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Chapter 7

Indices [Exponents]

Class - 9 Concise Mathematics Selina

Exercise 7(A)

Question 1(a)

(−2a2b3)4\Big(-\dfrac{2a^2}{b^3}\Big)^4 is equal to :

  1. −16a8b12-\dfrac{16a^8}{b^{12}}

  2. 16a8b12\dfrac{16a^8}{b^{12}}

  3. 2a8b12\dfrac{2a^8}{b^{12}}

  4. −2a8b12-\dfrac{2a^8}{b^{12}}

Answer

Simplifying the expression :

⇒(−2a2b3)4=(−2×a2b3)4=(−2)4×(a2b3)4=16×a2×4b3×4=16a8b12.\Rightarrow \Big(-\dfrac{2a^2}{b^3}\Big)^4 = \Big(-2 \times \dfrac{a^2}{b^3}\Big)^4 \\[1em] = (-2)^4 \times \Big(\dfrac{a^2}{b^3}\Big)^4 = 16 \times \dfrac{a^{2 \times 4}}{b^{3 \times 4}} \\[1em] = \dfrac{16a^8}{b^{12}}.

Hence, Option 2 is the correct option.

Question 1(b)

284÷74\sqrt[4]{28} ÷ \sqrt[4]{7} is equal to :

  1. 2\sqrt{2}

  2. 4\sqrt{4}

  3. 24\sqrt[4]{2}

  4. 23\sqrt[3]{2}

Answer

Simplifying the expression :

⇒284÷74=28474=(287)14=(4)14=(2)2×14=(2)12=2.\Rightarrow \sqrt[4]{28} ÷ \sqrt[4]{7} = \dfrac{\sqrt[4]{28}}{\sqrt[4]{7}} \\[1em] = \Big(\dfrac{28}{7}\Big)^{\dfrac{1}{4}} = (4)^{\dfrac{1}{4}} \\[1em] = (2)^{2 \times \dfrac{1}{4}} = (2)^{\dfrac{1}{2}} \\[1em] = \sqrt{2}.

Hence, Option 1 is the correct option.

Question 1(c)

(259)−32\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} is equal to :

  1. (53)3\Big(\dfrac{5}{3}\Big)^3

  2. (53)2\Big(\dfrac{5}{3}\Big)^2

  3. 27125\dfrac{27}{125}

  4. (35)−2\Big(\dfrac{3}{5}\Big)^{-2}

Answer

Simplifying the expression :

⇒(259)−32=[(53)2]−32=(53)−3=(35)3=27125.\Rightarrow \Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} = \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-\dfrac{3}{2}} \\[1em] = \Big(\dfrac{5}{3}\Big)^{-3} = \Big(\dfrac{3}{5}\Big)^3 \\[1em] = \dfrac{27}{125}.

Hence, Option 3 is the correct option.

Question 1(d)

2÷(243)−152 ÷ (243)^{-\dfrac{1}{5}} is equal to :

  1. 2 ÷ 3

  2. 16\dfrac{1}{6}

  3. 12

  4. 6

Answer

Simplifying the expression :

⇒2÷(243)−15=2÷(35)−15=2÷3−1=2÷13=2×3=6.\Rightarrow 2 ÷ (243)^{-\dfrac{1}{5}} = 2 ÷ (3^5)^{-\dfrac{1}{5}} \\[1em] = 2 ÷ 3^{-1} = 2 ÷ \dfrac{1}{3} \\[1em] = 2 \times 3 \\[1em] = 6.

Hence, Option 4 is the correct option.

Question 1(e)

(0.01)−12(0.01)^{-\dfrac{1}{2}} :

  1. 10

  2. 0.1

  3. (0.1)12(0.1)^{\frac{1}{2}}

  4. 0.0001

Answer

Simplifying the expression :

⇒(0.01)−12=(1100)−12=(1102)−12=(10−2)−12=101=10.\Rightarrow (0.01)^{-\dfrac{1}{2}} = \Big(\dfrac{1}{100}\Big)^{-\dfrac{1}{2}} \\[1em] = \Big(\dfrac{1}{10^2}\Big)^{-\dfrac{1}{2}} = (10^{-2})^{-\dfrac{1}{2}} \\[1em] = 10^1 = 10.

Hence, Option 1 is the correct option.

Question 1(f)

3×(32)25×703 \times (32)^{\dfrac{2}{5}} \times 7^0 is equal to :

  1. 0

  2. 12

  3. 1

  4. 9

Answer

Simplifying the expression :

⇒3×(32)25×70=3×(25)25×70=3×22×1=3×4=12.\Rightarrow 3 \times (32)^{\dfrac{2}{5}} \times 7^0 = 3 \times (2^5)^{\dfrac{2}{5}} \times 7^0 \\[1em] = 3 \times 2^2 \times 1 \\[1em] = 3 \times 4 \\[1em] = 12.

Hence, Option 2 is the correct option.

Question 2(i)

Evaluate :

33×(243)−23×(9)−133^3 \times (243)^{-\dfrac{2}{3}} \times (9)^{-\dfrac{1}{3}}

Answer

Simplifying the expression :

⇒33×(35)−23×(32)−13=33×(3)−103×(3)−23=(3)3+(−103)+(−23)=(3)9−10−23=(3)−33=3−1=13.\Rightarrow 3^3 \times (3^5)^{-\dfrac{2}{3}} \times (3^2)^{-\dfrac{1}{3}} \\[1em] = 3^3 \times (3)^{-\dfrac{10}{3}} \times (3)^{-\dfrac{2}{3}} \\[1em] = (3)^{3 + \Big(-\dfrac{10}{3}\Big) + \Big(-\dfrac{2}{3}\Big)} \\[1em] = (3)^{\dfrac{9 - 10 - 2}{3}} = (3)^{\dfrac{-3}{3}} \\[1em] = 3^{-1} = \dfrac{1}{3}.

Hence, 33×(243)−23×(9)−13=133^3 \times (243)^{-\dfrac{2}{3}} \times (9)^{-\dfrac{1}{3}} = \dfrac{1}{3}.

Question 2(ii)

Evaluate :

5−4×(125)53÷(25)−125^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}}

Answer

Simplifying the expression :

⇒5−4×(125)53÷(25)−12=5−4×(53)53÷(52)−12=5−4×55÷5−1=5−4×55÷15=5−4×55×51=5−4+5+1=52=25.\Rightarrow 5^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}} = 5^{-4} \times (5^3)^{\dfrac{5}{3}} ÷ (5^2)^{-\dfrac{1}{2}} \\[1em] = 5^{-4} \times 5^5 ÷ 5^{-1} = 5^{-4} \times 5^5 ÷ \dfrac{1}{5} \\[1em] = 5^{-4} \times 5^5 \times 5^1 \\[1em] = 5^{-4 + 5 + 1} = 5^2 \\[1em] = 25.

Hence, 5−4×(125)53÷(25)−12=25.5^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}} = 25.

Question 2(iii)

Evaluate :

(27125)23×(925)−32\Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}}

Answer

Simplifying the expression :

⇒(27125)23×(925)−32=[(35)3]23×[(35)2]−32=(35)2×(35)−3=(35)2+(−3)=(35)−1=53=123.\Rightarrow \Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}} = \Big[\Big(\dfrac{3}{5}\Big)^3\Big]^{\dfrac{2}{3}} \times \Big[\Big(\dfrac{3}{5}\Big)^2\Big]^{-\dfrac{3}{2}} \\[1em] = \Big(\dfrac{3}{5}\Big)^2 \times \Big(\dfrac{3}{5}\Big)^{-3} = \Big(\dfrac{3}{5}\Big)^{2 + (-3)} \\[1em] = \Big(\dfrac{3}{5}\Big)^{-1} = \dfrac{5}{3} = 1\dfrac{2}{3}.

Hence, (27125)23×(925)−32=123\Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}} = 1\dfrac{2}{3}.

Question 2(iv)

Evaluate :

70×(25)−32−5−37^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3}

Answer

Simplifying the expression :

⇒70×(25)−32−5−3=1×(52)−32−5−3=1×5−3−5−3=5−3−5−3=0.\Rightarrow 7^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3} = 1 \times (5^2)^{-\dfrac{3}{2}} - 5^{-3} \\[1em] = 1 \times 5^{-3} - 5^{-3} \\[1em] = 5^{-3} - 5^{-3} \\[1em] = 0.

Hence, 70×(25)−32−5−3=0.7^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3} = 0.

Question 2(v)

Evaluate :

(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

Simplifying the expression :

⇒(1681)−34×(499)32÷(343216)23=[(23)4]−34×[(73)2]32÷[(76)3]23=(23)4×−34×(73)2×32÷(76)3×23=(23)−3×(73)3÷(76)2=(32)3×(73)3×(67)2=33×73×6223×33×72=7×6223=2528=31.5\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}} = \Big[\Big(\dfrac{2}{3}\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] = \Big(\dfrac{2}{3}\Big)^{4 \times -\dfrac{3}{4}} \times \Big(\dfrac{7}{3}\Big)^{2 \times \dfrac{3}{2}} ÷ \Big(\dfrac{7}{6}\Big)^{3 \times \dfrac{2}{3}} \\[1em] = \Big(\dfrac{2}{3}\Big)^{-3} \times \Big(\dfrac{7}{3}\Big)^3 ÷ \Big(\dfrac{7}{6}\Big)^2 \\[1em] = \Big(\dfrac{3}{2}\Big)^3 \times \Big(\dfrac{7}{3}\Big)^3 \times \Big(\dfrac{6}{7}\Big)^2 \\[1em] = \dfrac{3^3 \times 7^3 \times 6^2}{2^3 \times 3^3 \times 7^2} \\[1em] = \dfrac{7 \times 6^2}{2^3} \\[1em] = \dfrac{252}{8} \\[1em] = 31.5

Hence, (1681)−34×(499)32÷(343216)23=31.5\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}} = 31.5

Question 3(i)

Simplify :

(8x3÷125y3)23(8x^3 ÷ 125y^3)^{\dfrac{2}{3}}

Answer

Simplifying the expression :

⇒(8x3125y3)23=[(2x)3(5y)3]23=[(2x5y)3]23=(2x5y)2=4x225y2.\Rightarrow \Big(\dfrac{8x^3}{125y^3}\Big)^{\dfrac{2}{3}} =\Big[\dfrac{(2x)^3}{(5y)^3}\Big]^{\dfrac{2}{3}} \\[1em] = \Big[\Big(\dfrac{2x}{5y}\Big)^3\Big]^{\dfrac{2}{3}} = \Big(\dfrac{2x}{5y}\Big)^2 \\[1em] = \dfrac{4x^2}{25y^2}.

Hence, (8x3÷125y3)23=4x225y2.(8x^3 ÷ 125y^3)^{\dfrac{2}{3}} = \dfrac{4x^2}{25y^2}.

Question 3(ii)

Simplify :

(a + b)-1.(a-1 + b-1)

Answer

Simplifying the expression :

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

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

Question 3(iii)

Simplify :

5n+3−6×5n+19×5n−5n×22\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2}

Answer

Simplifying the expression :

⇒5n+3−6×5n+19×5n−5n×22=5n.53−6×5n×515n(9−22)=5n(53−6×5)5n(9−4)=125−305=955=19.\Rightarrow \dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = \dfrac{5^n.5^3 - 6 \times 5^n \times 5^1}{5^n(9 - 2^2)} \\[1em] = \dfrac{5^n(5^3 - 6 \times 5)}{5^n(9 - 4)} = \dfrac{125 - 30}{5} = \dfrac{95}{5} \\[1em] = 19.

Hence, 5n+3−6×5n+19×5n−5n×22=19.\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = 19.

Question 3(iv)

Simplify :

(3x2)−3×(x9)23(3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}}

Answer

Simplifying the expression :

⇒(3x2)−3×(x9)23=(13x2)3×x9×23=127x6×x6=127.\Rightarrow (3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} = \Big(\dfrac{1}{3x^2}\Big)^3 \times x^{9 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{27x^6} \times x^6 \\[1em] = \dfrac{1}{27}.

Hence, (3x2)−3×(x9)23=127(3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} = \dfrac{1}{27}.

Question 4(i)

Evaluate :

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

Answer

Simplifying the expression :

⇒14+(0.01)−12−(27)23=12+(1100)−12−(33)23=12+(1102)−12−32=12+(10−2)−12−33×23=12+10−2×−12−32=12+10−9=12+1=1+22=32=112.\Rightarrow \sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}} = \dfrac{1}{2} + \Big(\dfrac{1}{100}\Big)^{-\dfrac{1}{2}} - (3^3)^{\dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + \Big(\dfrac{1}{10^2}\Big)^{-\dfrac{1}{2}} - 3^2 = \dfrac{1}{2} + (10^{-2})^{-\dfrac{1}{2}} - 3^{3 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + 10^{-2 \times -\dfrac{1}{2}} - 3^2 \\[1em] = \dfrac{1}{2} + 10 - 9 \\[1em] = \dfrac{1}{2} + 1 \\[1em] = \dfrac{1 + 2}{2} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.

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

Question 4(ii)

Evaluate :

(278)23−(14)−2+50\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0

Answer

Simplifying the expression :

⇒(278)23−(14)−2+50=[(32)3]23−(122)−2+50=(32)3×23−(22)2+1=(32)2−24+1=94−16+1=9−64+44=−514.\Rightarrow \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = \Big[\Big(\dfrac{3}{2}\Big)^3\Big]^{\dfrac{2}{3}} - \Big(\dfrac{1}{2^2}\Big)^{-2} + 5^0\\[1em] = \Big(\dfrac{3}{2}\Big)^{3 \times \dfrac{2}{3}} - (2^2)^2 + 1 \\[1em] = \Big(\dfrac{3}{2}\Big)^2 - 2^4 + 1 \\[1em] = \dfrac{9}{4} - 16 + 1 \\[1em] = \dfrac{9 - 64 + 4}{4} \\[1em] = -\dfrac{51}{4}.

Hence, (278)23−(14)−2+50=−514\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = -\dfrac{51}{4}.

Question 5(i)

Simplify the following and express with positive index :

(3−42−8)14\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}}

Answer

Simplifying the expression :

⇒(3−42−8)14=3−4×142−8×14=3−12−2=13122=223=43.\Rightarrow \Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}} = \dfrac{3^{-4\times \dfrac{1}{4}}}{2^{-8 \times \dfrac{1}{4}}} \\[1em] = \dfrac{3^{-1}}{2^{-2}} = \dfrac{\dfrac{1}{3}}{\dfrac{1}{2^2}} \\[1em] = \dfrac{2^2}{3} = \dfrac{4}{3}.

Hence, (3−42−8)14=43\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}} = \dfrac{4}{3}.

Question 5(ii)

Simplify the following and express with positive index :

(27−39−3)15\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}}

Answer

Simplifying the expression :

⇒(27−39−3)15=[(33)−3(32)−3]15=(33×−332×−3)15=(3−93−6)15=(3−9−(−6))15=(3−9+6)15=(3−3)15=(3)−35=(133)15=1335.\Rightarrow \Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \Big[\dfrac{(3^3)^{-3}}{(3^2)^{-3}}\Big]^{\dfrac{1}{5}}\\[1em] = \Big(\dfrac{3^{3 \times -3}}{3^{2 \times -3}}\Big)^{\dfrac{1}{5}} = \Big(\dfrac{3^{-9}}{3^{-6}}\Big)^{\dfrac{1}{5}} \\[1em] = (3^{-9 - (-6)})^{\dfrac{1}{5}} = (3^{-9 + 6})^{\dfrac{1}{5}} \\[1em] = (3^{-3})^{\dfrac{1}{5}} = (3)^{-\dfrac{3}{5}} \\[1em] = \Big(\dfrac{1}{3^3}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}.

Hence, (27−39−3)15=1335\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}.

Question 5(iii)

Simplify the following and express with positive index :

(32)−25÷(125)−23(32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}}

Answer

Simplifying the expression :

⇒(32)−25÷(125)−23=(25)−25÷(53)−23=(2)5×−25÷(5)3×−23=(2)−2÷5−2=122÷152=122×52=14×25=254=614.\Rightarrow (32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}} = (2^5)^{-\dfrac{2}{5}} ÷ (5^3)^{-\dfrac{2}{3}} \\[1em] = (2)^{5 \times -\dfrac{2}{5}} ÷ (5)^{3 \times -\dfrac{2}{3}} \\[1em] = (2)^{-2} ÷ 5^{-2} \\[1em] = \dfrac{1}{2^2} ÷ \dfrac{1}{5^2} \\[1em] = \dfrac{1}{2^2} \times 5^2 \\[1em] = \dfrac{1}{4} \times 25 \\[1em] = \dfrac{25}{4} = 6\dfrac{1}{4}.

Hence, (32)−25÷(125)−23=614(32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}} = 6\dfrac{1}{4}.

Question 5(iv)

Simplify the following and express with positive index :

[1 - {1 - (1 - n)-1}-1]-1

Answer

Simplifying the expression :

⇒[1−1−(1−n)−1−1]−1=[1−{1−11−n}−1]−1=[1−{1−n−11−n}−1]−1=[1−{−n1−n}−1]−1=[1−{−1−nn}]−1=[1+1−nn]−1=[n+1−nn]−1=[1n]−1=n.\Rightarrow [1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = \Big[1 - \text{\textbraceleft}1 - \dfrac{1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\dfrac{1 - n - 1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\dfrac{-n}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}-\dfrac{1 - n}{n}\text{\textbraceright}\Big]^{-1} \\[1em] = \Big[1 + \dfrac{1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{n + 1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{1}{n}\Big]^{-1} \\[1em] = n.

Hence, [1−1−(1−n)−1−1]−1=n[1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = n.

Question 6

If 2160 = 2a.3b.5c, find a, b and c. Hence, calculate the value of 3a × 2-b × 5-c.

Answer

Factorizing 2160, we get :

⇒ 2160 = 24 × 33 × 51

⇒ 2a.3b.5c = 24 × 33 × 51

⇒ a = 4, b = 3 and c = 1.

Substituting values of a, b and c in 3a × 2-b × 5-c, we get :

⇒34×2−3×5−1=81×123×15=81×18×15=8140=2140.\Rightarrow 3^4 \times 2^{-3} \times 5^{-1} = 81 \times \dfrac{1}{2^3} \times \dfrac{1}{5} \\[1em] = 81 \times \dfrac{1}{8} \times \dfrac{1}{5} \\[1em] = \dfrac{81}{40} = 2\dfrac{1}{40}.

Hence, a = 4, b = 3 and c = 1 and 3a × 2-b × 5-c = 21402\dfrac{1}{40}.

Question 7

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

Answer

Factorizing 1960, we get :

⇒ 1960 = 23.51.72

⇒ 2a.5b.7c = 23.51.72

⇒ a = 3, b = 1 and c = 2.

Substituting values of a, b and c in 2-a.7b.5-c, we get :

⇒2−3×71×5−2=123×7×152=18×7×125=7200.\Rightarrow 2^{-3} \times 7^1 \times 5^{-2} = \dfrac{1}{2^3} \times 7 \times \dfrac{1}{5^2} \\[1em] = \dfrac{1}{8} \times 7 \times \dfrac{1}{25} \\[1em] = \dfrac{7}{200}.

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

Question 8(i)

Simplify :

83a×25×22a4×211a×2−2a\dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}}

Answer

Simplifying the expression :

⇒83a×25×22a4×211a×2−2a=(23)3a×25×22a22×211a×2−2a=29a×25×22a22×211a×2−2a=29a+5+2a22+11a+(−2a)=211a+529a+2=2(11a+5)−(9a+2)=211a−9a+5−2=22a+3.\Rightarrow \dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = \dfrac{(2^3)^{3a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \dfrac{2^{9a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \dfrac{2^{9a + 5 + 2a}}{2^{2 + 11a + (-2a)}} \\[1em] = \dfrac{2^{11a + 5}}{2^{9a + 2}} \\[1em] = 2^{(11a + 5) - (9a + 2)} \\[1em] = 2^{11a - 9a + 5 - 2} \\[1em] = 2^{2a + 3}.

Hence, 83a×25×22a4×211a×2−2a=22a+3\dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = 2^{2a + 3}.

Question 8(ii)

Simplify :

3×27n+1+9×33n−18×33n−5×27n\dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n}

Answer

Simplifying the expression

⇒3×27n+1+9×33n−18×33n−5×27n=3×(33)n+1+(32)×33n−18×33n−5×(33)n=3×33(n+1)+32+3n−18×33n−5×33n=31+3(n+1)+33n+133n(8−5)=31+3n+3+33n+13×33n=33n+1.33+33n+133n+1=33n+1(33+1)33n+1=33+1=27+1=28.\Rightarrow \dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = \dfrac{3 \times (3^3)^{n + 1} + (3^2) \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times (3^3)^n}\\[1em] = \dfrac{3 \times 3^{3(n + 1)} + 3^{2 + 3n - 1}}{8 \times 3^{3n} - 5 \times 3^{3n}} \\[1em] = \dfrac{3^{1 + 3(n + 1)} + 3^{3n + 1}}{3^{3n}(8 - 5)} \\[1em] = \dfrac{3^{1 + 3n + 3} + 3^{3n + 1}}{3 \times 3^{3n}} \\[1em] = \dfrac{3^{3n + 1}.3^3 + 3^{3n + 1}}{3^{3n + 1}} \\[1em] = \dfrac{3^{3n + 1}(3^3 + 1)}{3^{3n + 1}} \\[1em] = 3^3 + 1 \\[1em] = 27 + 1 \\[1em] = 28.

Hence, 3×27n+1+9×33n−18×33n−5×27n=28\dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = 28.

Question 9

Show that :

(ama−n)m−n×(ana−l)n−l×(ala−m)l−m\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} = 1

Answer

Solving L.H.S. of the above equation :

⇒(ama−n)m−n×(ana−l)n−l×(ala−m)l−m⇒(am−(−n))m−n×(an−(−l))n−l×(al−(−m))l−m⇒(a(m+n))m−n×(a(n+l))n−l×(a(l+m))l−m⇒a(m+n)(m−n)×a(n+l)(n−l)×a(l+m)(l−m)⇒am2−n2×an2−l2×al2−m2⇒am2−n2+n2−l2+l2−m2⇒am2−m2−n2+n2−l2+l2⇒a0⇒1.\Rightarrow \Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} \\[1em] \Rightarrow (a^{m - (-n)})^{m - n} \times (a^{n - (-l)})^{n - l} \times (a^{l - (-m)})^{l - m} \\[1em] \Rightarrow (a^{(m + n)})^{m - n} \times (a^{(n + l)})^{n - l} \times (a^{(l + m)})^{l - m} \\[1em] \Rightarrow a^{(m + n)(m - n)} \times a^{(n + l)(n - l)} \times a^{(l + m)(l - m)} \\[1em] \Rightarrow a^{m^2 - n^2} \times a^{n^2 - l^2} \times a^{l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - n^2 + n^2 - l^2 +l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - m^2 - n^2 + n^2 - l^2 + l^2} \\[1em] \Rightarrow a^0 \\[1em] \Rightarrow 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that (ama−n)m−n×(ana−l)n−l×(ala−m)l−m\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} = 1.

Question 10

If a = xm + n.yl; b = xn + l.ym and c = xl + m.yn,

prove that : am - n.bn - l.cl - m = 1

Answer

Substituting values of a, b and c in L.H.S. of equation am - n.bn - l.cl - m = 1, we get :

⇒ am - n.bn - l.cl - m = (xm + n.yl)m - n.(xn + l.ym)n - l.(xl + m.yn)l- m

= (x(m + n)(m - n).yl(m - n)).(x(n + l)(n - l).ym(n - l)).(x(l + m)(l - m).yn(l - m))

= (xm2 - n2).(xn2 - l2).(xl2 - m2).(ylm - ln).(ymn - ml).(ynl - nm)

= xm2 - n2 + n2 - l2 + l2 - m2.ylm - ln + mn - ml + nl - nm

= x0.y0

= 1.1

= 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that am - n.bn - l.cl- m = 1.

Question 11(i)

Simplify :

(xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+ca+a2\Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2}

Answer

Simplifying the expression :

⇒(xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+ca+a2=(xa−b)a2+ab+b2×(xb−c)b2+bc+c2×(xc−a)c2+ca+a2=x(a−b)(a2+ab+b2)×x(b−c)(b2+bc+c2)×x(c−a)(c2+ca+a2)=xa3−b3×xb3−c3×xc3−a3=xa3−b3+b3−c3+c3−a3=x0=1.\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2} \\[1em] = (x^{a - b})^{a^2 + ab + b^2} \times (x^{b - c})^{b^2 + bc + c^2} \times (x^{c - a})^{c^2 + ca + a^2} \\[1em] = x^{(a - b)(a^2 + ab + b^2)} \times x^{(b - c)(b^2 + bc + c^2)} \times x^{(c - a)(c^2 + ca + a^2)} \\[1em] = x^{a^3 - b^3} \times x^{b^3 - c^3} \times x^{c^3 - a^3} \\[1em] = x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \\[1em] = x^0 \\[1em] = 1.

Hence, (xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+ca+a2\Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2} = 1.

Question 11(ii)

Simplify :

(xax−b)a2−ab+b2×(xbx−c)b2−bc+c2×(xcx−a)c2−ca+a2\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2}

Answer

Simplifying the expression :

⇒(xax−b)a2−ab+b2×(xbx−c)b2−bc+c2×(xcx−a)c2−ca+a2=(xa−(−b))a2−ab+b2×(xb−(−c))b2−bc+c2×(xc−(−a))c2−ca+a2=x(a+b)(a2−ab+b2)×x(b+c)(b2−bc+c2)×x(c+a)(c2−ca+a2)=xa3+b3×xb3+c3×xc3+a3=xa3+b3+b3+c3+c3+a3=x2a3+2b3+2c3=x2(a3+b3+c3).\Rightarrow \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} \\[1em] = (x^{a - (-b)})^{a^2 - ab + b^2} \times (x^{b - (-c)})^{b^2 - bc + c^2} \times (x^{c - (-a)})^{c^2 - ca + a^2} \\[1em] = x^{(a + b)(a^2 - ab + b^2)} \times x^{(b + c)(b^2 - bc + c^2)} \times x^{(c + a)(c^2 - ca + a^2)} \\[1em] = x^{a^3 + b^3} \times x^{b^3 + c^3} \times x^{c^3 + a^3} \\[1em] = x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] = x^{2a^3 + 2b^3 + 2c^3} \\[1em] = x^{2(a^3 + b^3 + c^3)}.

Hence, (xax−b)a2−ab+b2×(xbx−c)b2−bc+c2×(xcx−a)c2−ca+a2=x2(a3+b3+c3).\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} = x^{2(a^3 + b^3 + c^3)}.

Exercise 7(B)

Question 1(a)

a−1a−1+b−1\dfrac{a^{-1}}{a^{-1} + b^{-1}} is equal to :

  1. ba+b\dfrac{b}{a + b}

  2. a+ba\dfrac{a + b}{a}

  3. aa+b\dfrac{a}{a + b}

  4. a(a + b)

Answer

Simplifying the expression :

⇒a−1a−1+b−1=1a1a+1b=1ab+aab=aba(b+a)=ba+b.\Rightarrow \dfrac{a^{-1}}{a^{-1} + b^{-1}} = \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} + \dfrac{1}{b}} \\[1em] = \dfrac{\dfrac{1}{a}}{\dfrac{b + a}{ab}} \\[1em] = \dfrac{ab}{a(b + a)} \\[1em] = \dfrac{b}{a + b}.

Hence, Option 1 is the correct option.

Question 1(b)

If 42x=1324^{2x} = \dfrac{1}{32}, the value of x is :

  1. 1.25

  2. -1.25

  3. 1

  4. -1

Answer

Solving, the given expression :

⇒42x=132⇒(22)2x=125⇒24x=2−5⇒4x=−5⇒x=−54=−1.25\Rightarrow 4^{2x} = \dfrac{1}{32} \\[1em] \Rightarrow (2^2)^{2x} = \dfrac{1}{2^5} \\[1em] \Rightarrow 2^{4x} = 2^{-5} \\[1em] \Rightarrow 4x = -5 \\[1em] \Rightarrow x = -\dfrac{5}{4} = -1.25

Hence, Option 2 is the correct option.

Question 1(c)

3xy−1+2yx−1\dfrac{3x}{y^{-1}} + \dfrac{2y}{x^{-1}} is equal to :

  1. 6xy

  2. 3x2 + 2y2

  3. 5xy

  4. 6xy\dfrac{6}{xy}

Answer

Simplifying the expression :

⇒3xy−1+2yx−1=3x1y+2y1x=3xy+2xy=5xy.\Rightarrow \dfrac{3x}{y^{-1}} + \dfrac{2y}{x^{-1}} = \dfrac{3x}{\dfrac{1}{y}} + \dfrac{2y}{\dfrac{1}{x}} \\[1em] = 3xy + 2xy \\[1em] = 5xy.

Hence, Option 3 is the correct option.

Question 1(d)

(8−43÷2−2)\Big(8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big) is equal to :

  1. 14\dfrac{1}{4}

  2. −14-\dfrac{1}{4}

  3. −12-\dfrac{1}{2}

  4. 12\dfrac{1}{2}

Answer

Simplifying the expression :

⇒(8−43÷2−2)=[(23)−43÷122]=[2−4÷122]=124÷122=124×22=122=14.\Rightarrow \Big(8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big) = \Big[(2^3)^{-\dfrac{4}{3}} ÷ \dfrac{1}{2^2}\Big] \\[1em] = \Big[2^{-4} ÷ \dfrac{1}{2^2}\Big] \\[1em] = \dfrac{1}{2^4} ÷ \dfrac{1}{2^2} \\[1em] = \dfrac{1}{2^4} \times 2^2 \\[1em] = \dfrac{1}{2^2} \\[1em] = \dfrac{1}{4}.

Hence, Option 1 is the correct option.

Question 1(e)

If 82x + 5 = 1, value of x is :

  1. −52-\dfrac{5}{2}

  2. 52\dfrac{5}{2}

  3. 2

  4. 25\dfrac{2}{5}

Answer

Given,

⇒ 82x + 5 = 1

⇒ 82x + 5 = 80

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = −52-\dfrac{5}{2}.

Hence, Option 1 is the correct option.

Question 1(f)

If 3x + 1 = 9x - 2, the value of x is :

  1. -5

  2. 5

  3. 0

  4. 3

Answer

Given,

⇒ 3x + 1 = 9x - 2

⇒ 3x + 1 = (32)x - 2

⇒ 3x + 1 = 32x - 4

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, Option 2 is the correct option.

Question 1(g)

If ab=(ba)1−2x\sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x}, the value of x is :

  1. 34\dfrac{3}{4}

  2. 35\dfrac{3}{5}

  3. −34-\dfrac{3}{4}

  4. −35-\dfrac{3}{5}

Answer

Given,

⇒ab=(ba)1−2x⇒(ab)12=(ba)1−2x⇒(ab)12=(ab)−(1−2x)⇒(ab)12=(ab)(2x−1)⇒12=2x−1⇒2x=1+12⇒2x=2+12⇒x=32×2⇒x=34.\Rightarrow \sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{a}{b}\Big)^{-(1 - 2x)} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{a}{b}\Big)^{(2x - 1)} \\[1em] \Rightarrow \dfrac{1}{2} = 2x - 1 \\[1em] \Rightarrow 2x = 1 + \dfrac{1}{2} \\[1em] \Rightarrow 2x = \dfrac{2 + 1}{2} \\[1em] \Rightarrow x = \dfrac{3}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{3}{4}.

Hence, Option 1 is the correct option.

Question 2(i)

Solve for x :

22x + 1 = 8

Answer

Given,

⇒ 22x + 1 = 8

⇒ 22x + 1 = 23

⇒ 2x + 1 = 3

⇒ 2x = 3 - 1

⇒ 2x = 2

⇒ x = 22\dfrac{2}{2} = 1.

Hence, x = 1.

Question 2(ii)

Solve for x :

25x - 1 = 4 × 23x + 1

Answer

Given,

⇒ 25x - 1 = 4 × 23x + 1

⇒ 25x - 1 = 22 × 23x + 1

⇒ 25x - 1 = 22 + 3x + 1

⇒ 25x - 1 = 23x + 3

⇒ 5x - 1 = 3x + 3

⇒ 5x - 3x = 3 + 1

⇒ 2x = 4

⇒ x = 42\dfrac{4}{2} = 2.

Hence, x = 2.

Question 2(iii)

Solve for x :

34x + 1 = (27)x + 1

Answer

Given,

⇒ 34x + 1 = (27)x + 1

⇒ 34x + 1 = (33)x + 1

⇒ 34x + 1 = 33(x + 1)

⇒ 34x + 1 = 33x + 3

⇒ 4x + 1 = 3x + 3

⇒ 4x - 3x = 3 - 1

⇒ x = 2.

Hence, x = 2.

Question 2(iv)

Solve for x :

(49)x + 4 = 72 × (343)x + 1

Answer

Given,

⇒ (49)x + 4 = 72 × (343)x + 1

⇒ (72)x + 4 = 72 × (73)x + 1

⇒ 72(x + 4) = 72 × 73(x + 1)

⇒ 72x + 8 = 72 × 73x + 3

⇒ 72x + 8 = 72 + 3x + 3

⇒ 2x + 8 = 3x + 5

⇒ 3x - 2x = 8 - 5

⇒ x = 3.

Hence, x = 3.

Question 3(i)

Find x, if :

42x = 132\dfrac{1}{32}

Answer

Given,

⇒42x=132⇒(22)2x=125⇒24x=2−5⇒4x=−5⇒x=−54.\Rightarrow 4^{2x} = \dfrac{1}{32} \\[1em] \Rightarrow (2^2)^{2x} = \dfrac{1}{2^5} \\[1em] \Rightarrow 2^{4x} = 2^{-5} \\[1em] \Rightarrow 4x = -5 \\[1em] \Rightarrow x = -\dfrac{5}{4}.

Hence, x = −54-\dfrac{5}{4}.

Question 3(ii)

Find x, if :

2x+3=16\sqrt{2^{x + 3}} = 16

Answer

Given,

⇒ 2x+3=16\sqrt{2^{x + 3}} = 16

Squaring both sides, we get :

⇒ 2x + 3 = 162

⇒ 2x + 3 = (24)2

⇒ 2x + 3 = 28

⇒ x + 3 = 8

⇒ x = 8 - 3 = 5.

Hence, x = 5.

Question 3(iii)

Find x, if :

(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}

Answer

Given,

⇒(35)x+1=12527⇒(3512)x+1=(53)3⇒(35)x+12=(35)−3⇒x+12=−3⇒x+1=−6⇒x=−6−1=−7.\Rightarrow \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} \\[1em] \Rightarrow \Big(\dfrac{3}{5}^{\dfrac{1}{2}}\Big)^{x + 1} = \Big(\dfrac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3} \\[1em] \Rightarrow \dfrac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -6 - 1 = -7.

Hence, x = -7.

Question 3(iv)

Find x, if :

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

Answer

Given,

⇒(233)x−1=278⇒(2313)x−1=(32)3⇒(23)x−13=(23)−3⇒x−13=−3⇒x−1=3×−3⇒x−1=−9⇒x=−9+1=−8.\Rightarrow \Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8} \\[1em] \Rightarrow \Big(\dfrac{2}{3}^{\dfrac{1}{3}}\Big)^{x - 1} = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{2}{3}\Big)^{-3} \\[1em] \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 4(i)

Solve :

4x - 2 - 2x + 1 = 0

Answer

Given,

⇒ 4x - 2 - 2x + 1 = 0

⇒ (22)x - 2 - 2x + 1 = 0

⇒ 22(x - 2) = 2x + 1

⇒ 22x - 4 = 2x + 1

⇒ 2x - 4 = x + 1

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, x = 5.

Question 4(ii)

Solve :

3x2 : 3x = 9 : 1

Answer

Given,

⇒ 3x2 : 3x = 9 : 1

⇒3x23x=91⇒3x23x=3230⇒3x2−x=32−0⇒x2−x=2−0⇒x2−x=2⇒x2−x−2=0⇒x2−2x+x−2=0⇒x(x−2)+1(x−2)=0⇒(x+1)(x−2)=0⇒x+1=0 or x−2=0⇒x=−1 or x=2.\Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{9}{1} \\[1em] \Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{3^2}{3^0} \\[1em] \Rightarrow 3^{x^2 - x} = 3^{2- 0} \\[1em] \Rightarrow x^2 - x = 2 - 0 \\[1em] \Rightarrow x^2 - x = 2 \\[1em] \Rightarrow x^2 - x - 2 = 0 \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = 2.

Hence, x = -1 or x = 2.

Question 5(i)

Solve :

8 × 22x + 4 × 2x + 1 = 1 + 2x

Answer

Given,

⇒ 8 × 22x + 4 × 2x + 1 = 1 + 2x

⇒ 8 × 2(x)(2) + 4 × 2x.21 = 1 + 2x

Substituting 2x = a, in above equation, we get :

⇒ 8 × a2 + 4a × 2 = 1 + a

⇒ 8a2 + 8a = 1 + a

⇒ 8a2 + 8a - a - 1 = 0

⇒ 8a(a + 1) - 1(a + 1) = 0

⇒ (8a - 1)(a + 1) = 0

⇒ 8a - 1 = 0 or a + 1 = 0

⇒ 8a = 1 or a = -1

a cannot be negative as 2x, for any value of x is greater than 0.

⇒ 8(2x) = 1

⇒ 2x = 18\dfrac{1}{8}

⇒ 2x=1232^x = \dfrac{1}{2^3}

⇒ 2x = 2-3

⇒ x = -3.

Hence, x = -3.

Question 5(ii)

Solve :

22x + 2x + 2 - 4 × 23 = 0

Answer

Given,

⇒ 22x + 2x + 2 - 4 × 23 = 0

⇒ 2(x)(2) + 2x.22 - 4 × 8 = 0

Substituting 2x = a, we get :

⇒ a2 + 4a - 32 = 0

⇒ a2 + 8a - 4a - 32 = 0

⇒ a(a + 8) - 4(a + 8) = 0

⇒ (a - 4)(a + 8) = 0

⇒ a - 4 = 0 or a + 8 = 0

⇒ a = 4 or a = -8

a cannot be negative as 2x, for any value of x is greater than 0.

⇒ a = 4

⇒ 2x = 4

⇒ 2x = 22

⇒ x = 2.

Hence, x = 2.

Question 5(iii)

Solve :

(3)x−3=(34)x+1(\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1}

Answer

Given,

⇒(3)x−3=(34)x+1⇒(312)x−3=(314)x+1⇒3x−32=3x+14⇒x−32=x+14⇒4(x−3)=2(x+1)⇒4x−12=2x+2⇒4x−2x=2+12⇒2x=14⇒x=142=7.\Rightarrow (\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1} \\[1em] \Rightarrow (3^{\dfrac{1}{2}})^{x - 3} = (3^{\dfrac{1}{4}})^{x + 1} \\[1em] \Rightarrow 3^{\dfrac{x - 3}{2}} = 3^{\dfrac{x + 1}{4}} \\[1em] \Rightarrow \dfrac{x - 3}{2} = \dfrac{x + 1}{4}\\[1em] \Rightarrow 4(x - 3) = 2(x + 1) \\[1em] \Rightarrow 4x - 12 = 2x + 2 \\[1em] \Rightarrow 4x - 2x = 2 + 12 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \dfrac{14}{2} = 7.

Hence, x = 7.

Question 6

Find the values of m and n if :

42m = (163)−6n=(8)2(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2

Answer

Given,

42m = (163)−6n=(8)2(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2

Considering,

⇒42m=(8)2⇒(22)2m=(8)2⇒24m=8⇒24m=23⇒4m=3⇒m=34.\Rightarrow 4^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow 2^{4m} = 8 \\[1em] \Rightarrow 2^{4m} = 2^3 \\[1em] \Rightarrow 4m = 3 \\[1em] \Rightarrow m = \dfrac{3}{4}.

Considering,

⇒(163)−6n=(8)2⇒(16)13×(−6n)=8⇒(16)−2n=8⇒(24)−2n=23⇒(2)4×−2n=23⇒(2)−8n=23⇒−8n=3⇒n=−83.\Rightarrow (\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\dfrac{1}{3} \times \Big(-\dfrac{6}{n}\Big)} = 8 \\[1em] \Rightarrow (16)^{-\dfrac{2}{n}} = 8 \\[1em] \Rightarrow (2^4)^{-\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{4 \times -\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{-\dfrac{8}{n}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{n} = 3 \\[1em] \Rightarrow n = -\dfrac{8}{3}.

Hence, m=34 and n=−83m = \dfrac{3}{4} \text{ and } n = -\dfrac{8}{3}.

Question 7

Solve for x and y, if :

(32)x÷2y+1=1 and 8y−164−x2=0(\sqrt{32})^x ÷ 2^{y + 1} = 1 \text{ and } 8^y - 16^{4 - \dfrac{x}{2}} = 0

Answer

Given,

⇒(32)x÷2y+1=1⇒(32)x2y+1=1⇒(25)x2y+1=1⇒[(25)12]x=2y+1⇒25×12×x=2y+1⇒5x2=y+1⇒5x=2y+2⇒x=2y+25 ......(1)\Rightarrow (\sqrt{32})^x ÷ 2^{y + 1} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{32})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{2^5})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow [(2^5)^{\dfrac{1}{2}}]^x = 2^{y + 1} \\[1em] \Rightarrow 2^{5 \times \dfrac{1}{2} \times x} = 2^{y + 1} \\[1em] \Rightarrow \dfrac{5x}{2} = y + 1 \\[1em] \Rightarrow 5x = 2y + 2 \\[1em] \Rightarrow x = \dfrac{2y + 2}{5} \text{ ......(1)}

Given,

⇒8y−164−x2=0⇒8y=164−x2⇒(23)y=(24)4−x2⇒23y=(2)4(4−x2)⇒23y=216−2x⇒3y=16−2x⇒2x=16−3y ........(2)\Rightarrow 8^y - 16^{4 - \dfrac{x}{2}} = 0 \\[1em] \Rightarrow 8^y = 16^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow (2^3)^y = (2^4)^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow 2^{3y} = (2)^{4\Big(4 - \dfrac{x}{2}\Big)} \\[1em] \Rightarrow 2^{3y} = 2^{16 - 2x} \\[1em] \Rightarrow 3y = 16 - 2x \\[1em] \Rightarrow 2x = 16 - 3y \text{ ........(2)}

Substituting value of x from equation (1) in equation (2), we get :

⇒2(2y+25)=16−3y⇒4y+45=16−3y⇒4y+4=5(16−3y)⇒4y+4=80−15y⇒4y+15y=80−4⇒19y=76⇒y=7619=4.\Rightarrow 2\Big(\dfrac{2y + 2}{5}\Big) = 16 - 3y \\[1em] \Rightarrow \dfrac{4y + 4}{5} = 16 - 3y \\[1em] \Rightarrow 4y + 4 = 5(16 - 3y) \\[1em] \Rightarrow 4y + 4 = 80 - 15y \\[1em] \Rightarrow 4y + 15y = 80 - 4 \\[1em] \Rightarrow 19y = 76 \\[1em] \Rightarrow y = \dfrac{76}{19} = 4.

Substituting value of y in equation (1), we get :

⇒x=2y+25=2×4+25=8+25=105=2.\Rightarrow x = \dfrac{2y + 2}{5} = \dfrac{2 \times 4 + 2}{5} \\[1em] = \dfrac{8 + 2}{5} \\[1em] = \dfrac{10}{5} = 2.

Hence, x = 2 and y = 4.

Question 8(i)

Prove that :

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

Answer

Solving L.H.S. of the given equation :

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

Since, L.H.S. = R.H.S. = 1.

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

Question 8(ii)

Prove that :

xa(b−c)xb(a−c)÷(xbxa)c\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = 1

Answer

Solving L.H.S. of the given equation :

⇒xa(b−c)xb(a−c)÷(xbxa)c=xab−acxab−bc÷xbcxac=xab−ac−(ab−bc)÷xbc−ac=xab−ab−ac+bc÷xbc−ac=xbc−ac÷xbc−ac=xbc−acxbc−ac=1.\Rightarrow \dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = \dfrac{x^{ab - ac}}{x^{ab - bc}} ÷ \dfrac{x^{bc}}{x^{ac}} \\[1em] = x^{ab - ac - (ab - bc)} ÷ x^{bc - ac} \\[1em] = x^{ab - ab - ac + bc} ÷ x^{bc - ac} \\[1em] = x^{bc - ac} ÷ x^{bc - ac} \\[1em] = \dfrac{x^{bc - ac}}{x^{bc - ac}} \\[1em] = 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that xa(b−c)xb(a−c)÷(xbxa)c=1\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = 1.

Question 9

If ax = b, by = c and cz = a, prove that :

xyz = 1.

Answer

Given,

⇒ a = cz .......(1)

⇒ c = by .......(2)

⇒ b = ax ........(3)

Substituting value of c from equation (2) in (1), we get :

⇒ a = (by)z

⇒ a = byz

Substituting value of b from equation (3) in above equation, we get :

⇒ a = (ax)yz

⇒ a = axyz

⇒ xyz = 1.

Hence, proved that xyz = 1.

Question 10

If ax = by = cz and b2 = ac, prove that :

y = 2xzx+z\dfrac{2xz}{x + z}.

Answer

Given,

⇒ ax = by = cz = k (let)

⇒ ax = k

⇒ a = k1xk^{\dfrac{1}{x}} ........(1)

⇒ by = k

⇒ b = k1yk^{\dfrac{1}{y}} ........(2)

⇒ cz = k

⇒ c = k1zk^{\dfrac{1}{z}} ........(3)

Given,

⇒ b2 = ac

Substituting values of a, b and c in above equation, we get :

⇒(k1y)2=k1x×k1z⇒k2y=k1x+1z⇒2y=1x+1z⇒2y=z+xxz⇒y=2xzx+z.\Rightarrow (k^{\dfrac{1}{y}})^2 = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{y}} = k^{\dfrac{1}{x} + \dfrac{1}{z}} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{z + x}{xz} \\[1em] \Rightarrow y = \dfrac{2xz}{x + z}.

Hence, proved that y=2xzx+z.y = \dfrac{2xz}{x + z}.

Question 11

If 5-p = 4-q = 20r, show that :

1p+1q+1r=0\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.

Answer

Given,

⇒ 5-p = 4-q = 20r = k (let)

⇒ 5-p = k

⇒ 5 = k−1pk^{-\dfrac{1}{p}} .......(1)

⇒ 4-q = k

⇒ 4 = k−1qk^{-\dfrac{1}{q}} ......(2)

⇒ 20r = k

⇒ 20 = k1rk^{\dfrac{1}{r}} ............(3)

We know that,

⇒ 5 × 4 = 20

From equations (1), (2) and (3), we get :

⇒k−1p×k−1q=k1r⇒k−1p+(−1q)=k1r⇒k−1p−1q=k1r⇒−1p−1q=1r⇒1p+1q+1r=0.\Rightarrow k^{-\dfrac{1}{p}} \times k^{-\dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} + \Big(-\dfrac{1}{q}\Big)} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} - \dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow -\dfrac{1}{p} - \dfrac{1}{q} = \dfrac{1}{r} \\[1em] \Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.

Hence, proved that 1p+1q+1r=0.\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0..

Question 12

If m ≠ n and (m + n)-1 (m-1 + n-1) = mxny;

show that : x + y + 2 = 0

Answer

Given,

⇒(m+n)−1(m−1+n−1)=mxny⇒1(m+n)×(1m+1n)=mxny⇒1(m+n)×(n+mmn)=mxny⇒1mn=mxny⇒m−1n−1=mxny⇒x=−1 and y=−1.\Rightarrow (m + n)^{-1}(m^{-1} + n^{-1}) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{1}{m} + \dfrac{1}{n}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{n + m}{mn}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{mn} = m^xn^y \\[1em] \Rightarrow m^{-1}n^{-1} = m^xn^y \\[1em] \Rightarrow x = -1 \text{ and } y = -1.

Substituting value in L.H.S. of equation x + y + 2 = 0, we get :

⇒ x + y + 2 = (-1) + (-1) + 2 = -2 + 2 = 0.

Since, L.H.S. = R.H.S. = 0.

Hence, proved that x + y + 2 = 0.

Question 13

If 5x + 1 = 25x - 2; find the value of :

3x - 3 × 23 - x.

Answer

Given,

⇒ 5x + 1 = 25x - 2

⇒ 5x + 1 = (52)x - 2

⇒ 5x + 1 = 52(x - 2)

⇒ x + 1 = 2(x - 2)

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Substituting value of x in 3x - 3 × 23 - x, we get :

⇒ 3x - 3 × 23 - x = 35 - 3 × 23 - 5

= 32 × 2(-2)

= 9×122=94=2149 \times \dfrac{1}{2^2} = \dfrac{9}{4} = 2\dfrac{1}{4}.

Hence, 3x - 3 × 23 - x = 2142\dfrac{1}{4}.

Question 14

If 4x + 3 = 112 + 8 × 4x; find (18x)3x.

Answer

Given,

⇒ 4x + 3 = 112 + 8 × 4x

⇒ 4x.43 = 8(14 + 4x)

⇒ 64.4x = 8(14 + 4x)

⇒ 64.4x8\dfrac{64.4^x}{8} = 14 + 4x

⇒ 8.4x = 14 + 4x

⇒ 8.4x - 4x = 14

⇒ 4x(8 - 1) = 14

⇒ 7.4x = 14

⇒ 4x = 147\dfrac{14}{7}

⇒ (22)x = 2

⇒ 22x = 21

⇒ 2x = 1

⇒ x = 12\dfrac{1}{2}.

Substituting value of x in (18x)3x, we get :

⇒(18x)3x=(18×12)3×12=932=(32)32=32×32=33=27.\Rightarrow (18x)^{3x} = \Big(18 \times \dfrac{1}{2}\Big)^{3 \times \dfrac{1}{2}} \\[1em] = 9^{\dfrac{3}{2}} \\[1em] = (3^2)^{\dfrac{3}{2}} \\[1em] = 3^{2 \times \dfrac{3}{2}} \\[1em] = 3^3 \\[1em] = 27.

Hence, (18x)3x = 27.

Question 15(i)

Solve for x :

4x−1×(0.5)3−2x=(18)−x4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^{-x}

Answer

Given,

⇒4x−1×(0.5)3−2x=(18)−x⇒(22)x−1×(510)3−2x=(123)−x⇒22(x−1)×(12)3−2x=(2−3)−x⇒22x−2×(2−1)3−2x=23x⇒22x−2×2−1(3−2x)=23x⇒2(2x−2)+[−1(3−2x)]=23x⇒22x−2−3+2x=23x⇒24x−5=23x⇒4x−5=3x⇒4x−3x=5⇒x=5.\Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^{-x} \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\dfrac{5}{10}\Big)^{3 - 2x} = \Big(\dfrac{1}{2^3}\Big)^{-x} \\[1em] \Rightarrow 2^{2(x - 1)} \times \Big(\dfrac{1}{2}\Big)^{3 - 2x} = (2^{-3})^{-x} \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{-1(3 - 2x)} = 2^{3x} \\[1em] \Rightarrow 2^{(2x - 2) + [-1(3 - 2x)]} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2 - 3 + 2x} = 2^{3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{3x} \\[1em] \Rightarrow 4x - 5 = 3x \\[1em] \Rightarrow 4x - 3x = 5 \\[1em] \Rightarrow x = 5.

Hence, x = 5.

Question 15(ii)

Solve for x :

(a3x + 5)2.(ax)4 = a8x + 12

Answer

Given,

⇒ (a3x + 5)2.(ax)4 = a8x + 12

⇒ a2(3x + 5).a4x = a8x + 12

⇒ a6x + 10.a4x = a8x + 12

⇒ a6x + 10 + 4x = a8x + 12

⇒ a10x + 10 = a8x + 12

⇒ 10x + 10 = 8x + 12

⇒ 10x - 8x = 12 - 10

⇒ 2x = 2

⇒ x = 22\dfrac{2}{2} = 1.

Hence, x = 1.

Question 15(iii)

Solve for x :

(81)34−(132)−25+x(12)−1.20=27(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27

Answer

Given,

⇒(81)34−(132)−25+x(12)−1.20=27⇒(34)34−(125)−25+(2−1)−1x.1=27⇒34×34−(2−5)−25+2−1×−1.x=27⇒33−2−5×−25+2x=27⇒27−22+2x=27⇒27−4+2x=27⇒23+2x=27⇒2x=27−23⇒2x=4⇒x=42=2.\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27 \\[1em] \Rightarrow (3^4)^{\dfrac{3}{4}} - \Big(\dfrac{1}{2^5}\Big)^{-\dfrac{2}{5}} + (2^{-1})^{-1}x.1 = 27 \\[1em] \Rightarrow 3^{4 \times \dfrac{3}{4}} - (2^{-5})^{-\dfrac{2}{5}} + 2^{-1\times -1}.x = 27 \\[1em] \Rightarrow 3^3 - 2^{-5 \times -\dfrac{2}{5}} + 2x = 27 \\[1em] \Rightarrow 27 - 2^2 + 2x = 27 \\[1em] \Rightarrow 27 - 4 + 2x = 27 \\[1em] \Rightarrow 23 + 2x = 27 \\[1em] \Rightarrow 2x = 27 - 23 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} = 2.

Hence, x = 2.

Question 15(iv)

Solve for x :

23x + 3 = 23x + 1 + 48

Answer

Given,

⇒ 23x + 3 = 23x + 1 + 48

⇒ 23x.23 = 23x.21 + 48

⇒ 8.23x = 2.23x + 48

⇒ 8.23x - 2.23x = 48

⇒ 6.23x = 48

⇒ 23x = 486\dfrac{48}{6}

⇒ 23x = 8

⇒ 23x = 23

⇒ 3x = 3

⇒ x = 33\dfrac{3}{3} = 1.

Hence, x = 1.

Question 15(v)

Solve for x :

3(2x + 1) - 2x + 2 + 5 = 0

Answer

Given,

⇒ 3(2x + 1) - 2x + 2 + 5 = 0

⇒ 3.2x + 3 - 2x.22 + 5 = 0

⇒ 3.2x + 3 - 4.2x + 5 = 0

⇒ -2x + 8 = 0

⇒ 2x = 8

⇒ 2x = 23

⇒ x = 3.

Hence, x = 3.

Question 15(vi)

Solve for x :

9x + 2 = 720 + 9x

Answer

Given,

⇒ 9x + 2 = 720 + 9x

⇒ 9x.92 = 720 + 9x

⇒ 81.9x - 9x = 720

⇒ 9x(81 - 1) = 720

⇒ 9x.80 = 720

⇒ 9x = 72080\dfrac{720}{80}

⇒ 9x = 9

⇒ x = 1.

Hence, x = 1.

Test Yourself

Question 1(a)

(200 - 180) x 70 is equal to:

  1. 0

  2. 1

  3. 14

  4. none of these

Answer

Given, (200 - 180) x 70

As we know that a0 = 1

⇒ (1 - 1) x 1

⇒ 0 x 1

⇒ 0

Hence, option 1 is the correct option.

Question 1(b)

(-2)-1 ÷ (-2)-4 is equal to:

  1. 8

  2. 18\dfrac{1}{8}

  3. -8

  4. -18\dfrac{1}{8}

Answer

Given,

⇒ (-2)-1 ÷ (-2)-4

⇒ (-2)-1 ÷ 1(−2)4\dfrac{1}{(-2)^4}

⇒ 1(−2)÷1(−2)4\dfrac{1}{(-2)} ÷ \dfrac{1}{(-2)^4}

⇒ 1(−2)×(−2)4\dfrac{1}{(-2)} \times (-2)^4

⇒ (-2)3

⇒ -8.

Hence, option 3 is the correct option.

Question 1(c)

If ab=23÷(−23)0;then(ab)−2\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0; \text{then} \Big(\dfrac{a}{b}\Big)^{-2} is equal to:

  1. 49\dfrac{4}{9}

  2. −49-\dfrac{4}{9}

  3. 94\dfrac{9}{4}

  4. −94-\dfrac{9}{4}

Answer

Given, ab=23÷(−23)0\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0

As we know that a0 = 1

⇒ab=23÷1⇒ab=23⇒(ab)−2=(23)−2⇒(ab)−2=(32)2⇒(ab)−2=94.\Rightarrow \dfrac{a}{b} = \dfrac{2}{3} ÷ 1 \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{2}{3}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \dfrac{9}{4}.

Hence, option 3 is the correct option.

Question 1(d)

If 52x + 3 = 1, the value of x is:

  1. 32\dfrac{3}{2}

  2. −32-\dfrac{3}{2}

  3. 23\dfrac{2}{3}

  4. −23-\dfrac{2}{3}

Answer

Given, 52x + 3 = 1

As we know that a0 = 1,

⇒ 52x + 3 = 50

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = −32-\dfrac{3}{2}

Hence, option 2 is the correct option.

Question 1(e)

(2 + 3)-1 x (2-1 + 3-1) is equal to :

  1. 6

  2. -6

  3. 16\dfrac{1}{6}

  4. −16-\dfrac{1}{6}

Answer

Given,

⇒ (2 + 3)-1 x (2-1 + 3-1)

⇒(2+3)−1×(2−1+3−1)⇒(15)×[(12)+(13)]⇒15×[3+26]⇒15×56⇒16.\Rightarrow (2 + 3)^{-1} \times (2^{-1} + 3^{-1}) \\[1em] \Rightarrow \Big(\dfrac{1}{5}\Big) \times \Big[\Big(\dfrac{1}{2}\Big) + \Big(\dfrac{1}{3}\Big)\Big]\\[1em] \Rightarrow \dfrac{1}{5} \times \Big[\dfrac{3 + 2}{6}\Big]\\[1em] \Rightarrow \dfrac{1}{5} \times \dfrac{5}{6}\\[1em] \Rightarrow \dfrac{1}{6}.

Hence, option 3 is the correct option.

Question 1(f)

Statement 1: (34)−4×(34)−5=(34)3x\Big(\dfrac{3}{4}\Big)^{-4} \times \Big(\dfrac{3}{4}\Big)^{-5} = \Big(\dfrac{3}{4}\Big)^{3x}

⇒ x = -1

Statement 2: (34)−4−5=(34)3x\Big(\dfrac{3}{4}\Big)^{-4 - 5} = \Big(\dfrac{3}{4}\Big)^{3x}

⇒ 3x = -9

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Given,

⇒(34)−4×(34)−5=(34)3x⇒(34)−4+(−5)=(34)3x⇒(34)−4−5=(34)3x⇒(34)−9=(34)3x⇒3x=−9⇒x=−93⇒x=−3.\Rightarrow \Big(\dfrac{3}{4}\Big)^{-4} \times \Big(\dfrac{3}{4}\Big)^{-5} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-4 + (-5)} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-4 - 5} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-9} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow 3x = -9\\[1em] \Rightarrow x = -\dfrac{9}{3}\\[1em] \Rightarrow x = -3.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(g)

Statement 1: (58)−7×(85)−4=x\Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{8}{5}\Big)^{-4} = x

⇒ x = (58)3\Big(\dfrac{5}{8}\Big)^{3}

Statement 2: (58)−7×(58)4=x\Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{5}{8}\Big)^{4} = x

⇒ x = (85)3\Big(\dfrac{8}{5}\Big)^{3}

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Given,

⇒(58)−7×(85)−4=x⇒(58)−7×(58)4=x⇒(58)−7+4=x⇒(58)−3=x⇒x=(85)3\Rightarrow \Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{8}{5}\Big)^{-4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{5}{8}\Big)^{4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-7 + 4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-3} = x\\[1em] \Rightarrow x = \Big(\dfrac{8}{5}\Big)^{3}

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(h)

Assertion (A): (3-7 ÷ 3-10) x 3-5 = 19\dfrac{1}{9}.

Reason (R): 137×310×135\dfrac{1}{3^7} \times 3^{10} \times \dfrac{1}{3^5}.

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

(3-7 ÷ 3-10) x 3-5

⇒[(13)7÷(13)10]×(13)5⇒[(13)7×310]×135⇒31037×35⇒310312⇒132⇒19.\Rightarrow \Big[\Big(\dfrac{1}{3}\Big)^7 ÷ \Big(\dfrac{1}{3}\Big)^{10}\Big] \times \Big(\dfrac{1}{3}\Big)^5 \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{3}\Big)^7 \times 3^{10}\Big] \times \dfrac{1}{3^5} \\[1em] \Rightarrow \dfrac{3^{10}}{3^7 \times 3^5} \\[1em] \Rightarrow \dfrac{3^{10}}{3^{12}} \\[1em] \Rightarrow \dfrac{1}{3^2} \\[1em] \Rightarrow \dfrac{1}{9}.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Question 1(i)

Assertion (A): (13 + 23 + 33)12^\frac{1}{2} = x, then x = 1+8+27\sqrt{1} + \sqrt{8} + \sqrt{27}.

Reason (R): x = (1 + 8 + 27)12=3612=6^\frac{1}{2} = 36^\frac{1}{2} = 6

  1. A is true, but R is false.

  2. A is false, but R is true.

  3. Both A and R are true, and R is the correct reason for A.

  4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, x = (13 + 23 + 33)12^\frac{1}{2}

⇒(1+8+27)12⇒(36)12⇒36⇒6.\Rightarrow (1 + 8 + 27)^\frac{1}{2}\\[1em] \Rightarrow (36)^\frac{1}{2}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Question 2(i)

Evaluate :

952−3×80−(181)−129^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}

Answer

Simplifying the expression :

⇒952−3×80−(181)−12=(32)52−3×1−(134)−12=(3)2×52−3−(3−4)−12=35−3−3−4×−12=243−3−32=243−3−9=231.\Rightarrow 9^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} = (3^2)^{\dfrac{5}{2}} - 3 \times 1 - \Big(\dfrac{1}{3^4}\Big)^{-\dfrac{1}{2}} \\[1em] = (3)^{2 \times \dfrac{5}{2}} - 3 - (3^{-4})^{-\dfrac{1}{2}} \\[1em] = 3^5 - 3 - 3^{-4 \times -\dfrac{1}{2}} \\[1em] = 243 - 3 - 3^2 \\[1em] = 243 - 3 - 9 \\[1em] = 231.

Hence, 952−3×80−(181)−129^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} = 231.

Question 2(ii)

Evaluate :

(64)23−1253−12−5+(27)−23×(259)−12(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}}

Answer

Simplifying the expression :

⇒(64)23−1253−12−5+(27)−23×(259)−12=(26)23−(53)13−25+(33)−23×[(53)2]−12=(2)6×23−(5)3×13−32+(3)3×−23×(53)2×−12=24−51−32+3−2×(53)−1=16−5−32+132×35=−21+19×35=−21+345=−21+115=−315+115=−31415=−201415.\Rightarrow (64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} \\[1em] = (2^6)^{\dfrac{2}{3}} - (5^3)^{\dfrac{1}{3}} - 2^5 + (3^3)^{-\dfrac{2}{3}} \times \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-\dfrac{1}{2}}\\[1em] = (2)^{6 \times \dfrac{2}{3}} - (5)^{3 \times \dfrac{1}{3}} - 32 + (3)^{3 \times -\dfrac{2}{3}} \times \Big(\dfrac{5}{3}\Big)^{2 \times -\dfrac{1}{2}} \\[1em] = 2^4 - 5^1 - 32 + 3^{-2} \times \Big(\dfrac{5}{3}\Big)^{-1} \\[1em] = 16 - 5 - 32 + \dfrac{1}{3^2} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{1}{9} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{3}{45} \\[1em] = -21 + \dfrac{1}{15} \\[1em] = \dfrac{-315 + 1}{15} \\[1em] = \dfrac{-314}{15} \\[1em] = -20\dfrac{14}{15}.

Hence, (64)23−1253−12−5+(27)−23×(259)−12=−201415.(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} = -20\dfrac{14}{15}.

Question 2(iii)

Evaluate :

[(−23)−2]3×(13)−4×3−1×16\Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6}

Answer

Simplify the expression :

⇒[(−23)−2]3×(13)−4×3−1×16=(−23)−6×(3−1)−4×13×16=(32)6×34×118=36×3426×18=36+426×(2×3×3)=31026+1×32=310−227=3827=38÷27.\Rightarrow \Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6} \\[1em] = \Big(-\dfrac{2}{3}\Big)^{-6} \times (3^{-1})^{-4} \times \dfrac{1}{3} \times \dfrac{1}{6} \\[1em] = \Big(\dfrac{3}{2}\Big)^6 \times 3^4 \times \dfrac{1}{18} \\[1em] = \dfrac{3^6 \times 3^4}{2^6 \times 18} \\[1em] = \dfrac{3^{6 + 4}}{2^6 \times (2 \times 3 \times 3)} \\[1em] = \dfrac{3^{10}}{2^{6 + 1} \times 3^2} \\[1em] = \dfrac{3^{10 - 2}}{2^7} \\[1em] = \dfrac{3^8}{2^7} \\[1em] = 3^8 ÷ 2^7.

Hence, [(−23)−2]3×(13)−4×3−1×16=38÷27.\Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6} = 3^8 ÷ 2^7.

Question 3

Simplify :

3×9n+1−9×32n3×32n+3−9n+1\dfrac{3 \times 9^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}}

Answer

Simplify the expression :

⇒3×9n+1−9×32n3×32n+3−9n+1=3×(32)n+1−9×32n3×32n.33−(32)n+1=3×32(n+1)−9×32n81.32n−32(n+1)=3×32n+2−9×32n81.32n−32n+2=3×32n×32−9×32n81.32n−32n.32=32n(3×32−9)32n(81−32)=27−981−9=1872=14.\Rightarrow \dfrac{3 \times 9^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \dfrac{3 \times (3^2)^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n}.3^3 - (3^2)^{n + 1}} \\[1em] = \dfrac{3 \times 3^{2(n + 1)} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2(n + 1)}} \\[1em] = \dfrac{3 \times 3^{2n + 2} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n + 2}} \\[1em] = \dfrac{3 \times 3^{2n} \times 3^2 - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n}.3^2} \\[1em] = \dfrac{3^{2n}(3 \times 3^2 - 9)}{3^{2n}(81 - 3^2)} \\[1em] = \dfrac{27 - 9}{81 - 9} \\[1em] = \dfrac{18}{72} \\[1em] = \dfrac{1}{4}.

Hence, 3×9n+1−3×32n3×32n+3−9n+1=14\dfrac{3 \times 9^{n + 1} - 3 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \dfrac{1}{4}.

Question 4

Solve :

3x - 1 × 52y - 3 = 225

Answer

Solving the expression :

⇒3x−1×52y−3=225⇒3x.3−1×52y.5−3=32×52⇒3x31×52y53=32×52⇒3x×52y=32×52×31×53⇒3x×52y=32+1×52+3⇒3x×52y=33×55⇒x=3 and 2y=5⇒x=3 and y=52=212.\Rightarrow 3^{x - 1} \times 5^{2y - 3} = 225 \\[1em] \Rightarrow 3^x.3^{-1} \times 5^{2y}.5^{-3} = 3^2 \times 5^2 \\[1em] \Rightarrow \dfrac{3^x}{3^1} \times \dfrac{5^{2y}}{5^3} = 3^2 \times 5^2 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^2 \times 5^2 \times 3^1 \times 5^3 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^{2 + 1} \times 5^{2 + 3} \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^3 \times 5^5 \\[1em] \Rightarrow x = 3 \text{ and } 2y = 5 \\[1em] \Rightarrow x = 3 \text{ and } y = \dfrac{5}{2} = 2\dfrac{1}{2}.

Hence, x = 3 and y = 2122\dfrac{1}{2}.

Question 5

If (a−1b2a2b−4)7÷(a3b−5a−2b3)−5=ax.by\Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y, find x + y.

Answer

Given,

⇒(a−1b2a2b−4)7÷(a3b−5a−2b3)−5=ax.by⇒(a−1−2b2−(−4))7÷(a3−(−2)b−5−3)−5=ax.by⇒(a−3b6)7÷(a5b−8)−5=ax.by⇒(a−3×7.b6×7)÷(a5×−5.b−8×−5)=ax.by⇒(a−21.b42)÷(a−25.b40)=ax.by⇒a−21.b42a−25.b40=axby⇒a−21−(−25).b42−40=axby⇒a−21+25.b2=ax.by⇒a4.b2=ax.by⇒x=4 and y=2.\Rightarrow \Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-1 - 2}b^{2 - (-4)})^7 ÷ (a^{3 - (-2)}b^{-5 - 3})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3}b^6)^7 ÷ (a^5b^{-8})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3 \times 7}.b^{6 \times 7}) ÷ (a^{5 \times -5}.b^{-8 \times -5}) = a^x.b^y \\[1em] \Rightarrow (a^{-21}.b^{42}) ÷ (a^{-25}.b^{40}) = a^x.b^y \\[1em] \Rightarrow \dfrac{a^{-21}.b^{42}}{a^{-25}.b^{40}} = a^xb^y \\[1em] \Rightarrow a^{-21 - (-25)}.b^{42 - 40} = a^xb^y \\[1em] \Rightarrow a^{-21 + 25}.b^{2} = a^x.b^y \\[1em] \Rightarrow a^4.b^2 = a^x.b^y \\[1em] \Rightarrow x = 4 \text{ and } y = 2.

x + y = 4 + 2 = 6.

Hence, x + y = 6.

Question 6

If 3x + 1 = 9x - 3, find the value of 21 + x.

Answer

Given,

⇒ 3x + 1 = 9x - 3

⇒ 3x + 1 = (32)x - 3

⇒ 3x + 1 = 32(x - 3)

⇒ 3x + 1 = 32x - 6

⇒ x + 1 = 2x - 6

⇒ 2x - x = 1 + 6

⇒ x = 7.

Substituting value of x in 21 + x, we get :

⇒ 21 + 7 = 28 = 256.

Hence, 21 + x = 256.

Question 7

If 2x = 4y = 8z and 12x+14y+18z\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z} = 4, find the value of x.

Answer

Given,

⇒ 2x = 4y = 8z

⇒ 2x = (22)y = (23)z

⇒ 2x = 22y = 23z

⇒ x = 2y = 3z

⇒ x = 2y and x = 3z

⇒y=x2 and z=x3\Rightarrow y = \dfrac{x}{2} \text{ and } z = \dfrac{x}{3}.

Substituting value of y and z in 12x+14y+18z=4\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z} = 4, we get :

⇒12x+14×x2+18×x3=4⇒12x+12x+38x=4⇒22x+38x=4⇒8+38x=4⇒118x=4⇒x=118×4=1132.\Rightarrow \dfrac{1}{2x} + \dfrac{1}{4 \times \dfrac{x}{2}} + \dfrac{1}{8 \times \dfrac{x}{3}} = 4 \\[1em] \Rightarrow \dfrac{1}{2x} + \dfrac{1}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{2}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{8 + 3}{8x} = 4 \\[1em] \Rightarrow \dfrac{11}{8x} = 4 \\[1em] \Rightarrow x = \dfrac{11}{8 \times 4} = \dfrac{11}{32}.

Hence, x = 1132\dfrac{11}{32}.

Question 8

If 9n.32.3n−(27)n(3m.2)3=3−3\dfrac{9^n.3^2.3^n - (27)^n}{(3^m.2)^3} = 3^{-3}.

Show that : m - n = 1.

Answer

Given,

⇒9n.32.3n−(27)n(3m.2)3=3−3⇒(32)n.32.3n−(33)n(3m)3.(2)3=3−3⇒32n.32.3n−33n=3−3.(3m)3.(2)3⇒9.32n+n−33n=3−3.33m.8⇒9.33n−33n=8.33m−3⇒33n(9−1)=8.33m−3⇒8.33n=8.33(m−1)⇒33n=33(m−1)⇒3n=3(m−1)⇒n=m−1⇒m−n=1.\Rightarrow \dfrac{9^n.3^2.3^n - (27)^n}{(3^m.2)^3} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3^2)^n.3^2.3^n - (3^3)^n}{(3^m)^3.(2)^3} = 3^{-3} \\[1em] \Rightarrow 3^{2n}.3^2.3^n - 3^{3n} = 3^{-3}.(3^m)^3.(2)^3 \\[1em] \Rightarrow 9.3^{2n + n} - 3^{3n} = 3^{-3}.3^{3m}.8 \\[1em] \Rightarrow 9.3^{3n} - 3^{3n} = 8.3^{3m - 3} \\[1em] \Rightarrow 3^{3n}(9 - 1) = 8.3^{3m - 3} \\[1em] \Rightarrow 8.3^{3n} = 8.3^{3(m - 1)} \\[1em] \Rightarrow 3^{3n} = 3^{3(m - 1)} \\[1em] \Rightarrow 3n = 3(m - 1) \\[1em] \Rightarrow n = m - 1 \\[1em] \Rightarrow m - n = 1.

Hence, proved that m - n = 1.

Question 9

Solve for x : (13)x=44−34−6.(13)^{\sqrt{x}} = 4^4 - 3^4 - 6.

Answer

Given,

⇒(13)x=44−34−6⇒(13)x=256−81−6⇒(13)x=169⇒(13)x=132⇒x=2\Rightarrow (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 256 - 81 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 169 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 13^2 \\[1em] \Rightarrow \sqrt{x} = 2

Squaring both sides of the above equation, we get :

⇒(x)2=22⇒x=4.\Rightarrow (\sqrt{x})^2 = 2^2 \\[1em] \Rightarrow x = 4.

Hence, x = 4.

Question 10

If 34x = (81)-1 and (10)1y=0.0001(10)^{\dfrac{1}{y}} = 0.0001, find the value of 2-x × 16y.

Answer

Given,

⇒34x=(34)−1⇒34x=3−4⇒4x=−4⇒x=−44⇒x=−1.\Rightarrow 3^{4x} = (3^4)^{-1} \\[1em] \Rightarrow 3^{4x} = 3^{-4} \\[1em] \Rightarrow 4x = -4 \\[1em] \Rightarrow x = -\dfrac{4}{4} \\[1em] \Rightarrow x = -1.

Given,

⇒(10)1y=0.0001⇒(10)1y=1104⇒(10)1y=10−4⇒1y=−4⇒y=−14.\Rightarrow (10)^{\dfrac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = \dfrac{1}{10^4} \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \dfrac{1}{y} = -4 \\[1em] \Rightarrow y = -\dfrac{1}{4}.

Substituting value of x and y in 2-x × 16y, we get :

⇒2−x×16y=2−(−1)×16−14=21×(24)−14=2×24×−14=2×2−1=2×12=1.\Rightarrow 2^{-x} \times 16^y = 2^{-(-1)} \times 16^{-\dfrac{1}{4}} \\[1em] = 2^1 \times (2^4)^{-\dfrac{1}{4}} \\[1em] = 2 \times 2^{4 \times -\dfrac{1}{4}} \\[1em] = 2 \times 2^{-1} \\[1em] = 2 \times \dfrac{1}{2} \\[1em] = 1.

Hence, 2-x × 16y = 1.

Question 11

If (am)n = am.an, find the value of :

m(n - 1) - (n - 1)

Answer

Given,

⇒ (am)n = am.an

⇒ amn = am + n

⇒ mn = m + n

⇒ mn - m = n

⇒ m(n - 1) = n

⇒ m = nn−1\dfrac{n}{n - 1} .......(1)

Substituting value of m from equation (1) in m(n - 1) - (n - 1), we get :

⇒m(n−1)−(n−1)=nn−1×(n−1)−(n−1)=n−(n−1)=n−n+1=1.\Rightarrow m(n - 1) - (n - 1) = \dfrac{n}{n - 1} \times (n - 1) - (n - 1) \\[1em] = n - (n - 1) \\[1em] = n - n + 1 \\[1em] = 1.

Hence, m(n - 1) - (n - 1) = 1.

Question 12

If m = 153 and n=143\sqrt[3]{15} \text{ and } n = \sqrt[3]{14}, find the value of m - n - 1m2+mn+n2\dfrac{1}{m^2 + mn + n^2}.

Answer

Given,

⇒m=153 and n=143⇒m=(15)13 and n=(14)13\Rightarrow m = \sqrt[3]{15} \text{ and } n = \sqrt[3]{14} \\[1em] \Rightarrow m = (15)^{\dfrac{1}{3}} \text{ and } n = (14)^{\dfrac{1}{3}}

Cubing both sides, we get :

⇒m3=[(15)13]3 and n3=[(14)13]3⇒m3=(15)13×3 and n3=(14)13×3⇒m3=15 and n3=14.\Rightarrow m^3 = [(15)^{\dfrac{1}{3}}]^3 \text{ and } n^3 = [(14)^{\dfrac{1}{3}}]^3 \\[1em] \Rightarrow m^3 = (15)^{\dfrac{1}{3} \times 3} \text{ and } n^3 = (14)^{\dfrac{1}{3} \times 3} \\[1em] \Rightarrow m^3 = 15 \text{ and } n^3 = 14.

Simplifying the expression m−n−1m2+mn+n2m - n - \dfrac{1}{m^2 + mn + n^2}, we get :

⇒m(m2+mn+n2)−n(m2+mn+n2)−1m2+mn+n2⇒m3+m2n+mn2−nm2−mn2−n3−1m2+mn+n2⇒m3−n3−1m2+mn+n2\Rightarrow \dfrac{m(m^2 + mn + n^2) - n(m^2 + mn + n^2) - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 + m^2n + mn^2 - nm^2 - mn^2 - n^3 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 - n^3 - 1}{m^2 + mn + n^2} \\[1em]

Substituting value of m3 and n3 in above equation, we get :

⇒15−14−1m2+mn+n2⇒1−1m2+mn+n2⇒0m2+mn+n2⇒0.\Rightarrow \dfrac{15 - 14 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{1 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{0}{m^2 + mn + n^2} \\[1em] \Rightarrow 0.

Hence, m−n−1m2+mn+n2=0m - n - \dfrac{1}{m^2 + mn + n^2} = 0

Question 13

Evaluate :

(xqxr)1qr×(xrxp)1rp×(xpxq)1pq\Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}}

Answer

Simplifying the expression :

⇒(xqxr)1qr×(xrxp)1rp×(xpxq)1pq=(xq−r)1qr×(xr−p)1rp×(xp−q)1pq=(x)q−rqr×(x)r−prp×(x)p−qpq=x(q−rqr+r−prp+p−qpq)=x(p(q−r)+q(r−p)+r(p−q)pqr)=x(pq−pr+qr−qp+rp−rqpqr)=x(0pqr)=x0=1.\Rightarrow \Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}} \\[1em] = (x^{q - r})^{\dfrac{1}{qr}} \times (x^{r - p})^{\dfrac{1}{rp}} \times (x^{p - q})^{\dfrac{1}{pq}} \\[1em] = (x)^{\dfrac{q - r}{qr}} \times (x)^{\dfrac{r - p}{rp}} \times (x)^{\dfrac{p - q}{pq}} \\[1em] = x^{\Big(\dfrac{q - r}{qr} + \dfrac{r - p}{rp} + \dfrac{p - q}{pq}\Big)} \\[1em] = x^{\Big(\dfrac{p(q - r) + q(r - p) + r(p - q)}{pqr}\Big)} \\[1em] = x^{\Big(\dfrac{pq - pr + qr - qp + rp - rq}{pqr}\Big)} \\[1em] = x^{\Big(\dfrac{0}{pqr}\Big)} \\[1em] = x^0 \\[1em] = 1.

Hence, (xqxr)1qr×(xrxp)1rp×(xpxq)1pq=1\Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}} = 1.

Question 14(i)

Prove that :

a−1a−1+b−1+a−1a−1−b−1=2b2b2−a2\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}

Answer

To prove:

a−1a−1+b−1+a−1a−1−b−1=2b2b2−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. of the above equation, we get :

⇒a−1a−1+b−1+a−1a−1−b−1=1a1a+1b+1a1a−1b=1ab+aab+1ab−aab=aba(b+a)+aba(b−a)=bb+a+bb−a=b(b−a)+b(b+a)(b+a)(b−a)=b2−ba+b2+abb2−a2=2b2b2−a2.\Rightarrow \dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} + \dfrac{1}{b}} + \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} - \dfrac{1}{b}} \\[1em] = \dfrac{\dfrac{1}{a}}{\dfrac{b + a}{ab}} + \dfrac{\dfrac{1}{a}}{\dfrac{b - a}{ab}} \\[1em] = \dfrac{ab}{a(b + a)} + \dfrac{ab}{a(b - a)} \\[1em] = \dfrac{b}{b + a} + \dfrac{b}{b - a} \\[1em] = \dfrac{b(b - a) + b(b + a)}{(b + a)(b - a)} \\[1em] = \dfrac{b^2 - ba + b^2 + ab}{b^2 - a^2} \\[1em] = \dfrac{2b^2}{b^2 - a^2}.

Since, L.H.S. = R.H.S. = 2b2b2−a2\dfrac{2b^2}{b^2 - a^2}

Hence, proved that a−1a−1+b−1+a−1a−1−b−1=2b2b2−a2\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}.

Question 14(ii)

Prove that :

a+b+ca−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

Answer

To prove:

a+b+ca−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

Solving L.H.S. of the above equation, we get :

⇒a+b+ca−1b−1+b−1c−1+c−1a−1=a+b+c1ab+1bc+1ca=a+b+cc+a+babc=abc(a+b+c)a+b+c=abc.\Rightarrow \dfrac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = \dfrac{a + b + c}{\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}} \\[1em] = \dfrac{a + b + c}{\dfrac{c + a + b}{abc}} \\[1em] = \dfrac{abc(a + b + c)}{a + b + c} \\[1em] = abc.

Since, L.H.S. = R.H.S. = abc.

Hence, proved that a+b+ca−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.

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