} \times 3}\Big) \\[1em] \Rightarrow (x - x^{-1}) \\[1em] \Rightarrow x - \dfrac{1}{x}

Hence, (x1/3−x−1/3)(x2/3+1+x−2/3)=x−1x\Big(x^{1/3} - x^{-1/3}\Big)(x^{2/3} + 1 + x^{-2/3}\Big) = x - \dfrac{1}{x}.

Question 16(i)

Simplify :

a7 × a4 × a-6 × a0

Answer

Given,

a7 × a4 × a-6 × a0

Simplifying the expression :

⇒ a7 + 4 + (-6) × 1

⇒ a5 × 1

⇒ a5.

Hence, a7 × a4 × a-6 × a0 = a5.

Question 16(ii)

Simplify :

a43÷a−23a^{\dfrac{4}{3}} ÷ a^{\dfrac{-2}{3}}

Answer

Given,

a43÷a−23a^{\dfrac{4}{3}} ÷ a^{\dfrac{-2}{3}}

Simplifying the expression :

⇒a43÷a−23⇒a43a−23⇒a43−(−23)⇒a4−(−2)3⇒a4+23⇒a63⇒a2.\Rightarrow a^{\dfrac{4}{3}} ÷ a^{\dfrac{-2}{3}} \\[1em] \Rightarrow \dfrac{a^{\dfrac{4}{3}}}{a^{\dfrac{-2}{3}}} \\[1em] \Rightarrow a^{\dfrac{4}{3} - \Big(-\dfrac{2}{3}\Big)} \\[1em] \Rightarrow a^{\dfrac{4 - (-2)}{3}} \\[1em] \Rightarrow a^{\dfrac{4 + 2}{3}} \\[1em] \Rightarrow a^{\dfrac{6}{3}} \\[1em] \Rightarrow a^2.

Hence, a43÷a−23=a2a^{\dfrac{4}{3}} ÷ a^{\dfrac{-2}{3}}= a^2.

Question 16(iii)

Simplify :

(a-1 + b-1) ÷ (a-2 - b-2)

Answer

Given,

(a-1 + b-1) ÷ (a-2 - b-2)

Simplifying the expression :

⇒(a−1+b−1)÷(a−2−b−2)⇒(1a+1b)÷(1a2−1b2)⇒(b+aab)÷(b2−a2a2b2)⇒(b+aab)×a2b2b2−a2⇒ab(b+a)(b+a)(b−a)⇒ab(b−a).\Rightarrow (a^{-1} + b^{-1}) ÷ (a^{-2} - b^{-2}) \\[1em] \Rightarrow \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) ÷ \Big(\dfrac{1}{a^2} - \dfrac{1}{b^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{b + a}{ab}\Big) ÷ \Big(\dfrac{b^2 - a^2}{a^2b^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{b + a}{ab}\Big) \times \dfrac{a^2b^2}{b^2 - a^2}\\[1em] \Rightarrow \dfrac{ab(b + a)}{(b + a)(b - a)} \\[1em] \Rightarrow \dfrac{ab}{(b - a)}.

Hence, (a−1+b−1)÷(a−2−b−2)=ab(b−a)(a^{-1} + b^{-1}) ÷ (a^{-2} - b^{-2}) = \dfrac{ab}{(b - a)}.

Question 16(iv)

Simplify :

(a-1 + b-1) ÷ (ab)-1

Answer

Given,

(a-1 + b-1) ÷ (ab)-1

Simplifying the expression :

⇒(1a+1b)÷(1ab)⇒(b+aab)÷(1ab)⇒(a+bab)×ab⇒a+b.\Rightarrow \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) ÷ \Big(\dfrac{1}{ab}\Big) \\[1em] \Rightarrow \Big(\dfrac{b + a}{ab}\Big) ÷ \Big(\dfrac{1}{ab}\Big) \\[1em] \Rightarrow \Big(\dfrac{a + b}{ab}\Big) \times ab \\[1em] \Rightarrow a + b.

Hence, (a−1+b−1)÷(ab)−1=(a+b)(a^{-1} + b^{-1}) ÷ (ab)^{-1} = (a + b).

Question 16(v)

Simplify :

(a-1 × b-1) ÷ (a-1 + b-1)

Answer

Given,

(a-1 × b-1) ÷ (a-1 + b-1)

Simplifying the expression :

⇒(a−1×b−1)÷(a−1+b−1)⇒(1a×1b)÷(1a+1b)⇒1ab÷b+aab⇒1ab×aba+b⇒1a+b.\Rightarrow (a^{-1} \times b^{-1}) ÷ (a^{-1} + b^{-1}) \\[1em] \Rightarrow \Big(\dfrac{1}{a} \times \dfrac{1}{b}\Big) ÷ \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \\[1em] \Rightarrow \dfrac{1}{ab} ÷ \dfrac{b + a}{ab} \\[1em] \Rightarrow \dfrac{1}{ab} \times \dfrac{ab}{a + b} \\[1em] \Rightarrow \dfrac{1}{a + b}.

Hence, a−1×b−1÷a−1+b−1=1a+ba^{-1} \times b^{-1} ÷ a^{-1} + b^{-1} = \dfrac{1}{a + b}.

Question 16(vi)

Simplify :

(a + b)-1 × (a-1 + b-1)

Answer

Given,

(a + b)-1 × (a-1 + b-1)

Simplifying the expression :

⇒(a+b)−1×(a−1+b−1)⇒(1a+b)×(1a+1b)⇒(1a+b)×(b+aab)⇒1ab.\Rightarrow (a + b)^{-1} \times (a^{-1} + b^{-1}) \\[1em] \Rightarrow \Big(\dfrac{1}{a + b}\Big) \times \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \\[1em] \Rightarrow \Big(\dfrac{1}{a + b}\Big) \times \Big(\dfrac{b + a}{ab}\Big) \\[1em] \Rightarrow \dfrac{1}{ab}.

Hence, (a+b)−1×(a−1+b−1)=1ab(a + b)^{-1} \times (a{-1} + b{-1}) = \dfrac{1}{ab}.

Question 16(vii)

Simplify :

(a+b+c)(a−1b−1+b−1c−1+c−1a−1)\dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})}

Answer

Given,

(a+b+c)(a−1b−1+b−1c−1+c−1a−1)\dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})}

Simplifying the expression :

⇒(a+b+c)(a−1b−1+b−1c−1+c−1a−1)⇒(a+b+c)(1ab+1bc+1ac)⇒(a+b+c)(c+a+babc)⇒((a+b+c)×abca+b+c)⇒abc.\Rightarrow \dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} \\[1em] \Rightarrow \dfrac{(a + b +c)}{\Big(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}\Big)} \\[1em] \Rightarrow \dfrac{(a + b + c)}{\Big(\dfrac{c + a + b}{abc}\Big)} \\[1em] \Rightarrow {\Big(\dfrac{(a + b + c) \times abc}{a + b + c}\Big)} \\[1em] \Rightarrow abc.

Hence, (a+b+c)(a−1b−1+b−1c−1+c−1a−1)=abc\dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} = abc.

Question 17(i)

Prove that:

(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} = 1

Answer

Given,

(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} = 1

Solving L.H.S :

⇒(xaxb)a+b×(xbxc)b+c×(xcxa)c+a⇒(xa−b)a+b×(xb−c)b+c×(xc−a)c+a⇒(xa2−b2)×(xb2−c2)×(xc2−a2)⇒(xa2−b2+b2−c2+c2−a2)⇒x0⇒1.\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} \\[1em] \Rightarrow (x^{a - b})^{a + b} \times (x^{b - c})^{b + c} \times (x^{c - a})^{c + a} \\[1em] \Rightarrow (x^{a^2 - b^2}) \times (x^{b^2 - c^2}) \times (x^{c^2 - a^2}) \\[1em] \Rightarrow (x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}) \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.

Hence proved, (xaxb)a+b×(xbxc)b+c×(xcxa)c+a=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} = 1.

Question 17(ii)

Prove that:

(xaxb)1ab×(xbxc)1bc×(xcxa)1ac=1\Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} = 1

Answer

Given,

(xaxb)1ab×(xbxc)1bc×(xcxa)1ac=1\Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} = 1

Solving L.H.S :

⇒(xaxb)1ab×(xbxc)1bc×(xcxa)1ac⇒(xa−b)1ab×(xb−c)1bc×(xc−a)1ca⇒(x)a−bab×(x)b−cbc×(x)c−aac⇒(x)aab−bab×(x)bbc−cbc×(x)cac−aac⇒[(x)1b−1a]×[(x)1c−1b]×[(x)1a−1c]⇒[(x)1b−1a+1c−1b+1a−1c]⇒x0⇒1.\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} \\[1em] \Rightarrow (x^{a - b})^{\dfrac{1}{ab}} \times (x^{b - c})^{\dfrac{1}{bc}} \times (x^{c - a})^{\dfrac{1}{ca}} \\[1em] \Rightarrow (x)^{\dfrac{a - b}{ab}} \times (x)^{\dfrac{b - c}{bc}} \times (x)^{\dfrac{c - a}{ac}} \\[1em] \Rightarrow (x)^{\dfrac{a}{ab}- \dfrac{b}{ab}} \times (x)^{\dfrac{b}{bc} -\dfrac{c}{bc}} \times (x)^{\dfrac{c}{ac}- \dfrac{a}{ac}} \\[1em] \Rightarrow \Big[(x)^{\dfrac{1}{b} - \dfrac{1}{a}}\Big] \times \Big[(x)^{\dfrac{1}{c} - \dfrac{1}{b}}\Big] \times \Big[(x)^{\dfrac{1}{a} - \dfrac{1}{c}}\Big] \\[1em] \Rightarrow \Big[(x)^{\dfrac{1}{b} - \dfrac{1}{a} + \dfrac{1}{c} - \dfrac{1}{b} + \dfrac{1}{a} - \dfrac{1}{c}}\Big] \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.

Hence proved, (xaxb)ab×(xbxc)bc×(xcxa)ca=1\Big(\dfrac{x^a}{x^b}\Big)^{ab} \times \Big(\dfrac{x^b}{x^c}\Big)^{bc} \times \Big(\dfrac{x^c}{x^a}\Big)^{ca} = 1.

Question 17(iii)

Prove that:

(xaxb)a+b−c×(xbxc)b+c−a×(xcxa)c+a−b=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} = 1

Answer

Given,

(xaxb)a+b−c×(xbxc)b+c−a×(xcxa)c+a−b=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} = 1

Solving L.H.S :

⇒(xaxb)a+b−c×(xbxc)b+c−a×(xcxa)c+a−b⇒x(a−b)(a+b−c)×x(b−c)(b+c−a)×x(c−a)(c+a−b)⇒x(a2+ab−ac−ab−b2+bc)×x(b2+bc−ab−bc−c2+ac)×x(c2+ac−bc−ac−a2+ab)⇒x(a2−b2+bc−ac)×x(b2−c2+ac−ab)×x(c2−a2+ab−bc)⇒x(a2−b2+bc−ac)+(b2−c2+ac−ab)+(c2−a2+ab−bc)⇒x(a2−a2−b2+b2−c2+c2+bc−bc−ac+ac−ab+ab)⇒x0⇒1.\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} \\[1em] \Rightarrow x^{(a - b)(a + b - c)} \times x^{(b - c)(b + c - a)} \times x^{(c - a)(c + a - b)}\\[1em] \Rightarrow x^{(a^2 + ab − ac − ab − b^2 + bc)} \times x^{(b^2 + bc − ab − bc − c^2 + ac)} \times x^{(c^2 + ac − bc − ac − a^2 + ab)} \\[1em] \Rightarrow x^{(a^2 − b^2 + bc − ac)} \times x^{(b^2 − c^2 + ac − ab)} \times x^{(c^2 − a^2 + ab − bc)} \\[1em] \Rightarrow x^{(a^2 − b^2 + bc − ac) + (b^2 − c^2 + ac − ab) + (c^2 − a^2 + ab − bc)} \\[1em] \Rightarrow x^{(a^2 − a^2 − b^2 + b^2 − c^2 + c^2 + bc − bc − ac + ac − ab + ab )} \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.

Hence proved, (xaxb)a+b−c×(xbxc)b+c−a×(xcxa)c+a−b=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} = 1.

Question 18

Prove that:
a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}

Answer

Given,

a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}

Solving L.H.S :

⇒a−1(a−1+b−1)+a−1(a−1−b−1)⇒1a(1a+1b)+1a(1a−1b)⇒1a×1(b+aab)+1a×1(b−aab)⇒1a×abb+a+1a×abb−a⇒bb+a+bb−a⇒b(b−a)+b(b+a)(b)2−(a)2⇒b2−ba+b2+bab2−a2⇒2b2b2−a2.\Rightarrow \dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} \\[1em] \Rightarrow \dfrac{\dfrac{1}{a}}{\Big(\dfrac{1}{a} + \dfrac{1}{b}\Big)} + \dfrac{\dfrac{1}{a}}{\Big(\dfrac{1}{a} - \dfrac{1}{b}\Big)} \\[1em] \Rightarrow \dfrac{1}{a} \times \dfrac{1}{\Big(\dfrac{b + a}{ab}\Big)} + \dfrac{1}{a} \times \dfrac{1}{\Big(\dfrac{b - a}{ab}\Big)} \\[1em] \Rightarrow \dfrac{1}{a} \times \dfrac{ab}{b + a} + \dfrac{1}{a} \times \dfrac{ab}{b - a} \\[1em] \Rightarrow \dfrac{b}{b + a} + \dfrac{b}{b - a} \\[1em] \Rightarrow \dfrac{b(b - a) + b(b + a)}{(b)^2 - (a)^2} \\[1em] \Rightarrow \dfrac{b^2 - ba + b^2 + ba}{b^2 - a^2} \\[1em] \Rightarrow \dfrac{2b^2}{b^2 - a^2}.

Hence proved, a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}.

Question 19

Prove that:
11+xb−a+xc−a+11+xa−b+xc−b+11+xb−c+xa−c=1\dfrac{1}{1 + x^{b - a} + x^{c - a}} + \dfrac{1}{1 + x^{a - b} + x^{c - b}} + \dfrac{1}{1 + x^{b - c} + x^{a - c}} = 1

Answer

Given,

11+xb−a+xc−a+11+xa−b+xc−b+11+xb−c+xa−c=1\dfrac{1}{1 + x^{b - a} + x^{c - a}} + \dfrac{1}{1 + x^{a - b} + x^{c - b}} + \dfrac{1}{1 + x^{b - c} + x^{a - c}} = 1

Solving L.H.S :

Multiplying numerator and denominator of first term by xa, second term by xb, third term by xc we get,

⇒1×xa(1+xb−a+xc−a)×xa+1×xb(1+xa−b+xc−b)×xb+1×xc(1+xb−c+xa−c)×xc⇒1×xa(1×xa+xb−a×xa+xc−a×xa)+1×xb(1×xb+xa−b×xb+xc−b×xb)+1×xc(1×xc+xb−c×xc+xa−c×xc)⇒xa(xa+xb−a+a+xc−a+a)+xb(xb+xa−b+b+xc−b+b)+xc(xc+xb−c+c+xa−c+c)⇒xa(xa+xb+xc)+xb(xb+xa+xc)+xc(xc+xb+xa)⇒xa+xb+xc(xa+xb+xc)⇒1.\Rightarrow \dfrac{1 \times x^a}{(1 + x^{b - a} + x^{c - a})\times x^a} + \dfrac{1 \times x^b}{(1 + x^{a - b} + x^{c - b})\times x^b} + \dfrac{1 \times x^c}{(1 + x^{b - c} + x^{a - c})\times x^c} \\[1em] \Rightarrow \dfrac{1 \times x^a}{(1 \times x^a + x^{b - a} \times x^a + x^{c - a} \times x^a)} + \dfrac{1 \times x^b}{(1 \times x^b + x^{a - b} \times x^b + x^{c - b} \times x^b)} + \dfrac{1 \times x^c}{(1 \times x^c + x^{b - c} \times x^c + x^{a - c} \times x^c)} \\[1em] \Rightarrow \dfrac{x^a}{(x^a + x^{b - a + a} + x^{c - a + a})} + \dfrac{x^b}{(x^b + x^{a - b + b} + x^{c - b + b})} + \dfrac{x^c}{(x^c + x^{b - c + c} + x^{a - c + c})} \\[1em] \Rightarrow \dfrac{x^a}{(x^a + x^b + x^c)} + \dfrac{x^b}{(x^b + x^a + x^c)} + \dfrac{x^c}{(x^c + x^b + x^a)} \\[1em] \Rightarrow \dfrac{x^a + x^b + x^c}{(x^a + x^b + x^c)} \\[1em] \Rightarrow 1.

Hence proved, 11+xb−a+xc−a+11+xa−b+xc−b+11+xb−c+xa−c=1\dfrac{1}{1 + x^{b - a} + x^{c - a}} + \dfrac{1}{1 + x^{a - b} + x^{c - b}} + \dfrac{1}{1 + x^{b - c} + x^{a - c}} = 1.

Question 20

If abc = 1, prove that: 11+a+b−1+11+b+c−1+11+c+a−1=1\dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} = 1

Answer

Given,

11+a+b−1+11+b+c−1+11+c+a−1=1\dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} = 1

Solving L.H.S :

⇒11+a+b−1+11+b+c−1+11+c+a−1⇒11+a+1b+11+b+1c+11+c+1a\Rightarrow \dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} \\[1em] \Rightarrow \dfrac{1}{1 + a + \dfrac{1}{b}} + \dfrac{1}{1 + b + \dfrac{1}{c}} + \dfrac{1}{1 + c + \dfrac{1}{a}}

Multiplying numerator and denominator of first term by b, second term by c, and third term by a we get,

⇒1×b(1+a+1b)×b+1×c(1+b+1c)×c+1×a(1+c+1a)×a⇒b(b+ab+1)+c(c+bc+1)+a(a+ac+1)\Rightarrow \dfrac{1 \times b}{\Big(1 + a + \dfrac{1}{b}\Big) \times b} + \dfrac{1 \times c}{\Big(1 + b + \dfrac{1}{c}\Big) \times c} + \dfrac{1 \times a}{\Big(1 + c + \dfrac{1}{a}\Big) \times a} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{c}{(c + bc + 1)} + \dfrac{a}{(a + ac + 1)}

Since abc = 1, c = 1ab\dfrac{1}{ab}.

Substituting, c = 1ab\dfrac{1}{ab}, we get :

⇒b(b+ab+1)+1ab(1ab+b(1ab)+1)+a(a+a(1ab)+1)⇒b(b+ab+1)+1ab(1ab+1a+1)+a(a+1b+1)⇒b(b+ab+1)+1ab×(1+b+abab)+a(ab+1+bb)⇒b(b+ab+1)+1(1+b+ab)+ab(ab+1+b)⇒b+1+ab(b+ab+1)⇒1+b+ab(1+b+ab)⇒1.\Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{\dfrac{1}{ab}}{\Big(\dfrac{1}{ab} + b \Big(\dfrac{1}{ab}\Big) + 1\Big)} + \dfrac{a}{(a + a \Big(\dfrac{1}{ab}\Big) + 1)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{\dfrac{1}{ab}}{\Big(\dfrac{1}{ab} + \dfrac{1}{a} + 1\Big)} + \dfrac{a}{\Big(a + \dfrac{1}{b} + 1\Big)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{1}{ab \times \Big(\dfrac{1 + b + ab}{ab}\Big)} + \dfrac{a}{\Big(\dfrac{ab + 1 + b}{b}\Big)} \\[1em] \Rightarrow \dfrac{b}{(b + ab + 1)} + \dfrac{1}{\Big({1 + b + ab}\Big)} + \dfrac{ab}{\Big(ab + 1 + b\Big)} \\[1em] \Rightarrow \dfrac{b + 1 + ab}{(b + ab + 1)} \\[1em] \Rightarrow \dfrac{1 + b + ab}{(1 + b + ab)} \\[1em] \Rightarrow 1.

Hence proved, 11+a+b−1+11+b+c−1+11+c+a−1=1\dfrac{1}{1 + a + b^{-1}} + \dfrac{1}{1 + b + c^{-1}} + \dfrac{1}{1 + c + a^{-1}} = 1.

Question 21

If a, b, c are positive real numbers, show that:
a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1

Answer

Given,

a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1

Solving L.H.S :

⇒a−1b×b−1c×c−1a⇒(1a)b×(1b)c×(1c)a⇒(ba)×(cb)×(ac)⇒(ba)×(cb)×(ac)⇒1⇒1.\Rightarrow \sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{1}{a}\Big)b} \times \sqrt{\Big(\dfrac{1}{b}\Big)c} \times \sqrt{\Big(\dfrac{1}{c}\Big)a} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{b}{a}\Big)} \times \sqrt{\Big(\dfrac{c}{b}\Big)} \times \sqrt{\Big(\dfrac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{b}{a}\Big) \times \Big(\dfrac{c}{b}\Big) \times \Big(\dfrac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{1} \\[1em] \Rightarrow 1.

Hence proved, a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1.

Question 22

If 9n×32×3n−(27)n33m×23=3−3\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3} = 3^{-3}, prove that (m - n) = 1.

Answer

Given,

9n×32×3n−(27)n33m×23=3−3\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3}= 3^{-3}

Solving:

⇒9n×32×3n−(27)n33m×23=3−3⇒[(3)2]n×32×3n−[(3)3]n33m×8=3−3⇒(3)2n×32×3n−(3)3n33m×8=3−3⇒(3)2n+2+n−(3)3n33m×8=3−3⇒(3)3n+2−(3)3n33m×8=3−3⇒33n.32−33n33m×8=3−3⇒33n[(32)−1]33m×8=3−3⇒33n[9−1]33m×8=3−3⇒33n×833m×8=3−3⇒33n−3m=3−3⇒33(n−m)=3−3⇒3−3(m−n)=3−3\Rightarrow \dfrac{9^n \times 3^2 \times 3^n - (27)^n}{3^{3m} \times 2^3} = 3^{-3} \\[1em] \Rightarrow \dfrac{[(3)^2]^n \times 3^2 \times 3^n - [(3)^3]^n}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{2n} \times 3^2 \times 3^n - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{2n + 2 + n} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3)^{3n + 2} - (3)^{3n}}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}.3^2 - 3^{3n}}{3^{3m} \times 8} = 3^{-3}\\[1em] \Rightarrow \dfrac{3^{3n}[(3^2) - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n}[9 - 1]}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow \dfrac{3^{3n} \times 8}{3^{3m} \times 8} = 3^{-3} \\[1em] \Rightarrow 3^{3n - 3m} = 3^{-3} \\[1em] \Rightarrow 3^{3(n - m)} = 3^{-3} \\[1em] \Rightarrow 3^{-3(m - n)} = 3^{-3} \\[1em]

Equating the exponents,

⇒ -3(m - n) = -3

⇒ (m - n) = −3−3\dfrac{-3}{-3} = 1.

Hence proved, m - n = 1.

Question 23

If 21168 = x4 × y3 × z2, find the values of x, y, z.

Answer

Given,

21168 = x4 × y3 × z2

Factorizing 21168, we get :

⇒ 21168 = 24 × 33 × 72

⇒ 24 × 33 × 72 = x4 × y3 × z2

⇒ x = 2, y = 3, z = 7

Hence, x = 2, y = 3, z = 7.

Question 24

If 1960 = 2a × 5b × 7c, find the values of a, b, c. Hence, calculate the value of 2-a × 5-c × 7b

Answer

Given,

1960 = 2a × 5b × 7c

Factorizing 1960, we get :

⇒ 1960 = 23 × 51 × 72

Equating the exponents,

⇒ 23 × 51 × 72 = 2a × 5b × 7c

⇒ a = 3, b = 1, c = 2.

The value of 2-a × 5-c × 7b,

⇒12a×15c×7b⇒123×152×71⇒18×125×71⇒7200.\Rightarrow \dfrac{1}{2^a} \times \dfrac{1}{5^c} \times 7^b \\[1em] \Rightarrow \dfrac{1}{2^3} \times \dfrac{1}{5^2} \times 7^1 \\[1em] \Rightarrow \dfrac{1}{8} \times \dfrac{1}{25} \times 7^1 \\[1em] \Rightarrow \dfrac{7}{200}.

Hence, a = 3, b = 1, c = 2 and 2-a × 5-c × 7b = 7200\dfrac{7}{200}.

Solve for x:

Question 25

Solve for x:

3x=133^x = \dfrac{1}{3}

Answer

Given,

⇒3x=13⇒3x=13⇒3x=3−1\Rightarrow 3^x = \dfrac{1}{3} \\[1em] \Rightarrow 3^x = \dfrac{1}{3} \\[1em] \Rightarrow 3^x = 3^{-1}

Equating the exponents,

⇒ x = -1.

Hence, x = -1.

Question 26

Solve for x:

32x + 1 = 1

Answer

Given,

⇒ 3(2x + 1) = 1

⇒ 3(2x + 1) = 1

⇒ 3(2x + 1) = 30

Equating the exponents,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = −12-\dfrac{1}{2}

Hence, x = −12-\dfrac{1}{2}.

Question 27

Solve for x:

ab=(ba)1−3x\sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1-3x}

Answer

Given,

⇒ab=(ba)1−3x⇒(ab)12=(ba)1−3x⇒(ba)−12=(ba)1−3x\Rightarrow \sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\dfrac{b}{a}\Big)^{-\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x}

Equating the exponents,

⇒−12=1−3x⇒−1=2(1−3x)⇒−1=2−6x⇒6x=2+1⇒6x=3⇒x=36=12.\Rightarrow -\dfrac{1}{2} = 1 - 3x \\[1em] \Rightarrow -1 = 2(1 - 3x) \\[1em] \Rightarrow -1 = 2 - 6x \\[1em] \Rightarrow 6x = 2 + 1 \\[1em] \Rightarrow 6x = 3 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.

Hence, x = 12\dfrac{1}{2}.

Question 28

Solve for x:

(233)x−1=278\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x-1} = \dfrac{27}{8}

Answer

Given,

⇒(233)x−1=278⇒[(23)x−13]=3323⇒(23)x−13=(32)3⇒(23)x−13=(23)−3\Rightarrow \Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x-1} = \dfrac{27}{8} \\[1em] \Rightarrow \Big[\Big({\dfrac{2}{3}}\Big)^\dfrac{x - 1}{3}\Big] = \dfrac{3^3}{2^3} \\[1em] \Rightarrow \Big({\dfrac{2}{3}}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big({\dfrac{2}{3}}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{2}{3}\Big)^{-3}

Equating the exponents,

⇒x−13=−3⇒x−1=−3×3⇒x−1=−9⇒x=−9+1=−8.\Rightarrow \dfrac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.

Hence, x = -8.

Question 29

Solve for x:

(35)x−1=(27125)−1\Big(\sqrt{\dfrac{3}{5}}\Big)^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1}

Answer

Given,

(35)x−1=(27125)−1⇒[(35)12]x−1=(27125)−1⇒(35)x−12=(3353)−1⇒(35)x−12=(35)3×(−1)⇒(35)x−12=(35)−3\Big(\sqrt\dfrac{3}{5}\Big)^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big[\Big(\dfrac{3}{5}\Big)^{\dfrac{1}{2}}\Big]^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3^3}{5^3}\Big)^{-1} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3}{5}\Big)^{3 \times (-1)} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3}

Equating the exponents,

⇒x−12=−3⇒x−1=−3×2⇒x−1=−6⇒x=−6+1=−5.\Rightarrow \dfrac{x - 1}{2} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 2 \\[1em] \Rightarrow x - 1 = -6 \\[1em] \Rightarrow x = -6 + 1 = -5.

Hence, x = -5.

Question 30

Solve for x:

9 × 3x = (27)2x - 5

Answer

Given,

⇒ 9 × 3x = (27)2x - 5

⇒ 32 × 3x = (33)(2x - 5)

⇒ 3(x + 2) = 3(6x - 15)

Equating the exponents,

⇒ x + 2 = 6x - 15

⇒ 6x - x = 15 + 2

⇒ 5x = 17

⇒ x = 175=325\dfrac{17}{5} = 3\dfrac{2}{5}.

Hence, x = 3253\dfrac{2}{5}.

Question 31

Solve for x:

2(5x - 1) = 4 × 2(3x + 1)

Answer

Given,

⇒ 2(5x - 1) = 4 × 2(3x + 1)

⇒ 2(5x - 1) = 4 × 2(3x + 1)

⇒ 2(5x - 1) = 22 × 2(3x + 1)

⇒ 2(5x - 1) = 2(3x + 1 + 2)

⇒ 2(5x - 1) = 2(3x + 3)

Equating the exponents,

⇒ 5x - 1 = 3x + 3

⇒ 5x - 3x = 3 + 1

⇒ 2x = 4

⇒ x = 42=2\dfrac{4}{2} = 2

Hence, x = 2.

Question 32

Solve for x:

5(x - 3) × 3(2x - 8) = 225

Answer

Given,

⇒ 5(x - 3) × 3(2x - 8) = 225

⇒ 5(x - 3) × 3(2x - 8) = 52 × 32

Equating the exponents of 5,

⇒ x - 3 = 2

⇒ x = 2 + 3 = 5.

Equating the exponents of 3,

⇒ 2x - 8 = 2

⇒ 2x = 2 + 8

⇒ x = 102\dfrac{10}{2} = 5.

Hence, x = 5.

Question 33

Solve for n:

3n×9n+13n−1×27n−1=81\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n - 1}} = 81

Answer

Given,

3n×9n+13n−1×27n−1=81\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n - 1}} = 81

Solving for n,

⇒3n×9n+13n−1×27n−1=81⇒3n×[(3)2]n+13n−1×[(3)3]n−1=34⇒3n×32n+23n−1×(3)3n−3=34⇒32n+2+n3n−1+3n−3=34⇒33n+234n−4=34⇒33n+2−(4n−4)=34⇒33n+2−4n+4=34⇒36−n=34\Rightarrow \dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n-1}} = 81 \\[1em] \Rightarrow \dfrac{3^n \times [(3)^2]^{n + 1}}{3^{n - 1} \times [(3)^3]^{n-1}} = 3^4 \\[1em] \Rightarrow \dfrac{3^n \times 3^{2n + 2}}{3^{n - 1} \times (3)^{3n - 3}} = 3^4 \\[1em] \Rightarrow \dfrac{3^{2n + 2 + n}}{3^{n - 1 + 3n - 3}} = 3^4 \\[1em] \Rightarrow \dfrac{3^{3n + 2}}{3^{4n - 4}} = 3^4 \\[1em] \Rightarrow 3^{3n + 2 - (4n - 4)}= 3^4 \\[1em] \Rightarrow 3^{3n + 2 - 4n + 4}= 3^4 \\[1em] \Rightarrow 3^{6 - n}= 3^4

Equating the exponents,

⇒ 6 - n = 4

⇒ n = 6 - 4 = 2.

Hence, n = 2.

Question 34

If 2x = 3y = 12z, show that: 1z=1y+2x\dfrac{1}{z} = \dfrac{1}{y} + \dfrac{2}{x}

Answer

Given,

2x = 3y = 12z

Let 2x = 3y = 12z = k,

First term,

⇒ 2x = k

⇒ x = 1

⇒ 2 = k1xk^{\dfrac{1}{x}}

Second term,

⇒ 3y = k

⇒ y = 1

⇒ 3 = k1yk^{\dfrac{1}{y}}

Third term,

⇒ 12z = k

⇒ z = 1

⇒ 12 = k1zk^{\dfrac{1}{z}}

Factorizing 12, we get :

⇒ 12 = 22 × 3

We have 2=k1x,3=k1y,12=k1z2 = k^{\dfrac{1}{x}}, 3 = k^{\dfrac{1}{y}}, 12 = k^{\dfrac{1}{z}},

⇒k1z=(k1x)2+k1y⇒k1z=(k2x)+k1y Equating the exponents,⇒1z=2x+1y\Rightarrow k^{\dfrac{1}{z}} = \Big(k^{\dfrac{1}{x}}\Big)^2 + k^{\dfrac{1}{y}} \\[1em] \Rightarrow k^{\dfrac{1}{z}} = \Big(k^{\dfrac{2}{x}}\Big) + k^{\dfrac{1}{y}} \\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow \dfrac{1}{z} = \dfrac{2}{x} + \dfrac{1}{y}

Hence proved, 1z=1y+2x\dfrac{1}{z} = \dfrac{1}{y} + \dfrac{2}{x}.

Question 35

If 2x = 3y = 6-z, show that: 1x+1y+1z=0\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0

Answer

Given,

2x = 3y = 6-z

Let 2x = 3y = 6-z = k,

First term,

⇒ 2x = k

⇒ x = 1

⇒ 2 = k1xk^{\dfrac{1}{x}}

Second term,

⇒ 3y = k

⇒ y = 1

⇒ 3 = k1yk^{\dfrac{1}{y}}

Third term,

⇒ 6-z = k

⇒ z = -1

⇒ 6 = k−1zk^{-\dfrac{1}{z}}

Factorizing 6, we get :

⇒ 6 = 2 × 3

We have 2=k1x,3=k1y,6=k−1z2 = k^{\dfrac{1}{x}}, 3 = k^{\dfrac{1}{y}}, 6 = k^{-\dfrac{1}{z}},

⇒k−1z=k1x×k1y⇒k−1z=k1x×k1y Equating the exponents,⇒−1z=1x+1y⇒1x+1y+1z=0.\Rightarrow k^{-\dfrac{1}{z}} = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} \\[1em] \Rightarrow k^{-\dfrac{1}{z}} = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}}\\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow -\dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0.

Hence proved, 1x+1y+1z=0\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0.

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