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Chapter 2

Exponents (Powers)

Class - 8 Concise Mathematics Selina

Exercise 2(A)

Question 1(i)

(13)−3−(12)−3\Big(\dfrac{1}{3}\Big)^{-3} - \Big(\dfrac{1}{2}\Big)^{-3} is equal to :

  1. 1

  2. 127−18\dfrac{1}{27} - \dfrac{1}{8}

  3. 19

  4. -19

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

Hence,

(13)−3−(12)−3=(31)3−(21)3=(3×3×31×1×1)−(2×2×21×1×1)=271−81=19\Big(\dfrac{1}{3}\Big)^{-3} - \Big(\dfrac{1}{2}\Big)^{-3}\\[1em] = \Big(\dfrac{3}{1}\Big)^3 - \Big(\dfrac{2}{1}\Big)^3\\[1em] = \Big(\dfrac{3 \times 3 \times 3}{1 \times 1 \times 1}\Big) - \Big(\dfrac{2 \times 2 \times 2}{1 \times 1 \times 1}\Big) \\[1em] = \dfrac{27}{1} - \dfrac{8}{1} \\[1em] = 19

Hence, option 3 is the correct option.

Question 1(ii)

(23)3×(32)6\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{3}{2}\Big)^6 is equal to:

  1. 827\dfrac{8}{27}

  2. 278\dfrac{27}{8}

  3. 49\dfrac{4}{9}

  4. 94\dfrac{9}{4}

Answer

(23)3×(32)6=(2×2×23×3×3)×(3×3×3×3×3×32×2×2×2×2×2)=(827)×(72964)=(8×72927×64)=58321728=278\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{3}{2}\Big)^6\\[1em] = \Big(\dfrac{2 \times 2 \times 2}{3 \times 3 \times 3}\Big) \times \Big(\dfrac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2 \times 2}\Big)\\[1em] = \Big(\dfrac{8}{27}\Big) \times \Big(\dfrac{729}{64}\Big)\\[1em] = \Big(\dfrac{8 \times 729}{27 \times 64}\Big)\\[1em] = \dfrac{5832}{1728} \\[1em] = \dfrac{27}{8}

Hence, option 2 is the correct option.

Question 1(iii)

80+8−1+4−18^0 + 8^{-1} + 4^{-1} is equal to:

  1. 8388\dfrac{3}{8}

  2. 1381\dfrac{3}{8}

  3. 38\dfrac{3}{8}

  4. 2232\dfrac{2}{3}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

And a0=1a^0 = 1

Hence,

80+8−1+4−1=1+18+14=11+18+148^0 + 8^{-1} + 4^{-1}\\[1em] = 1 + \dfrac{1}{8} + \dfrac{1}{4} \\[1em] = \dfrac{1}{1} + \dfrac{1}{8} + \dfrac{1}{4}

LCM of 1, 8 and 4 is 2 x 2 x 2 = 8

=1×81×8+1×18×1+1×24×2=88+18+28=8+1+28=118=138= \dfrac{1 \times 8}{1 \times 8} + \dfrac{1 \times 1}{8 \times 1} + \dfrac{1 \times 2}{4 \times 2}\\[1em] = \dfrac{8}{8} + \dfrac{1}{8} + \dfrac{2}{8}\\[1em] = \dfrac{8 + 1 + 2}{8}\\[1em] = \dfrac{11}{8}\\[1em] = 1\dfrac{3}{8}

Hence, option 2 is the correct option.

Question 1(iv)

(−5)5×(−5)−3(-5)^5 \times (-5)^{-3} is equal to:

  1. 15\dfrac{1}{5}

  2. 5

  3. -25

  4. 25

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

(−5)5×(−5)−3=(−5)5×(−15)3=(−5×−5×−5×−5×−5)×(−1×−1×−15×5×5)=−3125×(−1125)=(−3125×−1125)=(3125125)=25(-5)^5 \times (-5)^{-3}\\[1em] = (-5)^5 \times \Big(\dfrac{-1}{5}\Big)^3\\[1em] = (-5 \times -5 \times -5 \times -5 \times -5) \times \Big(\dfrac{-1 \times -1 \times -1}{5 \times 5 \times 5}\Big)\\[1em] = -3125 \times \Big(\dfrac{-1}{125}\Big)\\[1em] = \Big(\dfrac{-3125 \times -1}{125}\Big)\\[1em] = \Big(\dfrac{3125}{125}\Big)\\[1em] = 25

Hence, option 4 is the correct option.

Question 2(i)

Evaluate:

(3−1×9−1)÷3−2(3^{-1} \times 9^{-1}) ÷ 3^{-2}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

Hence,

(3−1×9−1)÷3−2=(131×191)÷(13)2=(13×19)÷(1×13×3)=(1×13×9)÷(19)=(127)÷(19)=(127)×(91)=1×927×1=927=13(3^{-1} \times 9^{-1}) ÷ 3^{-2}\\[1em] = \Big(\dfrac{1}{3}^1 \times \dfrac{1}{9}^1\Big) ÷ \Big(\dfrac{1}{3}\Big)^2\\[1em] = \Big(\dfrac{1}{3} \times \dfrac{1}{9}\Big) ÷ \Big(\dfrac{1 \times 1}{3 \times 3}\Big)\\[1em] = \Big(\dfrac{1 \times 1}{3 \times 9}\Big) ÷ \Big(\dfrac{1}{9}\Big)\\[1em] = \Big(\dfrac{1}{27}\Big) ÷ \Big(\dfrac{1}{9}\Big)\\[1em] = \Big(\dfrac{1}{27}\Big) \times \Big(\dfrac{9}{1}\Big)\\[1em] = \dfrac{1 \times 9}{27 \times 1}\\[1em] = \dfrac{9}{27}\\[1em] = \dfrac{1}{3}

Hence, (3−1×9−1)÷3−2=13(3^{-1} \times 9^{-1}) ÷ 3^{-2} = \dfrac{1}{3}

Question 2(ii)

Evaluate:

(3−1×4−1)÷6−1(3^{-1} \times 4^{-1}) ÷ 6^{-1}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

Hence,

(3−1×4−1)÷6−1=(131×141)÷(16)1=(1×13×4)÷(16)=(112)÷(16)=(112)×(61)=(1×612×1)=612=12(3^{-1} \times 4^{-1}) ÷ 6^{-1}\\[1em] = \Big(\dfrac{1}{3}^1 \times \dfrac{1}{4}^1\Big) ÷ \Big(\dfrac{1}{6}\Big)^1\\[1em] = \Big(\dfrac{1 \times 1}{3 \times 4}\Big) ÷ \Big(\dfrac{1}{6}\Big)\\[1em] = \Big(\dfrac{1}{12}\Big) ÷ \Big(\dfrac{1}{6}\Big)\\[1em] = \Big(\dfrac{1}{12}\Big) \times \Big(\dfrac{6}{1}\Big)\\[1em] = \Big(\dfrac{1 \times 6}{12 \times 1}\Big)\\[1em] = \dfrac{6}{12}\\[1em] = \dfrac{1}{2}

Hence, (3−1×4−1)÷6−1=12(3^{-1} \times 4^{-1}) ÷ 6^{-1} = \dfrac{1}{2}

Question 2(iii)

Evaluate:

(2−1+3−1)3(2^{-1} + 3^{-1})^3

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

Hence,

(2−1+3−1)3=(121+131)3=(12+13)3(2^{-1} + 3^{-1})^3\\[1em] = \Big(\dfrac{1}{2}^1 + \dfrac{1}{3}^1\Big)^3\\[1em] = \Big(\dfrac{1}{2} + \dfrac{1}{3}\Big)^3

LCM of 2 and 3 is 2 x 3 = 6

=(1×32×3+1×23×2)3=(36+26)3=(3+26)3=(56)3=(5×5×56×6×6)=125216= \Big(\dfrac{1 \times 3}{2 \times 3} + \dfrac{1 \times 2}{3 \times 2}\Big)^3\\[1em] = \Big(\dfrac{3}{6} + \dfrac{2}{6}\Big)^3\\[1em] = \Big(\dfrac{3 + 2}{6}\Big)^3\\[1em] = \Big(\dfrac{5}{6}\Big)^3\\[1em] = \Big(\dfrac{5 \times 5 \times 5}{6 \times 6 \times 6}\Big)\\[1em] = \dfrac{125}{216}

Hence,(2−1+3−1)3=125216(2^{-1} + 3^{-1})^3 = \dfrac{125}{216}

Question 2(iv)

Evaluate:

(3−1÷4−1)2(3^{-1} ÷ 4^{-1})^2

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

Hence,

(3−1÷4−1)2=(131÷141)2=(13÷14)2=(13×41)2=(1×43×1)2=(43)2=(4×43×3)=(169)=1(79)(3^{-1} ÷ 4^{-1})^2\\[1em] = \Big(\dfrac{1}{3}^1 ÷ \dfrac{1}{4}^1\Big)^2\\[1em] = \Big(\dfrac{1}{3} ÷ \dfrac{1}{4}\Big)^2\\[1em] = \Big(\dfrac{1}{3} \times \dfrac{4}{1}\Big)^2\\[1em] = \Big(\dfrac{1 \times 4}{3 \times 1}\Big)^2\\[1em] = \Big(\dfrac{4}{3}\Big)^2\\[1em] = \Big(\dfrac{4 \times 4}{3 \times 3}\Big)\\[1em] = \Big(\dfrac{16}{9}\Big)\\[1em] = 1\Big(\dfrac{7}{9}\Big)

Hence, (3−1÷4−1)2=1(79)(3^{-1} ÷ 4^{-1})^2 = 1\Big(\dfrac{7}{9}\Big)

Question 2(v)

Evaluate:

(22+32)×(12)2(2^2 + 3^2) \times \Big(\dfrac{1}{2}\Big)^2

Answer

(22+32)×(12)2=(2×2+3×3)×(1×12×2)=(4+9)×(14)=(13)×(14)=(13×14)=(134)=3(14)(2^2 + 3^2) \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = (2 \times 2 + 3 \times 3) \times \Big(\dfrac{1 \times 1}{2 \times 2}\Big)\\[1em] = (4 + 9) \times \Big(\dfrac{1}{4}\Big)\\[1em] = (13) \times \Big(\dfrac{1}{4}\Big)\\[1em] = \Big(\dfrac{13 \times 1}{4}\Big)\\[1em] = \Big(\dfrac{13}{4}\Big)\\[1em] = 3\Big(\dfrac{1}{4}\Big)

Hence, (22+32)×(12)2=3(14)(2^2 + 3^2) \times \Big(\dfrac{1}{2}\Big)^2 = 3\Big(\dfrac{1}{4}\Big)

Question 2(vi)

Evaluate:

(52−32)×(23)−3(5^2 - 3^2) \times \Big(\dfrac{2}{3}\Big)^{-3}

Answer

(52−32)×(23)−3=(5×5−3×3)×(32)3=(25−9)×(3×3×32×2×2)=(16)×(278)=(16×278)=(4328)=54(5^2 - 3^2) \times \Big(\dfrac{2}{3}\Big)^{-3}\\[1em] = (5 \times 5 - 3 \times 3) \times \Big(\dfrac{3}{2}\Big)^3\\[1em] = (25 - 9) \times \Big(\dfrac{3 \times 3 \times 3}{2 \times 2 \times 2}\Big)\\[1em] = (16) \times \Big(\dfrac{27}{8}\Big)\\[1em] = \Big(\dfrac{16 \times 27}{8}\Big)\\[1em] = \Big(\dfrac{432}{8}\Big)\\[1em] = 54

Hence, (52−32)×(23)−3=54(5^2 - 3^2) \times \Big(\dfrac{2}{3}\Big)^{-3} = 54

Question 2(vii)

Evaluate:

[(14)−3−(13)−3]÷(16)−3\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

[(14)−3−(13)−3]÷(16)−3=[(41)3−(31)3]÷(61)3=[(4×4×4)−(3×3×3)]÷(6×6×6)=[64−27]÷(216)=[37]÷(216)=[37]×(1216)=(37×1216)=(37216)\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3}\\[1em] = \Big[\Big(\dfrac{4}{1}\Big)^3 - \Big(\dfrac{3}{1}\Big)^3\Big] ÷ \Big(\dfrac{6}{1}\Big)^3\\[1em] = [(4 \times 4 \times 4) - (3 \times 3 \times 3)] ÷ (6 \times 6 \times 6)\\[1em] = [64 - 27] ÷ (216)\\[1em] = [37] ÷ (216)\\[1em] = [37] \times \Big(\dfrac{1}{216}\Big)\\[1em] = \Big(\dfrac{37 \times 1}{216}\Big)\\[1em] = \Big(\dfrac{37}{216}\Big)

Hence, [(14)−3−(13)−3]÷(16)−3=37216\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3} = \dfrac{37}{216}.

Question 2(viii)

Evaluate:

[(−34)−2]2\Big[\Big(-\dfrac{3}{4}\Big)^{-2}\Big]^2

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

[(−34)−2]2=[(−43)2]2=[(−4×(−4)3×3)]2=[(169)]2=(16×169×9)=(25681)=31381\Big[\Big(-\dfrac{3}{4}\Big)^{-2}\Big]^2\\[1em] = \Big[\Big(\dfrac{-4}{3}\Big)^2\Big]^2\\[1em] = \Big[\Big(\dfrac{-4 \times (-4)}{3 \times 3}\Big)\Big]^2\\[1em] = \Big[\Big(\dfrac{16}{9}\Big)\Big]^2\\[1em] = \Big(\dfrac{16 \times 16}{9 \times 9}\Big)\\[1em] = \Big(\dfrac{256}{81}\Big)\\[1em] = 3\dfrac{13}{81}

Hence, [(−34)−2]2=31381\Big[\Big(-\dfrac{3}{4}\Big)^{-2}\Big]^2 = 3\dfrac{13}{81}

Question 2(ix)

Evaluate:

[(35)−2]−2\Big[\Big(\dfrac{3}{5}\Big)^{-2}\Big]^{-2}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

[(35)−2]−2=[(53)2]−2=[(5×53×3)]−2=(259)−2=(925)2=9×925×25=81625\Big[\Big(\dfrac{3}{5}\Big)^{-2}\Big]^{-2}\\[1em] = \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-2}\\[1em] = \Big[\Big(\dfrac{5 \times 5}{3 \times 3}\Big)\Big]^{-2}\\[1em] = \Big(\dfrac{25}{9}\Big)^{-2}\\[1em] = \Big(\dfrac{9}{25}\Big)^2\\[1em] = \dfrac{9 \times 9}{25 \times 25}\\[1em] = \dfrac{81}{625}

Hence, [(35)−2]−2=81625\Big[\Big(\dfrac{3}{5}\Big)^{-2}\Big]^{-2} = \dfrac{81}{625}

Question 2(x)

Evaluate:

(5−1×3−1)÷6−1(5^{-1} \times 3^{-1}) ÷ 6^{-1}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

(5−1×3−1)÷6−1=(151×131)÷161=(1×15×3)÷161=(115)÷16=(115)×61=(1×615×1)=615=25(5^{-1} \times 3^{-1}) ÷ 6^{-1}\\[1em] = \Big(\dfrac{1}{5}^1 \times \dfrac{1}{3}^1\Big) ÷ \dfrac{1}{6}^1\\[1em] = \Big(\dfrac{1 \times 1}{5 \times 3}\Big) ÷ \dfrac{1}{6}^1\\[1em] = \Big(\dfrac{1}{15}\Big) ÷ \dfrac{1}{6}\\[1em] = \Big(\dfrac{1}{15}\Big) \times \dfrac{6}{1}\\[1em] = \Big(\dfrac{1 \times 6}{15 \times 1}\Big)\\[1em] = \dfrac{6}{15}\\[1em] = \dfrac{2}{5}

Hence, (5−1×3−1)÷6−1=25(5^{-1} \times 3^{-1}) ÷ 6^{-1} = \dfrac{2}{5}

Question 3

If 1125 = 3m x 5n, find m and n.

Answer

As we know by Prime factorization of 1125, 1125 = 3 x 3 x 5 x 5 x 5

1125 = 32 x 53

3m x 5n = 32 x 53

Hence, m = 2 and n = 3.

Question 4

Find x, if 9×3x=(27)2x−39 \times 3^x = (27)^{2x-3}

Answer

9×3x=(27)2x−3⇒32×3x=(33)2x−3⇒32×3x=(3)3(2x−3)⇒32×3x=(3)6x−9⇒32=(3)6x−9÷3x⇒32=(3)6x−9−x [∵am÷an=am−n]⇒32=(3)5x−9⇒32=35x÷39⇒32×39=35x⇒32+9=35x [∵am×an=am+n]⇒311=35x⇒5x=11 [∵am=an⟹m=n]⇒x=115⇒x=2159 \times 3^x = (27)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3^3)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{3(2x - 3)}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{6x - 9}\\[1em] \Rightarrow 3^2 = (3)^{6x - 9} ÷ 3^x\\[1em] \Rightarrow 3^2 = (3)^{6x - 9 - x} \space [∵a^m ÷ a^n = a^{m-n}] \\[1em] \Rightarrow 3^2 = (3)^{5x - 9}\\[1em] \Rightarrow 3^2 = 3^{5x} ÷ 3^9\\[1em] \Rightarrow 3^2 \times 3^9 = 3^{5x}\\[1em] \Rightarrow 3^{2 + 9} = 3^{5x} \space [∵a^m \times a^n = a^{m+n}] \\[1em] \Rightarrow 3^{11} = 3^{5x}\\[1em] \Rightarrow 5x = 11 \space [∵a^m = a^n ⟹m = n] \\[1em] \Rightarrow x = \dfrac{11}{5} \\[1em] \Rightarrow x = 2\dfrac{1}{5}

If 9×3x=(27)2x−39 \times 3^x = (27)^{2x-3}, then x=215x = 2\dfrac{1}{5}.

Exercise 2(B)

Question 1(i)

If x=3mx = 3^m and y=3m+2,xyy = 3^{m + 2}, \dfrac{x}{y} is:

  1. 9

  2. 19\dfrac{1}{9}

  3. 6

  4. 9

Answer

x=3mx = 3^m and y=3m+2y = 3^{m + 2}. Hence,

xy=3m3m+2=3m3m×32=132=19\dfrac{x}{y} = \dfrac{3^m}{3^{m + 2}}\\[1em] = \dfrac{3^m}{3^m \times 3^2}\\[1em] = \dfrac{1}{3^2}\\[1em] = \dfrac{1}{9}

Hence, option 2 is the correct option.

Question 1(ii)

[x−23y−1]−1\Big[\dfrac{x^{-2}}{3y^{-1}}\Big]^{-1} is equal to:

  1. 3x2y\dfrac{3x^2}{y}

  2. x23y\dfrac{x^2}{3y}

  3. y3x2\dfrac{y}{3x^2}

  4. 3yx2\dfrac{3y}{x^2}

Answer

As we know, for any non-zero rational number a

a−n=1ana^{-n} = \dfrac{1}{a^n} and an=1a−na^{n} = \dfrac{1}{a^{-n}}.

[x−23y−1]−1=[y13x2]−1=[3x2y]\Big[\dfrac{x^{-2}}{3y^{-1}}\Big]^{-1}\\[1em] = \Big[\dfrac{y^1}{3x^2}\Big]^{-1}\\[1em] = \Big[\dfrac{3x^2}{y}\Big]

Hence, option 1 is the correct option.

Question 1(iii)

If [45]−3×[45]−5=[45]3x−2\Big[\dfrac{4}{5}\Big]^{-3} \times \Big[\dfrac{4}{5}\Big]^{-5} = \Big[\dfrac{4}{5}\Big]^{3x - 2}, the value of x is:

  1. 2

  2. 12\dfrac{1}{2}

  3. -2

  4. −12-\dfrac{1}{2}

Answer

According to product property,

am×an=am+na^m \times a^n = a^{m + n}

[45]−3×[45]−5=[45]3x−2⇒[45]−3+(−5)=[45]3x−2⇒[45]−8=[45]3x−2\Big[\dfrac{4}{5}\Big]^{-3} \times \Big[\dfrac{4}{5}\Big]^{-5} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em] ⇒ \Big[\dfrac{4}{5}\Big]^{-3 + (-5)} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em] ⇒ \Big[\dfrac{4}{5}\Big]^{-8} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em]

Using am=an⇒m=na^m = a^n \Rightarrow m = n

⇒−8=3x−2⇒−8+2=3x⇒−6=3x⇒x=−63⇒x=−2\Rightarrow -8 = 3x - 2\\[1em] \Rightarrow -8 + 2 = 3x\\[1em] \Rightarrow -6 = 3x\\[1em] \Rightarrow x = \dfrac{-6}{3}\\[1em] \Rightarrow x = -2

Hence, option 3 is the correct option.

Question 1(iv)

If [mn]x−1=[nm]x−5\Big[\dfrac{m}{n}\Big]^{x-1} = \Big[\dfrac{n}{m}\Big]^{x-5}, the value of x is:

  1. 3

  2. -3

  3. 13\dfrac{1}{3}

  4. −13-\dfrac{1}{3}

Answer

[mn]x−1=[nm]x−5⇒[mn]x−1=[mn]5−x\Big[\dfrac{m}{n}\Big]^{x-1} = \Big[\dfrac{n}{m}\Big]^{x-5}\\[1em] \Rightarrow \Big[\dfrac{m}{n}\Big]^{x-1} = \Big[\dfrac{m}{n}\Big]^{5-x}

According to the property,am=an⇒m=na^m = a^n \Rightarrow m = n

⇒(x−1)=(5−x)⇒x+x=5+1⇒2x=6⇒x=62⇒x=3\Rightarrow (x - 1) = (5 - x)\\[1em] \Rightarrow x + x = 5 + 1\\[1em] \Rightarrow 2x = 6\\[1em] \Rightarrow x = \dfrac{6}{2}\\[1em] \Rightarrow x = 3

Hence, option 1 is the correct option.

Question 1(v)

[17]−3×7−1×[149]\Big[\dfrac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\dfrac{1}{49}\Big] is equal to:

  1. -1

  2. 17\dfrac{1}{7}

  3. -7

  4. 1

Answer

[17]−3×7−1×[149]=[71]3×[17]1×[149]=[7×7×71×1×1]×[17]×[149]=[3431]×[17]×[149]=[343×1×11×7×49]=[343343]=1\Big[\dfrac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{7}{1}\Big]^3 \times \Big[\dfrac{1}{7}\Big]^1 \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{7 \times 7 \times 7}{1 \times 1 \times 1}\Big] \times \Big[\dfrac{1}{7}\Big] \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{343}{1}\Big] \times \Big[\dfrac{1}{7}\Big] \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{343 \times 1 \times 1}{1 \times 7 \times 49}\Big]\\[1em] = \Big[\dfrac{343}{343}\Big]\\[1em] = 1

Hence, option 4 is the correct option.

Question 2(i)

Compute:

18×30×53×221^8 \times 3^0 \times 5^3 \times 2^2

Answer

18×30×53×22=(1×1×1×1×1×1×1×1)×1×(5×5×5)×(2×2)=1×1×125×4=5001^8 \times 3^0 \times 5^3 \times 2^2\\[1em] = (1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1) \times 1 \times (5 \times 5 \times 5) \times (2 \times 2)\\[1em] = 1 \times 1 \times 125 \times 4 \\[1em] = 500

Hence, 18×30×53×22=5001^8 \times 3^0 \times 5^3 \times 2^2 = 500

Question 2(ii)

Compute:

(47)2×(4−3)4(4^7)^2 \times (4^{-3})^4

Answer

(47)2×(4−3)4=(4)7×2×[14]3×4=(4)14×[14]12=(4)14×(4)−12=(4)14−12=42=16(4^7)^2 \times (4^{-3})^4\\[1em] = (4)^{7 \times 2} \times \Big[\dfrac{1}{4}\Big]^{3 \times 4}\\[1em] = (4)^{14} \times \Big[\dfrac{1}{4}\Big]^{12}\\[1em] = (4)^{14} \times (4)^{-12}\\[1em] = (4)^{14-12} \\[1em] = 4^2 \\[1em] = 16

Hence, (47)2×(4−3)4=16(4^7)^2 \times (4^{-3})^4 = 16

Question 2(iii)

Compute:

(2−9÷2−11)3(2^{-9} ÷ 2^{-11})^3

Answer

According to quotient property,

am÷an=am−na^m ÷ a^n = a^{m - n}

(2−9÷2−11)3=(2−9−(−11))3=(2−9+11)3=(22)3=(2)2×3=26=2×2×2×2×2×2=64(2^{-9} ÷ 2^{-11})^3\\[1em] = (2^{-9 - (-11)})^3\\[1em] = (2^{-9 + 11})^3\\[1em] = (2^{2})^3\\[1em] = (2)^{2\times3}\\[1em] = 2^{6}\\[1em] = 2 \times 2 \times 2 \times 2 \times 2 \times 2\\[1em] = 64

Hence, (2−9÷2−11)3=64(2^{-9} ÷ 2^{-11})^3 = 64

Question 2(iv)

Compute:

[23]−4×[278]−2\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2}

Answer

[23]−4×[278]−2=[32]4×[827]2=[3424]×[2333]2=[3424]×[2636]=[2636]×[2−43−4]=[26−436−4]=[2232]=[49]\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2}\\[1em] = \Big[\dfrac{3}{2}\Big]^4 \times \Big[\dfrac{8}{27}\Big]^2\\[1em] = \Big[\dfrac{3^4}{2^4}\Big] \times \Big[\dfrac{2^3}{3^3}\Big]^2\\[1em] = \Big[\dfrac{3^4}{2^4}\Big] \times \Big[\dfrac{2^6}{3^6}\Big]\\[1em] = \Big[\dfrac{2^6}{3^6}\Big] \times \Big[\dfrac{2^{-4}}{3^{-4}}\Big] \\[1em] = \Big[\dfrac{2^{6-4}}{3^{6-4}}\Big]\\[1em] = \Big[\dfrac{2^2}{3^2}\Big]\\[1em] = \Big[\dfrac{4}{9}\Big]

Hence, [23]−4×[278]−2=[49]\Big[\dfrac{2}{3}\Big]^{-4} \times \Big[\dfrac{27}{8}\Big]^{-2} = \Big[\dfrac{4}{9}\Big]

Question 2(v)

Compute:

[5628]0÷[25]3×1625\Big[\dfrac{56}{28}\Big]^0 ÷ \Big[\dfrac{2}{5}\Big]^3 \times \dfrac{16}{25}

Answer

[5628]0÷[25]3×1625=1÷[25]3×1625 [∵a0=1]=1×[52]3×1625=[52]3×[2452]=[5323]×[2452]=53×2−3×24×5−2=(5)3−2×(2)4−3=(5)1×(2)1=10\Big[\dfrac{56}{28}\Big]^0 ÷ \Big[\dfrac{2}{5}\Big]^3 \times \dfrac{16}{25}\\[1em] = 1 ÷ \Big[\dfrac{2}{5}\Big]^3 \times \dfrac{16}{25} \space [∵ a^0 = 1] \\[1em] = 1 \times \Big[\dfrac{5}{2}\Big]^3 \times \dfrac{16}{25}\\[1em] = \Big[\dfrac{5}{2}\Big]^3 \times \Big[\dfrac{2^4}{5^2}\Big]\\[1em] = \Big[\dfrac{5^3}{2^3}\Big] \times \Big[\dfrac{2^4}{5^2}\Big]\\[1em] = 5^3 \times 2^{-3} \times 2^4 \times 5^{-2} \\[1em] = (5)^{3-2} \times (2)^{4-3}\\[1em] = (5)^{1} \times (2)^{1}\\[1em] = 10

[5628]0÷[25]3×1625=10\Big[\dfrac{56}{28}\Big]^0 ÷ \Big[\dfrac{2}{5}\Big]^3 \times \dfrac{16}{25} = 10

Question 2(vi)

Compute:

(12)−2×33(12)^{-2} \times 3^3

Answer

(12)−2×33=(22×3)−2×33=(22)−2×(3)−2×33=(2)−4×(3)−2+3=[12]4×31=116×31=316(12)^{-2} \times 3^3\\[1em] = (2^2 \times 3)^{-2} \times 3^3\\[1em] = (2^2)^{-2} \times (3)^{-2} \times 3^3\\[1em] = (2)^{-4} \times (3)^{-2+3}\\[1em] = \Big[\dfrac{1}{2}\Big]^4 \times 3^1\\[1em] = \dfrac{1}{16} \times 3^1\\[1em] = \dfrac{3}{16}

(12)−2×33=316(12)^{-2} \times 3^3 = \dfrac{3}{16}

Question 2(vii)

Compute:

(−5)4×(−5)6÷(−5)9(-5)^4 \times (-5)^6 ÷ (-5)^9

Answer

(−5)4×(−5)6÷(−5)9=(−5)4+6−9=(−5)1=−5(-5)^4 \times (-5)^6 ÷ (-5)^9\\[1em] = (-5)^{4+6-9}\\[1em] = (-5)^1\\[1em] = -5

(−5)4×(−5)6÷(−5)9=−5(-5)^4 \times (-5)^6 ÷ (-5)^9 = -5

Question 2(viii)

Compute:

[−13]4÷[−13]8×[−13]5\Big[-\dfrac{1}{3}\Big]^4 ÷ \Big[-\dfrac{1}{3}\Big]^8 \times \Big[-\dfrac{1}{3}\Big]^5

Answer

[−13]4÷[−13]8×[−13]5=[−13]4−8+5=[−13]1=−13\Big[-\dfrac{1}{3}\Big]^4 ÷ \Big[-\dfrac{1}{3}\Big]^8 \times \Big[-\dfrac{1}{3}\Big]^5\\[1em] = \Big[-\dfrac{1}{3}\Big]^{4-8+5}\\[1em] = \Big[-\dfrac{1}{3}\Big]^1\\[1em] = -\dfrac{1}{3}

[−13]4÷[−13]8×[−13]5=−13\Big[-\dfrac{1}{3}\Big]^4 ÷ \Big[-\dfrac{1}{3}\Big]^8 \times \Big[-\dfrac{1}{3}\Big]^5 = -\dfrac{1}{3}

Question 2(ix)

Compute:

90×4−1÷2−49^0 \times 4^{-1} ÷ 2^{-4}

Answer

90×4−1÷2−4=1×4−1÷2−4 [∵a0=1]=1×2−2÷2−4=1×2−2−(−4)=1×2−2+4=1×22=1×4=49^0 \times 4^{-1} ÷ 2^{-4}\\[1em] = 1 \times 4^{-1} ÷ 2^{-4} \space [∵a^0 = 1] \\[1em] = 1 \times 2^{-2} ÷ 2^{-4}\\[1em] = 1 \times 2^{-2-(-4)}\\[1em] = 1 \times 2^{-2+4}\\[1em] = 1 \times 2^{2}\\[1em] = 1 \times 4\\[1em] = 4

90×4−1÷2−4=49^0 \times 4^{-1} ÷ 2^{-4} = 4

Question 2(x)

Compute:

(625)−34(625)^{-\dfrac{3}{4}}

Answer

(625)−34=(5)−4×34=(5)−3=[15]3=1125(625)^{-\dfrac{3}{4}}\\[1em] = (5)^{-4 \times \dfrac{3}{4}}\\[1em] = (5)^{-3}\\[1em] = \Big[\dfrac{1}{5}\Big]^3\\[1em] = \dfrac{1}{125}

(625)−34=1125(625)^{-\dfrac{3}{4}} = \dfrac{1}{125}

Question 2(xi)

Compute:

[2764]−23\Big[\dfrac{27}{64}\Big]^{-\dfrac{2}{3}}

Answer

[2764]−23=[3343]−23=[34]−3×23=[34]−2=[43]2=[4×43×3]=169=179\Big[\dfrac{27}{64}\Big]^{-\dfrac{2}{3}}\\[1em] = \Big[\dfrac{3^3}{4^3}\Big]^{-\dfrac{2}{3}}\\[1em] = \Big[\dfrac{3}{4}\Big]^{-3 \times \dfrac{2}{3}}\\[1em] = \Big[\dfrac{3}{4}\Big]^{-2}\\[1em] = \Big[\dfrac{4}{3}\Big]^2\\[1em] = \Big[\dfrac{4 \times 4}{3 \times 3}\Big]\\[1em] = \dfrac{16}{9}\\[1em] = 1\dfrac{7}{9}

[2764]−23=179\big[\dfrac{27}{64}\Big]^{-\dfrac{2}{3}} = 1\dfrac{7}{9}

Question 2(xii)

Compute:

[132]−25\Big[\dfrac{1}{32}\Big]^{-\dfrac{2}{5}}

Answer

[132]−25=[1525]−25=[12]−5×25=[12]−2=[21]2=[2×21×1]=41=4\Big[\dfrac{1}{32}\Big]^{-\dfrac{2}{5}}\\[1em] = \Big[\dfrac{1^5}{2^5}\Big]^{-\dfrac{2}{5}}\\[1em] = \Big[\dfrac{1}{2}\Big]^{-5 \times \dfrac{2}{5}}\\[1em] = \Big[\dfrac{1}{2}\Big]^{-2}\\[1em] = \Big[\dfrac{2}{1}\Big]^2\\[1em] = \Big[\dfrac{2 \times 2}{1 \times 1}\Big]\\[1em] = \dfrac{4}{1}\\[1em] = 4

[132]−25=4\big[\dfrac{1}{32}\Big]^{-\dfrac{2}{5}} = 4

Question 2(xiii)

Compute:

(125)−23÷(8)23(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}}

Answer

(125)−23÷(8)23=(53)−23÷(23)23=(5)−3×23÷(2)3×23=(5)−2÷(2)2=(5)−2×(12)2=(5)−2×14=(15)2×14=125×14=1100(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}}\\[1em] = (5^3)^{-\dfrac{2}{3}} ÷ (2^3)^{\dfrac{2}{3}}\\[1em] = (5)^{-3\times\dfrac{2}{3}} ÷ (2)^{3\times\dfrac{2}{3}}\\[1em] = (5)^{-2} ÷ (2)^{2}\\[1em] = (5)^{-2} \times \Big(\dfrac{1}{2}\Big)^{2}\\[1em] = (5)^{-2} \times \dfrac{1}{4}\\[1em] = \Big(\dfrac{1}{5}\Big)^2 \times \dfrac{1}{4}\\[1em] = \dfrac{1}{25} \times \dfrac{1}{4}\\[1em] = \dfrac{1}{100}

(125)−23÷(8)23=1100(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}} = \dfrac{1}{100}

Question 2(xiv)

Compute:

(243)25÷(32)−25(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}

Answer

(243)25÷(32)−25=(35)25÷(25)−25=(3)5×25÷(2)−5×25=(3)2÷(2)−2=(3)2×(12)−2=(3)2×212=(9)×(4)=36(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}\\[1em] = (3^5)^{\dfrac{2}{5}} ÷ (2^5)^{-\dfrac{2}{5}}\\[1em] = (3)^{5\times\dfrac{2}{5}} ÷ (2)^{-5\times\dfrac{2}{5}}\\[1em] = (3)^{2} ÷ (2)^{-2}\\[1em] = (3)^{2} \times \Big(\dfrac{1}{2}\Big)^{-2}\\[1em] = (3)^{2} \times \dfrac{2}{1}^{2}\\[1em] = (9) \times (4)\\[1em] = 36

(243)25÷(32)−25=36(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}} = 36

Question 2(xv)

Compute:

(−3)4−(34)0×(−2)5÷(64)23(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}}

Answer

As we know for any rational number,

a0=1a^0 = 1

(−3)4−(34)0×(−2)5÷(64)23=(−3×−3×−3×−3)−1×(−2)5÷(43)23=81−(−2×−2×−2×−2×−2)÷(4)3×23=81−(−32)÷(4)2=81+32÷(4×4)=81+32÷16=81+2=83(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}}\\[1em] =(-3 \times -3 \times -3 \times -3) - 1 \times (-2)^5 ÷ (4^3)^{\dfrac{2}{3}}\\[1em] = 81 - (-2 \times -2 \times -2 \times -2 \times -2) ÷ (4)^{3\times\dfrac{2}{3}}\\[1em] = 81 - (-32) ÷ (4)^{2}\\[1em] = 81 + 32 ÷ (4 \times 4)\\[1em] = 81 + 32 ÷ 16\\[1em] = 81 + 2\\[1em] = 83

(−3)4−(34)0×(−2)5÷(64)23=83(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}} = 83

Question 2(xvi)

Compute:

(27)23÷(8116)−14(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}}

Answer

(27)23÷(8116)−14=(33)23÷(3424)−14=(3)3×23÷(32)−4×14=(3)2÷(32)−1=(3)2÷(23)=(3)2×(32)=9×(32)=9×32=272=1312(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}}\\[1em] = (3^3)^{\dfrac{2}{3}} ÷ \Big(\dfrac{3^4}{2^4}\Big)^{-\dfrac{1}{4}}\\[1em] = (3)^{3\times \dfrac{2}{3}} ÷ \Big(\dfrac{3}{2}\Big)^{-4\times\dfrac{1}{4}}\\[1em] = (3)^2 ÷ \Big(\dfrac{3}{2}\Big)^{-1}\\[1em] = (3)^2 ÷ \Big(\dfrac{2}{3}\Big)\\[1em] = (3)^2 \times \Big(\dfrac{3}{2}\Big)\\[1em] = 9 \times \Big(\dfrac{3}{2}\Big)\\[1em] = \dfrac{9 \times 3}{2}\\[1em] = \dfrac{27}{2}\\[1em] = 13\dfrac{1}{2}

(27)23÷(8116)−14=1312(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}} = 13\dfrac{1}{2}

Question 3(i)

Simplify:

843+2532−(127)−238^{\dfrac{4}{3}} + 25^{\dfrac{3}{2}} - \Big(\dfrac{1}{27}\Big)^{-\dfrac{2}{3}}

Answer

843+2532−(127)−23=(23)43+(52)32−(1333)−23=(2)3×43+(5)2×32−(13)−3×23=(2)4+(5)3−(13)−2=(2)4+(5)3−(31)2=16+125−9=1328^{\dfrac{4}{3}} + 25^{\dfrac{3}{2}} - \Big(\dfrac{1}{27}\Big)^{-\dfrac{2}{3}}\\[1em] = (2^3)^{\dfrac{4}{3}} + (5^2)^{\dfrac{3}{2}} - \Big(\dfrac{1^3}{3^3}\Big)^{-\dfrac{2}{3}}\\[1em] = (2)^{3\times\dfrac{4}{3}} + (5)^{2\times\dfrac{3}{2}} - \Big(\dfrac{1}{3}\Big)^{-3\times\dfrac{2}{3}}\\[1em] = (2)^4 + (5)^3 - \Big(\dfrac{1}{3}\Big)^{-2}\\[1em] = (2)^4 + (5)^3 - \Big(\dfrac{3}{1}\Big)^2\\[1em] = 16 + 125 - 9\\[1em] = 132

843+2532−(127)−23=1328^{\dfrac{4}{3}} + 25^{\dfrac{3}{2}} - \Big(\dfrac{1}{27}\Big)^{-\dfrac{2}{3}} = 132

Question 3(ii)

Simplify:

[(64)−2]−3÷[{(−8)2}3]2[(64)^{-2}]^{-3} ÷ [\lbrace(-8)^2\rbrace^3]^2

Answer

[(64)−2]−3÷[{(−8)2}3]2=[(26)−2]−3÷[{(−23)2}3]2=[(2)−6×2]−3÷[{(−2)3×2}3]2=[(2)−12]−3÷[{(−2)6}3]2=[(2)]−12×(−3)÷[{(−2)}6×3]2=(2)36÷[(−2)18]2=(2)36÷(−2)18×2=(2)36÷(−2)36=236−236=(2)36−36×136−136=(2)0×(−1)36=1[(64)^{-2}]^{-3} ÷ [\lbrace (-8)^2\rbrace^3]^2\\[1em] = [(2^6)^{-2}]^{-3} ÷ [\lbrace (-2^3)^2\rbrace^3]^2\\[1em] = [(2)^{-6 \times 2}]^{-3} ÷ [\lbrace (-2)^{3 \times 2}\rbrace^3]^2\\[1em] = [(2)^{-12}]^{-3} ÷ [\lbrace (-2)^{6}\rbrace^3]^2\\[1em] = [(2)]^{-12 \times (-3)} ÷ [\lbrace (-2)\rbrace^{6 \times 3}]^2\\[1em] = (2)^{36} ÷ [(-2)^{18}]^2\\[1em] = (2)^{36} ÷ (-2)^{18 \times 2}\\[1em] = (2)^{36} ÷ (-2)^{36}\\[1em] = \dfrac{2^ {36}}{-2^{36}}\\[1em] = (2)^{36-36}\times\dfrac{1^ {36}}{-1^{36}}\\[1em] = (2)^0\times(-1)^{36}\\[1em] = 1

[(64)−2]−3÷[{(−8)2}3]2=1[(64)^{-2}]^{-3} ÷ [\lbrace (-8)^2\rbrace^3]^2 = 1

Question 3(iii)

Simplify:

(2−3−2−4)(2−3+2−4)(2^{-3} - 2^{-4})(2^{-3} + 2^{-4})

Answer

(2−3−2−4)(2−3+2−4)=(123−124)(123+124)=(18−116)(18+116)(2^{-3} - 2^{-4})(2^{-3} + 2^{-4})\\[1em] = \Big(\dfrac{1}{2}^3 - \dfrac{1}{2}^4\Big)\Big(\dfrac{1}{2}^3 + \dfrac{1}{2}^4\Big)\\[1em] = \Big(\dfrac{1}{8} - \dfrac{1}{16}\Big)\Big(\dfrac{1}{8} + \dfrac{1}{16}\Big)

LCM of 8 and 16 is 2 x 2 x 2 x 2 = 16

=(1×28×2−1×116×1)(1×28×2+1×116×1)=(216−116)(216+116)=(2−116)(2+116)=(116)(316)=(3256)= \Big(\dfrac{1 \times 2}{8 \times 2} - \dfrac{1 \times 1}{16 \times 1}\Big)\Big(\dfrac{1 \times 2}{8 \times 2} + \dfrac{1 \times 1}{16 \times 1}\Big)\\[1em] = \Big(\dfrac{2}{16} - \dfrac{1}{16}\Big)\Big(\dfrac{2}{16} + \dfrac{1}{16}\Big)\\[1em] = \Big(\dfrac{2-1}{16}\Big)\Big(\dfrac{2+1}{16}\Big)\\[1em] = \Big(\dfrac{1}{16}\Big)\Big(\dfrac{3}{16}\Big)\\[1em] = \Big(\dfrac{3}{256}\Big)

(2−3−2−4)(2−3+2−4)=(3256)(2^{-3} - 2^{-4})(2^{-3} + 2^{-4}) = \Big(\dfrac{3}{256}\Big)

Question 4(i)

Evaluate:

(−5)0(-5)^0

Answer

According to property of exponents, a0=1a^0 = 1

Then, (−5)0=1(-5)^0 = 1

Question 4(ii)

Evaluate:

80+40+208^0 + 4^0 + 2^0

Answer

According to property of exponents, a0=1a^0 = 1

80+40+20=1+1+1=38^0 + 4^0 + 2^0 \\[1em] = 1 + 1 + 1 \\[1em] = 3

Hence, 80+40+20=38^0 + 4^0 + 2^0 = 3

Question 4(iii)

Evaluate:

(8+4+2)0(8 + 4 + 2)^0

Answer

According to property of exponents, a0=1a^0 = 1

(8+4+2)0=(14)0=1(8 + 4 + 2)^0\\[1em] = (14)^0\\[1em] = 1

Hence, (8+4+2)0=1(8 + 4 + 2)^0 = 1

Question 4(iv)

Evaluate:

4x04x^0

Answer

According to property of exponents, a0=1a^0 = 1

4x0=4×1=44x^0\\[1em] = 4 \times 1\\[1em] = 4

Hence, 4x0=44x^0 = 4

Question 4(v)

Evaluate:

(4x)0(4x)^0

Answer

According to property of exponents, a0=1a^0 = 1

Then, (4x)0=1(4x)^0 = 1

Question 4(vi)

Evaluate:

[(103)0]5[(10^3)^0]^5

Answer

According to property of exponents,

a0=1a^0 = 1

[(103)0]5=[(10×10×10)0]5=[(1000)0]5=[1]5=[1×1×1×1×1]=1[(10^3)^0]^5\\[1em] = [(10 \times 10 \times 10)^0]^5\\[1em] = [(1000)^0]^5\\[1em] = [1]^5\\[1em] = [1 \times 1 \times 1 \times 1 \times 1]\\[1em] = 1

[(103)0]5=1[(10^3)^0]^5 = 1

Question 4(vii)

Evaluate:

(7x0)2(7x^0)^2

Answer

According to property of exponents,

a0=1a^0 = 1

(7x0)2=(7×1)2=(7)2=(7×7)=49(7x^0)^2\\[1em] = (7 \times 1)^2\\[1em] = (7)^2\\[1em] = (7\times 7)\\[1em] = 49

(7x0)2=49(7x^0)^2 = 49

Question 4(viii)

Evaluate:

90+9−1−9−2+912−9−129^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}}

Answer

90+9−1−9−2+912−9−12=1+191−192+(32)12−1912=1+191−192+(3)2×12−123212=1+191−192+(3)1−132×12=1+19−192+(3)−131=1+19−19×9+(3)−13=4+19−181−13=41+19−181−139^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3^2)^{\dfrac{1}{2}} -\dfrac{1}{9}^{\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3)^{2\times\dfrac{1}{2}} -\dfrac{1^2}{3^2}^{\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3)^1 -\dfrac{1}{3}^{2\times\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9} - \dfrac{1}{9}^2 + (3) -\dfrac{1}{3}^1\\[1em] = 1 + \dfrac{1}{9} - \dfrac{1}{9\times 9} + (3) -\dfrac{1}{3}\\[1em] = 4 + \dfrac{1}{9} - \dfrac{1}{81} -\dfrac{1}{3}\\[1em] = \dfrac{4}{1} + \dfrac{1}{9} - \dfrac{1}{81} -\dfrac{1}{3}

LCM of 1, 9, 81 and 3 is 3 x 3 x 3 x 3 =81

=4×811×81+1×99×9−1×181×1−1×273×27=32481+981−181−2781=324+9−1−2781=30581=36281= \dfrac{4 \times 81}{1 \times 81} + \dfrac{1 \times 9}{9 \times 9} - \dfrac{1 \times 1}{81 \times 1} -\dfrac{1 \times 27}{3 \times 27}\\[1em] = \dfrac{324}{81} + \dfrac{9}{81} - \dfrac{1}{81} -\dfrac{27}{81}\\[1em] = \dfrac{324 + 9 -1 -27}{81}\\[1em] = \dfrac{305}{81}\\[1em] = 3\dfrac{62}{81}

90+9−1−9−2+912−9−12=362819^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}} = 3\dfrac{62}{81}

Question 5(i)

Simplify:

a5b2a2b−3\dfrac{a^5b^2}{a^2b^{-3}}

Answer

a5b2a2b−3=a5−2b2−(−3)=a3b2+3=a3b5\dfrac{a^5b^2}{a^2b^{-3}}\\[1em] = a^{5-2}b^{2-(-3)}\\[1em] = a^{3}b^{2+3}\\[1em] = a^{3}b^{5}

a5b2a2b−3=a3b5\dfrac{a^5b^2}{a^2b^{-3}}= a^{3}b^{5}

Question 5(ii)

Simplify:

15y8÷3y315y^8 ÷ 3y^3

Answer

15y8÷3y3=153y8−3=5y515y^8 ÷ 3y^3\\[1em] = \dfrac{15}{3}y^{8-3}\\[1em] = 5y^5

15y8÷3y3=5y515y^8 ÷ 3y^3 = 5y^5

Question 5(iii)

Simplify:

x10y6÷x3y−2x^{10}y^6 ÷ x^3y^{-2}

Answer

x10y6÷x3y−2=x10−3y6−(−2)=x7y6+2=x7y8x^{10}y^6 ÷ x^3y^{-2}\\[1em] = x^{10-3}y^{6-(-2)}\\[1em] = x^{7}y^{6+2}\\[1em] = x^{7}y^{8}

x10y6÷x3y−2=x7y8x^{10}y^6 ÷ x^3y^{-2} = x^{7}y^{8}

Question 5(iv)

Simplify:

5z16÷15z−115z^{16} ÷ 15z^{-11}

Answer

5z16÷15z−11=515z16−(−11)=13z16+11=13z275z^{16} ÷ 15z^{-11}\\[1em] = \dfrac{5}{15}z^{16-(-11)}\\[1em] = \dfrac{1}{3}z^{16+11}\\[1em] = \dfrac{1}{3}z^{27}

5z16÷15z−11=13z275z^{16} ÷ 15z^{-11} = \dfrac{1}{3}z^{27}

Question 5(v)

Simplify:

(36x2)12(36x^2)^{\dfrac{1}{2}}

Answer

(36x2)12=[(6x)2]12=(6x)2×12=(6x)22=6x(36x^2)^{\dfrac{1}{2}}\\[1em] = [(6x)^2]^{\dfrac{1}{2}}\\[1em] = (6x)^{2 \times \dfrac{1}{2}}\\[1em] = (6x)^{\dfrac{2}{2}}\\[1em] = 6x

(36x2)12=6x(36x^2)^{\dfrac{1}{2}} = 6x

Question 5(vi)

Simplify:

(125x−3)13(125x^{-3})^{\dfrac{1}{3}}

Answer

(125x−3)13=(125x3)13=(53x3)13=(5x)3×13=(5x)33=5x=5x−1(125x^{-3})^{\dfrac{1}{3}}\\[1em] = \Big(\dfrac{125}{x^3}\Big)^{\dfrac{1}{3}}\\[1em] = \Big(\dfrac{5^3}{x^3}\Big)^{\dfrac{1}{3}}\\[1em] = \Big(\dfrac{5}{x}\Big)^{3 \times \dfrac{1}{3}}\\[1em] = \Big(\dfrac{5}{x}\Big)^{\dfrac{3}{3}}\\[1em] = \dfrac{5}{x} = 5x^{-1}

(125x−3)13=5x−1(125x^{-3})^{\dfrac{1}{3}} = 5x^{-1}

Question 5(vii)

Simplify:

(2x2y−3)−2(2x^2y^{-3})^{-2}

Answer

(2x2y−3)−2=(2x2y3)−2=(y32x2)2=y64x4=14y6x−4(2x^2y^{-3})^{-2}\\[1em] = \Big(\dfrac{2x^2}{y^3}\Big)^{-2}\\[1em] = \Big(\dfrac{y^3}{2x^2}\Big)^2\\[1em] = \dfrac{y^6}{4x^4}\\[1em] = \dfrac{1}{4}y^6x^{-4}

(2x2y−3)−2=14y6x−4(2x^2y^{-3})^{-2} = \dfrac{1}{4}y^6x^{-4}

Question 5(viii)

Simplify:

(27x−3y6)23(27x^{-3}y^6)^{\dfrac{2}{3}}

Answer

(27x−3y6)23=(27y6x3)23=(33y2×3x3)23=(3y2x)3×23=(3y2x)63=(3y2x)2=9y4x2=9y4x−2(27x^{-3}y^6)^{\dfrac{2}{3}}\\[1em] = \Big(\dfrac{27y^6}{x^3}\Big)^{\dfrac{2}{3}}\\[1em] = \Big(\dfrac{3^3y^{2\times3}}{x^3}\Big)^{\dfrac{2}{3}}\\[1em] = \Big(\dfrac{3y^2}{x}\Big)^{3\times\dfrac{2}{3}}\\[1em] = \Big(\dfrac{3y^2}{x}\Big)^{\dfrac{6}{3}}\\[1em] = \Big(\dfrac{3y^2}{x}\Big)^2\\[1em] = \dfrac{9y^4}{x^2}\\[1em] = 9y^4x^{-2}

(27x−3y6)23=9y4x−2(27x^{-3}y^6)^{\dfrac{2}{3}} = 9y^4x^{-2}

Question 5(ix)

Simplify:

(−2x23y−32)6(-2x^{\dfrac{2}{3}}y^{\dfrac{-3}{2}})^6

Answer

(−2x23y−32)6=(−26x6×23y6×−32)=(64x123y−182)=(64x4y−9)=64x4y9(-2x^{\dfrac{2}{3}}y^{\dfrac{-3}{2}})^6\\[1em] = \Big(-2^6x^{6\times\dfrac{2}{3}}y^{6\times\dfrac{-3}{2}}\Big)\\[1em] = \Big(64x^{\dfrac{12}{3}}y^{\dfrac{-18}{2}}\Big)\\[1em] = \Big(64x^4y^{-9}\Big)\\[1em] = \dfrac{64x^4}{y^9}

(−2x23y−32)6=64x4y9(-2x^{\dfrac{2}{3}}y^{\dfrac{-3}{2}})^6 = \dfrac{64x^4}{y^9}

Question 6

Simplify: (xa+b)a−b.(xb+c)b−c.(xc+a)c−a(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a}

Answer

(xa+b)a−b.(xb+c)b−c.(xc+a)c−a=(x)(a+b)×(a−b).(x)(b+c)×(b−c).(x)(c+a)×(c−a)=(x)a2−b2.(x)b2−c2.(x)c2−a2=(x)a2−b2+b2−c2+c2−a2=(x)0=1(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a}\\[1em] = (x)^{(a+b)\times(a-b)}.(x)^{(b+c)\times(b-c)}.(x)^{(c+a)\times(c-a)}\\[1em] = (x)^{a^2-b^2}.(x)^{b^2-c^2}.(x)^{c^2-a^2}\\[1em] = (x)^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] = (x)^{0}\\[1em] = 1

(xa+b)a−b.(xb+c)b−c.(xc+a)c−a=1(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a} = 1

Question 7(i)

Simplify:

(x20y−10z55)÷x3y3(\sqrt[5]{x^{20}y^{-10}z^5}) ÷ \dfrac{x^3}{y^3}

Answer

(x20y−10z55)÷x3y3=(x205y−105z55)÷x3y3=(x4y−2z1)÷x3y3=(x4y−2z1)×y3x3=(x4−3y−2+3z)=(x1y1z)=xyz(\sqrt[5]{x^{20}y^{-10}z^5}) ÷ \dfrac{x^3}{y^3}\\[1em] = (x^{\dfrac{20}{5}}y^{\dfrac{-10}{5}}z^{\dfrac{5}{5}}) ÷ \dfrac{x^3}{y^3}\\[1em] = (x^4y^{-2}z^1) ÷ \dfrac{x^3}{y^3}\\[1em] = (x^4y^{-2}z^1) \times \dfrac{y^3}{x^3}\\[1em] = (x^{4-3}y^{-2+3}z)\\[1em] = (x^1y^1z)\\[1em] = xyz

(x20y−10z55)÷x3y3=xyz(\sqrt[5]{x^{20}y^{-10}z^5}) ÷ \dfrac{x^3}{y^3} = xyz

Question 7(ii)

Simplify:

[256a1681b4]−34\Big[\dfrac{256a^{16}}{81b^4}\Big]^{\dfrac{-3}{4}}

Answer

[256a1681b4]−34=[28a1634b4]−34=[22a43b]4×−34=[22a43b]−3=[3b22a4]3=[33b326a12]=[27b364a12]=[2764]a−12b3\Big[\dfrac{256a^{16}}{81b^4}\Big]^{\dfrac{-3}{4}}\\[1em] = \Big[\dfrac{2^{8}a^{16}}{3^4b^4}\Big]^{\dfrac{-3}{4}}\\[1em] = \Big[\dfrac{2^{2}a^{4}}{3b}\Big]^{4\times\dfrac{-3}{4}}\\[1em] = \Big[\dfrac{2^{2}a^{4}}{3b}\Big]^{-3}\\[1em] = \Big[\dfrac{3b}{2^{2}a^{4}}\Big]^{3}\\[1em] = \Big[\dfrac{3^3b^3}{2^{6}a^{12}}\Big]\\[1em] = \Big[\dfrac{27b^3}{64a^{12}}\Big]\\[1em] = \Big[\dfrac{27}{64}\Big]a^{-12}b^3

[256a1681b4]−34=[2764]a−12b3\Big[\dfrac{256a^{16}}{81b^4}\Big]^{\dfrac{-3}{4}} = \Big[\dfrac{27}{64}\Big]a^{-12}b^3

Question 8(i)

Simplify and express as positive indices:

(a−2b)−2.(ab)−3(a^{-2}b)^{-2}.(ab)^{-3}

Answer

(a−2b)−2.(ab)−3=(a−2×(−2)b−2).(a−3b−3)=(a4b−2).(a−3b−3)=(a4+(−3)b−2+(−3))=(a4−3b−2−3)=(a1b−5)(a^{-2}b)^{-2}.(ab)^{-3}\\[1em] = (a^{-2\times(-2)}b^{-2}).(a^{-3}b^{-3})\\[1em] = (a^{4}b^{-2}).(a^{-3}b^{-3})\\[1em] = (a^{4+(-3)}b^{-2+(-3)})\\[1em] = (a^{4-3}b^{-2-3})\\[1em] = (a^{1}b^{-5})

(a−2b)−2.(ab)−3=(a1b−5)=ab5(a^{-2}b)^{-2}.(ab)^{-3} = (a^{1}b^{-5}) = \dfrac{a}{b^5}

Question 8(ii)

Simplify and express as positive indices:

(xny−m)4×(x3y−2)−n(x^ny^{-m})^4\times(x^3y^{-2})^{-n}

Answer

(xny−m)4×(x3y−2)−n=(xny−m)4×(x3y−2)−n=(x4ny−4m)×(x−3ny2n)=(x4n−3ny−4m+2n)=(xny−4my2n)=(xny2ny4m)(x^ny^{-m})^4\times(x^3y^{-2})^{-n}\\[1em] = (x^ny^{-m})^4\times(x^3y^{-2})^{-n}\\[1em] = (x^{4n}y^{-4m})\times(x^{-3n}y^{2n})\\[1em] = (x^{4n-3n}y^{-4m+2n})\\[1em] = (x^{n}y^{-4m}y^{2n})\\[1em] = (\dfrac{x^{n}y^{2n}}{y^{4m}})

(xny−m)4×(x3y−2)−n=(xny2ny4m)(x^ny^{-m})^4\times(x^3y^{-2})^{-n} = (\dfrac{x^{n}y^{2n}}{y^{4m}})

Question 8(iii)

Simplify and express as positive indices:

[125a−3y6]−13\Big[\dfrac{125a^{-3}}{y^6}\Big]^{\dfrac{-1}{3}}

Answer

[125a−3y6]−13=[53a−3y6]−13=[5a−1y2]3×−13=[5a−1y2]−1=[y25a−1]1=[ay25]\Big[\dfrac{125a^{-3}}{y^6}\Big]^{\dfrac{-1}{3}}\\[1em] = \Big[\dfrac{5^{3}a^{-3}}{y^6}\Big]^{\dfrac{-1}{3}}\\[1em] = \Big[\dfrac{5a^{-1}}{y^2}\Big]^{3\times\dfrac{-1}{3}}\\[1em] = \Big[\dfrac{5a^{-1}}{y^2}\Big]^{-1}\\[1em] = \Big[\dfrac{y^2}{5a^{-1}}\Big]^{1}\\[1em] = \Big[\dfrac{ay^2}{5}\Big]

[125a−3y6]−13=ay25\Big[\dfrac{125a^{-3}}{y^6}\Big]^{\dfrac{-1}{3}} = \dfrac{ay^2}{5}

Question 8(iv)

Simplify and express as positive indices:

[32x−5243y−5]−15\Big[\dfrac{32x^{-5}}{243y^{-5}}\Big]^{\dfrac{-1}{5}}

Answer

[32x−5243y−5]−15=[25x−535y−5]−15=[2x−13y−1]5×−15=[2x−13y−1]−1=[2y3x]−1=[3x2y]\Big[\dfrac{32x^{-5}}{243y^{-5}}\Big]^{\dfrac{-1}{5}}\\[1em] = \Big[\dfrac{2^{5}x^{-5}}{3^5y^{-5}}\Big]^{\dfrac{-1}{5}}\\[1em] = \Big[\dfrac{2x^{-1}}{3y^{-1}}\Big]^{5\times\dfrac{-1}{5}}\\[1em] = \Big[\dfrac{2x^{-1}}{3y^{-1}}\Big]^{-1}\\[1em] = \Big[\dfrac{2y}{3x}\Big]^{-1}\\[1em] = \Big[\dfrac{3x}{2y}\Big]

[32x−5243y−5]−15=[3x2y]\Big[\dfrac{32x^{-5}}{243y^{-5}}\Big]^{\dfrac{-1}{5}}= \Big[\dfrac{3x}{2y}\Big]

Question 8(v)

Simplify and express as positive indices:

(a−2b)12×(ab−3)13(a^{-2}b)^{\dfrac{1}{2}} \times (ab^{-3})^{\dfrac{1}{3}}

Answer

(a−2b)12×(ab−3)13=(a−2×12b12)×(a13b−3×13)=(a−1b12)×(a13b−1)=(a−1+13b12+(−1))=(a−33+13b12+−22)=(a−3+13b1−22)=(a−23b−12)=1a23b12(a^{-2}b)^{\dfrac{1}{2}} \times (ab^{-3})^{\dfrac{1}{3}}\\[1em] = \Big(a^{-2\times\dfrac{1}{2}}b^{\dfrac{1}{2}}\Big) \times \Big(a^{\dfrac{1}{3}}b^{-3\times\dfrac{1}{3}}\Big)\\[1em] = \Big(a^{-1}b^{\dfrac{1}{2}}\Big) \times \Big(a^{\dfrac{1}{3}}b^{-1}\Big)\\[1em] = \Big(a^{-1 + \dfrac{1}{3}}b^{\dfrac{1}{2} + (-1)}\Big)\\[1em] = \Big(a^{\dfrac{-3}{3} + \dfrac{1}{3}}b^{\dfrac{1}{2} + \dfrac{-2}{2}}\Big)\\[1em] = \Big(a^{\dfrac{-3+1}{3}}b^{\dfrac{1-2}{2}}\Big)\\[1em] = \Big(a^{\dfrac{-2}{3}}b^{\dfrac{-1}{2}}\Big)\\[1em] = \dfrac{1}{a^{\dfrac{2}{3}}b^{\dfrac{1}{2}}}

(a−2b)12×(ab−3)13=1a23b12(a^{-2}b)^{\dfrac{1}{2}} \times (ab^{-3})^{\dfrac{1}{3}} = \dfrac{1}{a^{\dfrac{2}{3}}b^{\dfrac{1}{2}}}

Question 8(vi)

Simplify and express as positive indices:

(xy)m−n.(yz)n−l.(zx)l−m(xy)^{m-n}.(yz)^{n-l}.(zx)^{l-m}

Answer

(xy)m−n.(yz)n−l.(zx)l−m=(xm−nym−n).(yn−lzn−l).(zl−mxl−m)=x(m−n)+(l−m)y(m−n)+(n−l)z(n−l)+(l−m)=x−n+lym−lzn−m=xlymznxnylzm(xy)^{m-n}.(yz)^{n-l}.(zx)^{l-m}\\[1em] = (x^{m-n}y^{m-n}).(y^{n-l}z^{n-l}).(z^{l-m}x^{l-m})\\[1em] = x^{(m-n)+(l-m)}y^{(m-n)+(n-l)}z^{(n-l)+(l-m)}\\[1em] = x^{-n+l}y^{m-l}z^{n-m}\\[1em] = \dfrac{x^ly^mz^n}{x^ny^lz^m}

(xy)m−n.(yz)n−l.(zx)l−m=x−n+lym−lzn−m(xy)^{m-n}.(yz)^{n-l}.(zx)^{l-m} = x^{-n+l}y^{m-l}z^{n-m}

Question 9

Show that: [xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

Answer

To prove: [xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

Taking LHS:

[xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=[xa(a−b)x−b(a−b)].[xb(b−c)x−c(b−c)].[xc(c−a)x−a(c−a)]=[xa2−abx−ab+b2].[xb2−bcx−bc+c2].[xc2−acx−ac+a2]=[x(a2−ab)−(−ab+b2)].[x(b2−bc)−(−bc+c2)].[x(c2−ac)−(−ac+a2)]=[xa2−ab+ab−b2].[xb2−bc+bc−c2].[xc2−ac+ac−a2]=[xa2−b2].[xb2−c2].[xc2−a2]=x(a2−b2)+(b2−c2)+(c2−a2)=xa2−b2+b2−c2+c2−a2=x0=1=RHS\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a}\\[1em] = \Big[\dfrac{x^{a(a-b)}}{x^{-b(a-b)}}\Big].\Big[\dfrac{x^{b(b-c)}}{x^{-c(b-c)}}\Big].\Big[\dfrac{x^{c(c-a)}}{x^{-a(c-a)}}\Big]\\[1em] = \Big[\dfrac{x^{a^2-ab}}{x^{-ab+b^2}}\Big].\Big[\dfrac{x^{b^2-bc}}{x^{-bc+c^2}}\Big].\Big[\dfrac{x^{c^2-ac}}{x^{-ac+a^2}}\Big]\\[1em] = \Big[{x^{(a^2-ab)-(-ab+b^2)}}\Big].\Big[x^{(b^2-bc)-(-bc+c^2)}\Big].\Big[x^{(c^2-ac)-(-ac+a^2)}\Big]\\[1em] = \Big[{x^{a^2-ab+ab-b^2}}\Big].\Big[x^{b^2-bc+bc-c^2}\Big].\Big[x^{c^2-ac+ac-a^2}\Big]\\[1em] = \Big[x^{a^2-b^2}\Big].\Big[x^{b^2-c^2}\Big].\Big[x^{c^2-a^2}\Big]\\[1em] = x^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}\\[1em] = x^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] = x^0\\[1em] = 1 \\[1em] = \text{RHS}

∴ LHS = RHS

[xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

Question 10

Evaluate: x5+n×(x2)3n+1x7n−2\dfrac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}}

Answer

x5+n×(x2)3n+1x7n−2=x5+n×(x)2(3n+1)x7n−2=x5+n×(x)6n+2x7n−2=x(5+n)+(6n+2)x7n−2=x5+n+6n+2x7n−2=x(7+7n)−(7n−2)=x7+7n−7n+2=x9\dfrac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}}\\[1em] = \dfrac{x^{5+n}\times(x)^{2(3n+1)}}{x^{7n-2}}\\[1em] = \dfrac{x^{5+n}\times(x)^{6n+2}}{x^{7n-2}}\\[1em] = \dfrac{x^{(5+n)+(6n+2)}}{x^{7n-2}}\\[1em] = \dfrac{x^{5+n+6n+2}}{x^{7n-2}}\\[1em] = x^{(7+7n)-(7n-2)}\\[1em] = x^{7+7n-7n+2}\\[1em] = x^{9}

x5+n×(x2)3n+1x7n−2=x9\dfrac{x^{5+n}\times(x^2)^{3n+1}}{x^{7n-2}} = x^{9}

Question 11

Evaluate: a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}}

Answer

a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3=a2n+1×a4n2−1a4n2−n×a4n+6=a(2n+1)+(4n2−1)a(4n2−n)+(4n+6)=a2n+4n2a4n2+3n+6=a(2n+4n2)−(4n2+3n+6)=a2n+4n2−4n2−3n−6=a−n−6=1an+6\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}}\\[1em] = \dfrac{a^{2n+1}\times a^{4n^2-1}}{a^{4n^2-n}\times a^{4n+6}}\\[1em] = \dfrac{a^{(2n+1)+(4n^2-1)}}{a^{(4n^2-n)+(4n+6)}}\\[1em] = \dfrac{a^{2n+4n^2}}{a^{4n^2+3n+6}}\\[1em] = a^{(2n+4n^2)-(4n^2+3n+6)}\\[1em] = a^{2n+4n^2-4n^2-3n-6}\\[1em] = a^{-n-6}\\[1em] = \dfrac{1}{a^{n+6}}

a2n+1×a(2n+1)(2n−1)an(4n−1)×(a2)2n+3=1an+6\dfrac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times (a^2)^{2n+3}} = \dfrac{1}{a^{n+6}}

Question 12

Prove that: (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}

Answer

To Prove: (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}

Taking LHS:

(m+n)−1(m−1+n−1)=(1m+n)(1m+1n)=(1m+n)(nmn+mnm)=(1m+n)(n+mmn)=1×(n+m)(m+n)×mn=1mn=(mn)−1=RHS(m + n)^{-1}(m^{-1} + n^{-1})\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{1}{m} + \dfrac{1}{n}\Big)\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{n}{mn} + \dfrac{m}{nm}\Big)\\[1em] = \Big(\dfrac{1}{m + n}\Big)\Big(\dfrac{n + m}{mn}\Big)\\[1em] = \dfrac{1 \times (n + m)}{(m + n) \times mn}\\[1em] = \dfrac{1}{mn}\\[1em] = (mn)^{-1} \\[1em] = \text{RHS}

∴ LHS = RHS

Hence, (m+n)−1(m−1+n−1)=(mn)−1(m + n)^{-1}(m^{-1} + n^{-1}) = (mn)^{-1}

Question 13(i)

Prove that:

[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

Answer

To prove:[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

Taking LHS:

[xaxb]1ab[xbxc]1bc[xcxa]1ca=[xa−b]1ab[xb−c]1bc[xc−a]1ca=[x]a−bab[x]b−cbc[x]c−aca=[x]aab−bab[x]bbc−cbc[x]cca−aca=[x]1b−1a[x]1c−1b[x]1a−1c=[x]1b−1a+1c−1b+1a−1c=[x]1b−1b−1a+1a+1c−1c=x0=1=RHS\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}}\\[1em] = \Big[x^{a-b}\Big]^{\dfrac{1}{ab}}\Big[x^{b-c}\Big]^{\dfrac{1}{bc}}\Big[x^{c-a}\Big]^{\dfrac{1}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{a-b}{ab}}\Big[x\Big]^{\dfrac{b-c}{bc}}\Big[x\Big]^{\dfrac{c-a}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{a}{ab}-\dfrac{b}{ab}}\Big[x\Big]^{\dfrac{b}{bc}-\dfrac{c}{bc}}\Big[x\Big]^{\dfrac{c}{ca}-\dfrac{a}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{a}}\Big[x\Big]^{\dfrac{1}{c}-\dfrac{1}{b}}\Big[x\Big]^{\dfrac{1}{a}-\dfrac{1}{c}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{c}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{c}}\\[1em] = x^0\\[1em] = 1 \\[1em] = \text{RHS}

∴ LHS = RHS

Hence,[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

Question 13(ii)

Prove that:

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

Answer

To prove: 11+xa−b+11+xb−a=1\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=1

Taking LHS:

11+xa−b+11+xb−a=11+xaxb+11+xbxa=xbxb+xa+xaxa+xb=xb+xaxb+xa=1=RHS\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}\\[1em] = \dfrac{1}{1+\dfrac{x^a}{x^b}}+\dfrac{1}{1+\dfrac{x^b}{x^a}}\\[1em] = \dfrac{{x^b}}{{x^b}+{x^a}}+\dfrac{{x^a}}{{x^a}+{x^b}}\\[1em] = \dfrac{{x^b}+{x^a}}{{x^b}+{x^a}}\\[1em] = 1 \\[1em] = \text{RHS}

∴ LHS = RHS

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

Question 14(i)

Find the value of n, when:

12−5×122n+1=1213÷12712^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7

Answer

12−5×122n+1=1213÷127⇒12(−5)+(2n+1)=1213−7⇒12−5+2n+1=126⇒12−4+2n=126⇒−4+2n=6⇒2n=6+4⇒2n=10⇒n=102⇒n=512^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7\\[1em] \Rightarrow 12^{(-5)+(2n+1)} = 12^{13 - 7}\\[1em] \Rightarrow 12^{-5+2n+1} = 12^{6}\\[1em] \Rightarrow 12^{-4+2n} = 12^{6}\\[1em] \Rightarrow -4+2n = 6\\[1em] \Rightarrow 2n = 6+4\\[1em] \Rightarrow 2n = 10\\[1em] \Rightarrow n = \dfrac{10}{2}\\[1em] \Rightarrow n = 5\\[1em]

If 12−5×122n+1=1213÷12712^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7 then n = 5.

Question 14(ii)

Find the value of n, when:

a2n−3×(a2)n+1(a4)−3=(a3)3÷(a6)−3\dfrac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}

Answer

a2n−3×(a2)n+1(a4)−3=(a3)3÷(a6)−3⇒a2n−3×(a)2(n+1)(a)4×(−3)=(a)3×3÷(a)6×(−3)⇒a2n−3×(a)2n+2(a)−12=a9÷a−18⇒a(2n−3)+(2n+2)(a)−12=a9−(−18)⇒a2n−3+2n+2(a)−12=a9+18⇒a4n−1(a)−12=a27⇒a(4n−1)−(−12)=a27⇒a4n−1+12=a27⇒a4n+11=a27⇒4n+11=27⇒4n=27−11⇒4n=16⇒n=164⇒n=4\dfrac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}\\[1em] \Rightarrow \dfrac{a^{2n-3}\times(a)^{2(n+1)}}{(a)^{4\times(-3)}} = (a)^{3\times3} ÷ (a)^{6\times(-3)}\\[1em] \Rightarrow \dfrac{a^{2n-3}\times(a)^{2n+2}}{(a)^{-12}} = a^9 ÷ a^{-18}\\[1em] \Rightarrow \dfrac{a^{(2n-3)+(2n+2)}}{(a)^{-12}} = a^{9-(-18)}\\[1em] \Rightarrow \dfrac{a^{2n-3+2n+2}}{(a)^{-12}} = a^{9+18}\\[1em] \Rightarrow \dfrac{a^{4n-1}}{(a)^{-12}} = a^{27}\\[1em] \Rightarrow a^{(4n-1)-(-12)} = a^{27}\\[1em] \Rightarrow a^{4n-1+12} = a^{27}\\[1em] \Rightarrow a^{4n+11} = a^{27}\\[1em] \Rightarrow 4n+11 = 27\\[1em] \Rightarrow 4n = 27-11\\[1em] \Rightarrow 4n = 16\\[1em] \Rightarrow n = \dfrac{16}{4}\\[1em] \Rightarrow n = 4

If a2n−3×(a2)n+1(a4)−3=(a3)3÷(a6)−3\dfrac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}, then n = 4

Question 15(i)

Simplify:

a2n+3.a(2n+1)(n+2)(a3)2n+1.an(2n+1)\dfrac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}}

Answer

a2n+3.a(2n+1)(n+2)(a3)2n+1.an(2n+1)=a2n+3.a2n2+4n+n+2a3(2n+1).a2n2+n=a2n+3.a2n2+5n+2a6n+3.a2n2+n=a(2n+3)+(2n2+5n+2)a(6n+3)+(2n2+n)=a2n+3+2n2+5n+2a6n+3+2n2+n=a2n2+7n+5a2n2+7n+3=a(2n2+7n+5)−(2n2+7n+3)=a2n2+7n+5−2n2−7n−3=a2\dfrac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}}\\[1em] = \dfrac{a^{2n+3}.a^{2n^2+4n+n+2}}{a^{3(2n+1)}.a^{2n^2+n}}\\[1em] = \dfrac{a^{2n+3}.a^{2n^2+5n+2}}{a^{6n+3}.a^{2n^2+n}}\\[1em] = \dfrac{a^{(2n+3)+(2n^2+5n+2)}}{a^{(6n+3)+(2n^2+n)}}\\[1em] = \dfrac{a^{2n+3+2n^2+5n+2}}{a^{6n+3+2n^2+n}}\\[1em] = \dfrac{a^{2n^2+7n+5}}{a^{2n^2+7n+3}}\\[1em] = a^{(2n^2+7n+5)-(2n^2+7n+3)}\\[1em] = a^{2n^2+7n+5-2n^2-7n-3}\\[1em] = a^{2}\\[1em]

a2n+3.a(2n+1)(n+2)(a3)2n+1.an(2n+1)=a2\dfrac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}} = a^2

Question 15(ii)

Simplify:

x2n+7.(x2)3n+2x4(2n+3)\dfrac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}}

Answer

x2n+7.(x2)3n+2x4(2n+3)=x2n+7.x2(3n+2)x4(2n+3)=x2n+7.x6n+4x8n+12=x(2n+7)+(6n+4)x8n+12=x2n+7+6n+4x8n+12=x8n+11x8n+12=x(8n+11)−(8n+12)=x8n+11−8n−12=x−1=1x\dfrac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}}\\[1em] = \dfrac{x^{2n+7}.x^{2(3n+2)}}{x^{4(2n+3)}}\\[1em] = \dfrac{x^{2n+7}.x^{6n+4}}{x^{8n+12}}\\[1em] = \dfrac{x^{(2n+7)+(6n+4)}}{x^{8n+12}}\\[1em] = \dfrac{x^{2n+7+6n+4}}{x^{8n+12}}\\[1em] = \dfrac{x^{8n+11}}{x^{8n+12}}\\[1em] = x^{(8n+11)-(8n+12)}\\[1em] = x^{8n+11-8n-12}\\[1em] = x^{-1}\\[1em] = \dfrac{1}{x}

x2n+7.(x2)3n+2x4(2n+3)=1x\dfrac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}} = \dfrac{1}{x}

Question 16(i)

Evaluate:

(2−3+3−2)×70(2^{-3} + 3^{-2})\times 7^0

Answer

(2−3+3−2)×70=(123+132)×1=(18+19)(2^{-3} + 3^{-2})\times 7^0\\[1em] = \Big(\dfrac{1}{2^{3}} + \dfrac{1}{3^{2}}\Big)\times 1\\[1em] = \Big(\dfrac{1}{8} + \dfrac{1}{9}\Big)

LCM of 8 and 9 is 2 x 2 x 2 x 3 x 3 = 72

=(1×98×9+1×89×8)=(972+872)=(9+872)=(1772)= \Big(\dfrac{1 \times 9}{8 \times 9} + \dfrac{1 \times 8}{9 \times 8}\Big)\\[1em] = \Big(\dfrac{9}{72} + \dfrac{8}{72}\Big)\\[1em] = \Big(\dfrac{9 + 8}{72}\Big)\\[1em] = \Big(\dfrac{17}{72}\Big)

(2−3+3−2)×70=(1772)(2^{-3} + 3^{-2})\times 7^0 = \Big(\dfrac{17}{72}\Big)

Question 16(ii)

Evaluate:

(80+2−1)×32(8^0 + 2^{-1})\times 3^2

Answer

(80+2−1)×32=(1+121)×32=(11+12)×9(8^0 + 2^{-1})\times 3^2\\[1em] = \Big(1 + \dfrac{1}{2^1}\Big)\times 3^2\\[1em] = \Big(\dfrac{1}{1} + \dfrac{1}{2}\Big)\times 9

LCM of 1 and 2 is 2

=(1×21×2+1×12×1)×9=(22+12)×9=(2+12)×9=(32)×9=(3×92)=272=1312= \Big(\dfrac{1 \times 2}{1 \times 2} + \dfrac{1 \times 1}{2 \times 1}\Big)\times 9\\[1em] = \Big(\dfrac{2}{2} + \dfrac{1}{2}\Big)\times 9\\[1em] = \Big(\dfrac{2+1}{2}\Big)\times 9\\[1em] = \Big(\dfrac{3}{2}\Big)\times 9\\[1em] = \Big(\dfrac{3 \times 9}{2}\Big)\\[1em] = \dfrac{27}{2}\\[1em] = 13\dfrac{1}{2}

(80+2−1)×32=1312(8^0 + 2^{-1})\times 3^2 = 13\dfrac{1}{2}

Question 16(iii)

Evaluate:

[(16)−1−(15)−1]−2\Big[\Big(\dfrac{1}{6}\Big)^{-1} - \Big(\dfrac{1}{5}\Big)^{-1}\Big]^{-2}

Answer

[(16)−1−(15)−1]−2=[61−51]−2=[6−5]−2=[1]−2=112=12=1\Big[\Big(\dfrac{1}{6}\Big)^{-1} - \Big(\dfrac{1}{5}\Big)^{-1}\Big]^{-2}\\[1em] = [6^1 - 5^1]^{-2}\\[1em] = [6 - 5]^{-2}\\[1em] = [1]^{-2}\\[1em] = \dfrac{1}{1}^2\\[1em] = 1^2\\[1em] = 1

[(16)−1−(15)−1]−2=1\Big[\Big(\dfrac{1}{6}\Big)^{-1} - \Big(\dfrac{1}{5}\Big)^{-1}\Big]^{-2} = 1

Question 16(iv)

Evaluate:

[{(−13)−2}2]−1\Big[\Big\lbrace\Big(-\dfrac{1}{3}\Big)^{-2}\Big\rbrace^2\Big]^{-1}

Answer

[{(−13)−2}2]−1=[{(−3)2}2]−1=[{9}2]−1=[81]−1=181\Big[\Big\lbrace\Big(-\dfrac{1}{3}\Big)^{-2}\Big\rbrace^2\Big]^{-1}\\[1em] = [\lbrace (-3)^2\rbrace^2]^{-1}\\[1em] = [\lbrace 9\rbrace^2]^{-1}\\[1em] = [81]^{-1}\\[1em] = \dfrac{1}{81}

[{(−13)−2}2]−1=181\Big[\Big\lbrace\Big(-\dfrac{1}{3}\Big)^{-2}\Big\rbrace^2\Big]^{-1} = \dfrac{1}{81}

Question 16(v)

Evaluate:

5n+2−5n+15n+1\dfrac{5^{n+2}-5^{n+1}}{5^{n+1}}

Answer

5n+2−5n+15n+1=5n+25n+1−5n+15n+1=5(n+2)−(n+1)−5(n+1)−(n+1)=5n+2−n−1−5n+1−n−1=51−50=5−1=4\dfrac{5^{n+2}-5^{n+1}}{5^{n+1}}\\[1em] = \dfrac{5^{n+2}}{5^{n+1}} -\dfrac{5^{n+1}}{5^{n+1}}\\[1em] = 5^{(n+2)-(n+1)} -5^{(n+1)-(n+1)}\\[1em] = 5^{n+2-n-1} -5^{n+1-n-1}\\[1em] = 5^{1} -5^{0}\\[1em] = 5 -1\\[1em] = 4

5n+2−5n+15n+1=4\dfrac{5^{n+2}-5^{n+1}}{5^{n+1}} = 4

Question 17(i)

Find the value of x; if:

1(125)x−7=52x−1\dfrac{1}{(125)^{x-7}}=5^{2x-1}

Answer

1(125)x−7=52x−1⇒1(53)x−7=52x−1⇒1(5)3(x−7)=52x−1⇒153x−21=52x−1⇒(15)3x−21=52x−1⇒5−(3x−21)=52x−1⇒5−3x+21=52x−1⇒−3x+21=2x−1⇒1+21=2x+3x⇒22=5x⇒x=225\dfrac{1}{(125)^{x-7}}=5^{2x-1}\\[1em] \Rightarrow \dfrac{1}{(5^3)^{x-7}}=5^{2x-1}\\[1em] \Rightarrow \dfrac{1}{(5)^{3(x-7)}}=5^{2x-1}\\[1em] \Rightarrow \dfrac{1}{5^{3x-21}}=5^{2x-1}\\[1em] \Rightarrow \Big(\dfrac{1}{5}\Big)^{3x-21}=5^{2x-1}\\[1em] \Rightarrow 5^{-(3x-21)}=5^{2x-1}\\[1em] \Rightarrow 5^{-3x+21}=5^{2x-1}\\[1em] \Rightarrow -3x + 21 = 2x - 1\\[1em] \Rightarrow 1 + 21 = 2x + 3x\\[1em] \Rightarrow 22 = 5x\\[1em] \Rightarrow x = \dfrac{22}{5}

If 1(125)x−7=52x−1\dfrac{1}{(125)^{x-7}}=5^{2x-1} then x=225x = \dfrac{22}{5}

Question 17(ii)

Find the value of x; if:

(23)3×(23)−4=(23)2x+1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}

Answer

(23)3×(23)−4=(23)2x+1⇒(23)3+(−4)=(23)2x+1⇒3−4=2x+1⇒−1=2x+1⇒−1−1=2x⇒−2=2x⇒x=−22⇒x=−1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}\\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{3+(-4)} = \Big(\dfrac{2}{3}\Big)^{2x+1}\\[1em] \Rightarrow 3 - 4 = 2x + 1\\[1em] \Rightarrow -1 = 2x + 1\\[1em] \Rightarrow -1 - 1 = 2x\\[1em] \Rightarrow -2 = 2x\\[1em] \Rightarrow x = \dfrac{-2}{2}\\[1em] \Rightarrow x = -1

If (23)3×(23)−4=(23)2x+1\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{2}{3}\Big)^{-4} = \Big(\dfrac{2}{3}\Big)^{2x+1}, then x=−1x = -1.

Question 17(iii)

Find the value of x; if:

4n÷4−3=454^n ÷ 4^{-3} = 4^5

Answer

4n÷4−3=45⇒4n−(−3)=45⇒n−(−3)=5⇒n+3=5⇒n=5−3⇒n=24^n ÷ 4^{-3} = 4^5\\[1em] \Rightarrow 4^{n-(-3)} = 4^5\\[1em] \Rightarrow n - (-3) = 5\\[1em] \Rightarrow n + 3 = 5\\[1em] \Rightarrow n = 5 - 3\\[1em] \Rightarrow n = 2

If 4n÷4−3=454^n ÷ 4^{-3} = 4^5, then n=2n = 2.

Question 18

Simplify:81×3n+1−9×3n81×3n+2−9×3n+1\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}}

Answer

81×3n+1−9×3n81×3n+2−9×3n+1=34×3n+1−32×3n34×3n+2−32×3n+1=34+(n+1)−32+n34+(n+2)−32+(n+1)=3n+5−32+n3n+6−3n+3=(3n+2.33)−32+n(3n+3.33)−3n+3=(3n+2)(3n+3)33−133−1=(3(n+2)−(n+3))33−133−1=(3n+2−n−3)=(3−1)=13\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}}\\[1em] = \dfrac{3^4\times3^{n+1}-3^2\times3^n}{3^4\times3^{n+2}-3^2\times3^{n+1}}\\[1em] = \dfrac{3^{4+(n+1)}-3^{2+n}}{3^{4+(n+2)}-3^{2+(n+1)}}\\[1em] = \dfrac{3^{n+5}-3^{2+n}}{3^{n+6}-3^{n+3}}\\[1em] = \dfrac{(3^{n+2}.3^3)-3^{2+n}}{(3^{n+3}.3^3)-3^{n+3}}\\[1em] = \dfrac{(3^{n+2})}{(3^{n+3})}\dfrac{3^3-1}{3^{3}-1}\\[1em] = (3^{(n+2)-(n+3)})\dfrac{\cancel {3^3-1}}{\cancel {3^3-1}}\\[1em] = (3^{n+2-n-3})\\[1em] = (3^{-1})\\[1em] = \dfrac{1}{3}

81×3n+1−9×3n81×3n+2−9×3n+1=13\dfrac{81\times3^{n+1}-9\times3^n}{81\times3^{n+2}-9\times3^{n+1}} = \dfrac{1}{3}

Question 19

If 2n−7×5n−4=12502^{n-7} \times 5^{n-4} = 1250, find n.

Answer

Finding prime factors of 1250,

2n−7×5n−4=1250⇒2n−7×5n−4=21×542^{n-7} \times 5^{n-4} = 1250\\[1em] \Rightarrow 2^{n-7} \times 5^{n-4} = 2^1 \times 5^4

On comparing the exponent of 2 or 5, we get

n−7=1⇒n=1+7⇒n=8n - 7 = 1 \\[1em] \Rightarrow n = 1 + 7\\[1em] \Rightarrow n = 8

OR

n−4=4⇒n=4+4⇒n=8n - 4 = 4\\[1em] \Rightarrow n = 4 + 4\\[1em] \Rightarrow n = 8

If 2n−7×5n−4=12502^{n-7} \times 5^{n-4} = 1250, then n=8n = 8.

Test Yourself

Question 1(i)

The multiplicative inverse of (80+50)(80−50)(8^0 + 5^0)(8^0 - 5^0) is:

  1. 0

  2. 49

  3. 1

  4. undefined

Answer

(80+50)(80−50)=(1+1)(1−1)=2×0=0(8^0 + 5^0)(8^0 - 5^0)\\[1em] = (1 + 1)(1 - 1)\\[1em] = 2 \times 0\\[1em] = 0

The multiplicative inverse of 0 = 10\dfrac{1}{0}

And as we know 10\dfrac{1}{0} is not defined.

Hence, option 4 is the correct option.

Question 1(ii)

The value of [(14)−2+(13)−2]÷(15)−2\Big[\Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{3}\Big)^{-2}\Big] ÷ \Big(\dfrac{1}{5}\Big)^{-2} is:

  1. 125\dfrac{1}{25}

  2. 1

  3. 0

  4. 625

Answer

[(14)−2+(13)−2]÷(15)−2=[(4)2+(3)2]÷(5)2=[16+9]÷25=25÷25=1\Big[\Big(\dfrac{1}{4}\Big)^{-2} + \Big(\dfrac{1}{3}\Big)^{-2}\Big] ÷ \Big(\dfrac{1}{5}\Big)^{-2}\\[1em] = [(4)^2 + (3)^2] ÷ (5)^2\\[1em] = [16 + 9] ÷ 25\\[1em] = 25 ÷ 25\\[1em] = 1

Hence, option 2 is the correct option.

Question 1(iii)

If 34×93=9n3^4 \times 9^3 = 9^n, then the value of n is:

  1. 5

  2. 7

  3. 10

  4. none of the above

Answer

34×93=9n⇒34×(32)3=(32)n⇒34×36=32n⇒34+6=32n⇒310=32n⇒10=2n⇒n=102⇒n=53^4 \times 9^3 = 9^n\\[1em] \Rightarrow 3^4 \times (3^2)^3 = (3^2)^n\\[1em] \Rightarrow 3^4 \times 3^6 = 3^{2n}\\[1em] \Rightarrow 3^{4+6} = 3^{2n}\\[1em] \Rightarrow 3^{10} = 3^{2n}\\[1em] \Rightarrow 10 = 2n\\[1em] \Rightarrow n = \dfrac{10}{2}\\[1em] \Rightarrow n = 5

Hence, option 1 is the correct option.

Question 1(iv)

If (56)5×(65)−4=(56)3x\Big(\dfrac{5}{6}\Big)^5 \times \Big(\dfrac{6}{5}\Big)^{-4} = \Big(\dfrac{5}{6}\Big)^{3x}, then the value of xx is:

  1. 13\dfrac{1}{3}

  2. 203\dfrac{20}{3}

  3. -3

  4. 3

Answer

(56)5×(65)−4=(56)3x⇒(56)5×(56)4=(56)3x⇒(56)5+4=(56)3x⇒(56)9=(56)3x⇒9=3x⇒x=93⇒x=3\Big(\dfrac{5}{6}\Big)^5 \times \Big(\dfrac{6}{5}\Big)^{-4} = \Big(\dfrac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{5}{6}\Big)^5 \times \Big(\dfrac{5}{6}\Big)^4 = \Big(\dfrac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{5}{6}\Big)^{5+4} = \Big(\dfrac{5}{6}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{5}{6}\Big)^9 = \Big(\dfrac{5}{6}\Big)^{3x}\\[1em] \Rightarrow 9 = 3x\\[1em] \Rightarrow x = \dfrac{9}{3}\\[1em] \Rightarrow x = 3

Hence, option 4 is the correct option.

Question 1(v)

(25)−8÷(25)5\Big(\dfrac{2}{5}\Big)^{-8} ÷ \Big(\dfrac{2}{5}\Big)^5 is equal to:

  1. (25)−3\Big(\dfrac{2}{5}\Big)^{-3}

  2. (25)−13\Big(\dfrac{2}{5}\Big)^{-13}

  3. (25)13\Big(\dfrac{2}{5}\Big)^{13}

  4. (52)−13\Big(\dfrac{5}{2}\Big)^{-13}

Answer

(25)−8÷(25)5⇒(25)−8−5⇒(25)−13\Big(\dfrac{2}{5}\Big)^{-8} ÷ \Big(\dfrac{2}{5}\Big)^5\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{-8 - 5}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{-13}

Hence, option 2 is the correct option.

Question 1(vi)

Statement 1: (x0 + y0)(x + y)0 = 1, x, y ≠ 0.

Statement 2: (1 + 1)(1 - 1) = 2 x 0 = 0.

Which of the following options is correct?

  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

We know that,

For any number a ≠ 0, a0 = 1.

a0 = 1

Statement 1 : (x0 + y0)(x + y)0 = 1

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

⇒ (1 + 1) x 1

⇒ 2 x 1

⇒ 2.

R.H.S. = 1

Since, L.H.S. ≠ R.H.S.

So, statement 1 is false.

Statement 2 : (1 + 1)(1 - 1) = 2 x 0 = 0

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

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0.

R.H.S. = 0

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

So, statement 2 is true.

Hence, option 4 is the correct option.

Question 1(vii)

Assertion (A) : (-100)3 = -10,00,000

Reason (R) : (-p)q = pq; if q is even.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

According to assertion : (-100)3 = -10,00,000

Solving L.H.S.,

⇒ (-100)3

⇒ (-100) x (-100) x (-100)

⇒ -10,00,000

As, L.H.S. = R.H.S.

So, assertion (A) is true.

We know that,

(-p)q = pq; if q is even because the negative sign is raised to an even power and thus becomes positive.

So, reason (R) is true but reason does not explains assertion.

Hence, option 2 is the correct option.

Question 1(viii)

Assertion (A) : (70 + 20) (70 - 20) = 0.

Reason (R) : Every non-zero number raised to the power zero (0) is always equal to 1.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Using the property:

⇒ a0 = 1 for any a ≠ 0

Thus, every non-zero number raised to the power zero (0) is always equal to 1.

So, reason (R) is true.

According to assertion : (70 + 20)(70 - 20) = 0

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

⇒ (70 + 20)(70 - 20)

⇒ (1 + 1)(1 - 1)

⇒ 2 x 0

⇒ 0.

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

So, assertion (A) is true and reason clearly explains assertion.

Hence, option 1 is the correct option.

Question 1(ix)

Assertion (A) : (15)−5×(12)−5=(10)−5\Big(\dfrac{1}{5}\Big)^{-5} \times \Big(\dfrac{1}{2}\Big)^{-5} = (10)^{-5}.

Reason (R) : p−q=1pq and 1p−q=pqp^{-q} = \dfrac{1}{p^q} \text{ and } \dfrac{1}{p^{-q}} = p^q, p ≠ 0.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

According to assertion : (15)−5×(12)−5=(10)−5\Big(\dfrac{1}{5}\Big)^{-5} \times \Big(\dfrac{1}{2}\Big)^{-5} = (10)^{-5}

Solving L.H.S.,

⇒(15)−5×(12)−5⇒55×25⇒(5×2)5⇒105.\Rightarrow \Big(\dfrac{1}{5}\Big)^{-5} \times \Big(\dfrac{1}{2}\Big)^{-5}\\[1em] \Rightarrow 5^5 \times 2^5 \\[1em] \Rightarrow (5 \times 2)^5 \\[1em] \Rightarrow 10^5.

Since, L.H.S. ≠ R.H.S.

So, assertion (A) is false.

We know that,

p−q=1pq and 1p−q=pqp^{-q} = \dfrac{1}{p^q} \text{ and } \dfrac{1}{p^{-q}} = p^q, for p ≠ 0

So, reason (R) is true.

Hence, option 4 is the correct option.

Question 1(x)

Assertion (A) : (p - q)-1 (p-1 - q-1) = -(pq)-1.

Reason (R) : a-1 and a1 are multiplicative inverses of each other.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

According to assertion : (p - q)-1(p-1 - q-1) = -(pq)-1

Solving L.H.S.,

⇒(p−q)−1(p−1−q−1)⇒(1p−q)(1p−1q)⇒(1p−q)(q−ppq)⇒−(1p−q)(p−qpq)⇒−(p−q)pq(p−q)⇒−1pq⇒−(pq)−1\Rightarrow (p - q)^{-1}(p^{-1} - q^{-1})\\[1em] \Rightarrow \Big(\dfrac{1}{p - q}\Big)\Big(\dfrac{1}{p} - \dfrac{1}{q}\Big)\\[1em] \Rightarrow \Big(\dfrac{1}{p - q}\Big)\Big(\dfrac{q - p}{pq}\Big)\\[1em] \Rightarrow -\Big(\dfrac{1}{p - q}\Big)\Big(\dfrac{p - q}{pq} \Big)\\[1em] \Rightarrow -\dfrac{(p - q)}{pq(p - q)}\\[1em] \Rightarrow -\dfrac{1}{pq}\\[1em] \Rightarrow -(pq)^{-1}

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

So, assertion (A) is true.

Multiplying a-1 and a1, we get :

⇒ a-1 x a1

⇒ 1a\dfrac{1}{a} x a

⇒ 1.

Thus, a-1 and a1 are multiplicative inverses of each other.

So, reason is true but it does not explains assertion.

Hence, option 2 is the correct option.

Question 2(i)

Evaluate:

(−35)3\Big(-\dfrac{3}{5}\Big)^3

Answer

(−35)3=(−3×3×35×5×5)=−27125\Big(-\dfrac{3}{5}\Big)^3\\[1em] = \Big(-\dfrac{3 \times 3 \times 3}{5 \times 5 \times 5}\Big)\\[1em] = -\dfrac{27}{125}

Hence, (−35)3=−27125\Big(-\dfrac{3}{5}\Big)^3 = -\dfrac{27}{125}.

Question 2(ii)

Evaluate:

(27)−2\Big(\dfrac{2}{7}\Big)^{-2}

Answer

(27)−2=(72)2=(7×72×2)=494=2414\Big(\dfrac{2}{7}\Big)^{-2}\\[1em] = \Big(\dfrac{7}{2}\Big)^2\\[1em] = \Big(\dfrac{7 \times 7}{2 \times 2}\Big)\\[1em] = \dfrac{49}{4}\\[1em] = 24\dfrac{1}{4}

Hence, (27)−2=2414\Big(\dfrac{2}{7}\Big)^{-2} = 24\dfrac{1}{4}.

Question 3(i)

Evaluate:

[−25]4×[−52]2\Big[-\dfrac{2}{5}\Big]^4 \times \Big[-\dfrac{5}{2}\Big]^2

Answer

[−25]4×[−52]2=[−2454]×[−5222]=[−2454]×[−2−25−2]=[24−254−2]=[2252]=[425]\Big[-\dfrac{2}{5}\Big]^4 \times \Big[-\dfrac{5}{2}\Big]^2\\[1em] = \Big[\dfrac{-2^4}{5^4}\Big] \times \Big[\dfrac{-5^2}{2^2}\Big]\\[1em] = \Big[\dfrac{-2^4}{5^4}\Big] \times \Big[\dfrac{-2^{-2}}{5^{-2}}\Big]\\[1em] = \Big[\dfrac{2^{4-2}}{5^{4-2}}\Big]\\[1em] = \Big[\dfrac{2^2}{5^2}\Big]\\[1em] = \Big[\dfrac{4}{25}\Big]

Hence,[−25]4×[−52]2=[425]\Big[-\dfrac{2}{5}\Big]^4 \times \Big[-\dfrac{5}{2}\Big]^2 = \Big[\dfrac{4}{25}\Big].

Question 3(ii)

Evaluate:

[−37]−5×[73]2\Big[-\dfrac{3}{7}\Big]^{-5} \times \Big[\dfrac{7}{3}\Big]^2

Answer

[−37]−5×[73]2=[−73]5×[73]2=[−7535]×[7232]=[−75+235+2]=[−7737]\Big[-\dfrac{3}{7}\Big]^{-5} \times \Big[\dfrac{7}{3}\Big]^2\\[1em] = \Big[-\dfrac{7}{3}\Big]^5 \times \Big[\dfrac{7}{3}\Big]^2\\[1em] = \Big[-\dfrac{7^5}{3^5}\Big] \times \Big[\dfrac{7^2}{3^2}\Big]\\[1em] = \Big[-\dfrac{7^{5+2}}{3^{5+2}}\Big]\\[1em] = \Big[-\dfrac{7^{7}}{3^{7}}\Big]

Hence,[−37]−5×[73]2=[−7737]\Big[-\dfrac{3}{7}\Big]^{-5} \times \Big[\dfrac{7}{3}\Big]^2 = \Big[-\dfrac{7^{7}}{3^{7}}\Big].

Question 4(i)

Evaluate:

{(−12)−2}−3\Big\lbrace\Big(-\dfrac{1}{2}\Big)^{-2}\Big\rbrace^{-3}

Answer

{(−12)−2}−3={(−21)2}−3={(−2×21×1)}−3={41}−3={14}3={1×1×14×4×4}=164\Big\lbrace\Big(-\dfrac{1}{2}\Big)^{-2}\Big\rbrace^{-3}\\[1em] = \Big\lbrace\Big(-\dfrac{2}{1}\Big)^2\Big\rbrace^{-3}\\[1em] = \Big\lbrace\Big(-\dfrac{2 \times 2}{1 \times 1}\Big)\Big\rbrace^{-3}\\[1em] = \Big\lbrace\dfrac{4}{1}\Big\rbrace^{-3}\\[1em] = \Big\lbrace\dfrac{1}{4}\Big\rbrace^3\\[1em] = \Big\lbrace\dfrac{1 \times 1 \times 1 }{4 \times 4 \times 4}\Big\rbrace\\[1em] = \dfrac{1}{64}

{(−12)−2}−3=164\Big\lbrace\Big(-\dfrac{1}{2}\Big)^{-2}\Big\rbrace^{-3} = \dfrac{1}{64}

Question 4(ii)

Evaluate:

[{(−15)2}−2]−1\Big[\Big\lbrace\Big(-\dfrac{1}{5}\Big)^2\Big\rbrace^{-2}\Big]^{-1}

Answer

[{(−15)2}−2]−1=[{(−1×−15×5)}−2]−1=[{125}−2]−1=[{251}2]−1=[{25×251×1}]−1=[6251]−1=1625\Big[\Big\lbrace\Big(-\dfrac{1}{5}\Big)^2\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\Big(\dfrac{-1 \times -1}{5 \times 5}\Big)\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\dfrac{1}{25}\Big\rbrace^{-2}\Big]^{-1}\\[1em] = \Big[\Big\lbrace\dfrac{25}{1}\Big\rbrace^2\Big]^{-1}\\[1em] = \Big[\Big\lbrace\dfrac{25 \times 25}{1 \times 1}\Big\rbrace\Big]^{-1}\\[1em] = \Big[\dfrac{625}{1}\Big]^{-1}\\[1em] = \dfrac{1}{625}

[{(−15)2}−2]−1=1625\Big[\Big\lbrace\Big(-\dfrac{1}{5}\Big)^2\Big\rbrace^{-2}\Big]^{-1} = \dfrac{1}{625}

Question 5(i)

Evaluate:

(7−1−8−1)−(3−1−4−1)−1(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1}

Answer

(7−1−8−1)−(3−1−4−1)−1=(17−18)−(13−14)−1(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1}\\[1em] = \Big(\dfrac{1}{7} - \dfrac{1}{8}\Big) - \Big(\dfrac{1}{3} - \dfrac{1}{4}\Big)^{-1}

LCM of 7 and 8 is 2 x 2 x 2 x 7 = 56

And

LCM of 3 and 4 is 2 x 2 x 3 = 12

=(1×87×8−1×78×7)−(1×43×4−1×34×3)−1=(856−756)−(412−312)−1=(8−756)−(4−312)−1=(156)−(112)−1=(156)−(121)1= \Big(\dfrac{1 \times 8}{7 \times 8} - \dfrac{1 \times 7}{8 \times 7}\Big) - \Big(\dfrac{1 \times 4}{3 \times 4} - \dfrac{1 \times 3}{4 \times 3}\Big)^{-1}\\[1em] = \Big(\dfrac{8}{56} - \dfrac{7}{56}\Big) - \Big(\dfrac{4}{12} - \dfrac{3}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{8 - 7}{56}\Big) - \Big(\dfrac{4 - 3}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{1}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{12}{1}\Big)^1

LCM of 56 and 1 is 56

=(156)−(12×561×56)=(156)−(67256)=(1−67256)=−67156=−115556= \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{12 \times 56}{1 \times 56}\Big)\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{672}{56}\Big)\\[1em] = \Big(\dfrac{1 - 672}{56}\Big)\\[1em] = -\dfrac{671}{56}\\[1em] = -11\dfrac{55}{56}

(7−1−8−1)−(3−1−4−1)−1=−115556(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1} = -11\dfrac{55}{56}

Question 5(ii)

Evaluate:

5−7÷5−10×5−55^{-7} ÷ 5^{-10} \times 5^{-5}

Answer

5−7÷5−10×5−5=5(−7)−(−10)+(−5)=5−7+10−5=53−5=5−2=152=1255^{-7} ÷ 5^{-10} \times 5^{-5}\\[1em] = 5^{(-7) - (-10) + (-5)}\\[1em] = 5^{-7 + 10 - 5}\\[1em] = 5^{3 - 5}\\[1em] = 5^{-2}\\[1em] = \dfrac{1}{5}^2\\[1em] = \dfrac{1}{25}

5−7÷5−10×5−5=1255^{-7} ÷ 5^{-10} \times 5^{-5} = \dfrac{1}{25}

Question 6

By what number should (−5)−1(-5)^{-1} be divided to give the quotient (−25)−1(-25)^{-1}.

Answer

Let the number be xx. So,

(−5)−1÷x=(−25)−1⇒−15÷x=−125⇒−15×1x=−125⇒−15x=−125⇒−5x=25⇒x=255⇒x=5(-5)^{-1} ÷ x = (-25)^{-1}\\[1em] \Rightarrow \dfrac{-1}{5} ÷ x = \dfrac{-1}{25}\\[1em] \Rightarrow \dfrac{-1}{5} \times \dfrac{1}{x} = \dfrac{-1}{25}\\[1em] \Rightarrow -\dfrac{1}{5x} = -\dfrac{1}{25}\\[1em] \Rightarrow -5x = 25 \\[1em] \Rightarrow x = \dfrac{25}{5} \\[1em] \Rightarrow x = 5

Hence, (-5)-1 should be divided by 5 to give the quotient as (-25)-1.

Question 7

Find nn so that 811÷85=8−3×82n−18^{11} ÷ 8^5 = 8^{-3} \times 8^{2n-1}.

Answer

811÷85=8−3×82n−1⇒811−5=8−3+(2n−1)⇒86=8−3+2n−1⇒86=8−4+2n⇒6=−4+2n⇒6+4=2n⇒10=2n⇒n=102⇒n=58^{11} ÷ 8^5 = 8^{-3} \times 8^{2n-1}\\[1em] \Rightarrow 8^{11 - 5} = 8^{-3 + (2n-1)}\\[1em] \Rightarrow 8^{6} = 8^{-3 + 2n - 1}\\[1em] \Rightarrow 8^{6} = 8^{-4 + 2n}\\[1em] \Rightarrow 6 = -4 + 2n\\[1em] \Rightarrow 6 + 4 = 2n\\[1em] \Rightarrow 10 = 2n\\[1em] \Rightarrow n = \dfrac{10}{2} \\[1em] \Rightarrow n = 5

If 811÷85=8−3×82n−18^{11} ÷ 8^5 = 8^{-3} \times 8^{2n-1}, then n=5n = 5.

Question 8

Find nn so that 9n+2=240+9n9^{n+2} = 240 + 9^n.

Answer

9n+2=240+9n⇒9n+2−9n=240⇒9n[92−1]=240⇒9n[81−1]=240⇒9n[80]=240⇒9n=24080⇒9n=3⇒32n=31⇒2n=1⇒n=129^{n+2} = 240 + 9^n\\[1em] \Rightarrow 9^{n+2} - 9^{n} = 240\\[1em] \Rightarrow 9^{n}[9^2 - 1] = 240\\[1em] \Rightarrow 9^n[81 - 1] = 240\\[1em] \Rightarrow 9^n[80] = 240\\[1em] \Rightarrow 9^n = \dfrac{240}{80}\\[1em] \Rightarrow 9^n = 3\\[1em] \Rightarrow 3^{2n} = 3^1 \\[1em] \Rightarrow 2n = 1 \\[1em] \Rightarrow n = \dfrac{1}{2}

If 9n+2=240+9n9^{n+2} = 240 + 9^n, then n=12n = \dfrac{1}{2}.

Question 9(i)

Find xx, if:

32x−1=(27)x−33^{2x-1} = (27)^{x-3}

Answer

32x−1=(27)x−3⇒32x−1=(33)x−3⇒32x−1=(3)3(x−3)⇒32x−1=(3)3x−9⇒2x−1=3x−9⇒2x−3x=−9+1⇒−x=−8⇒x=83^{2x-1} = (27)^{x-3}\\[1em] \Rightarrow 3^{2x-1} = (3^3)^{x-3}\\[1em] \Rightarrow 3^{2x-1} = (3)^{3(x-3)}\\[1em] \Rightarrow 3^{2x-1} = (3)^{3x-9}\\[1em] \Rightarrow 2x - 1 = 3x - 9\\[1em] \Rightarrow 2x - 3x = -9 + 1\\[1em] \Rightarrow -x = -8\\[1em] \Rightarrow x = 8

If 32x−1=(27)x−33^{2x-1} = (27)^{x-3}, then x=8x = 8.

Question 9(ii)

Find xx, if:

25x×55×(125)35×(625)4=125\dfrac{25^x \times 5^5 \times (125)^3}{5 \times (625)^4} = 125

Answer

25x×55×(125)35×(625)4=125⇒52x×55×(53)35×(54)4=53⇒52x×55×(5)95×(5)16=53⇒52x+5+951+16=53⇒52x+14517=53⇒52x+14−17=53⇒52x−3=53⇒2x−3=3⇒2x=3+3⇒2x=6⇒x=62⇒x=3\dfrac{25^x \times 5^5 \times (125)^3}{5 \times (625)^4} = 125\\[1em] \Rightarrow \dfrac{5^{2x} \times 5^5 \times (5^3)^3}{5 \times (5^4)^4} = 5^3\\[1em] \Rightarrow \dfrac{5^{2x} \times 5^5 \times (5)^9}{5 \times (5)^{16}} = 5^3\\[1em] \Rightarrow \dfrac{5^{2x+5+9}}{5^{1+16}} = 5^3\\[1em] \Rightarrow \dfrac{5^{2x+14}}{5^{17}} = 5^3\\[1em] \Rightarrow 5^{2x+14-17} = 5^3\\[1em] \Rightarrow 5^{2x-3} = 5^3\\[1em] \Rightarrow 2x - 3 = 3\\[1em] \Rightarrow 2x = 3 + 3\\[1em] \Rightarrow 2x = 6\\[1em] \Rightarrow x = \dfrac{6}{2}\\[1em] \Rightarrow x = 3

If 25x×55×(125)35×(625)4=125\dfrac{25^x \times 5^5 \times (125)^3}{5 \times (625)^4} = 125, then x=3x = 3.

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