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Chapter 5

Exponents — Exercise 5(B)

Class - 7 Concise Mathematics Selina



Exercise 5(B)

Question 1

Fill in the blanks :

(i) In 52 = 25, base = .................... and index = ........................

(ii) If index = 3x and base = 2y, the number = ........................

Answer

As we know, in the expression an, a is called the base and n is called the index (or exponent).

(i) In 52 = 25, base = 5 and index = 2

(ii) If index = 3x and base = 2y, the number = (2y)3x

Question 2(i)

Evaluate :

28 ÷ 23

Answer

Solving,

28 ÷ 23

= 2823\dfrac{2^8}{2^3}

= 2(8 - 3)

= 25

= 32

Hence, 28 ÷ 23 = 25 = 32.

Question 2(ii)

Evaluate :

23 ÷ 28

Answer

Solving,

23÷28=2328=1283=125=132\Rightarrow 2^3 ÷ 2^8 \\[1em] = \dfrac{2^3}{2^8} \\[1em] = \dfrac{1}{2^{8-3}} \\[1em] = \dfrac{1}{2^5} \\[1em] = \dfrac{1}{32}

Hence, 23÷28=125=1322^3 ÷ 2^8 = \dfrac{1}{2^5} = \dfrac{1}{32}.

Question 2(iii)

Evaluate :

(26)0

Answer

Solving,

As any non-zero base raised to the power zero is equal to 1,

(26)0=1\Rightarrow (2^6)^0 \\[1em] = 1

Hence, (26)0 = 1.

Question 2(iv)

Evaluate :

(30)6

Answer

Solving,

(30)6=16=1\Rightarrow (3^0)^6 \\[1em] = 1^6 \\[1em] = 1

Hence, (30)6 = 1.

Question 2(v)

Evaluate :

83 × 8-5 × 84

Answer

Solving,

83×85×84=83+(5)+4=82=64\Rightarrow 8^3 \times 8^{-5} \times 8^4 \\[1em] = 8^{3 + (-5) + 4} \\[1em] = 8^2 \\[1em] = 64

Hence, 83 × 8-5 × 84 = 82 = 64.

Question 2(vi)

Evaluate :

54 × 53 ÷ 55

Answer

Solving,

54×53÷55=54×5355=5755=575=52=25\Rightarrow 5^4 \times 5^3 \div 5^5 \\[1em] = \dfrac{5^4 \times 5^3}{5^5} \\[1em] = \dfrac{5^7}{5^5} \\[1em] = 5^{7 - 5} \\[1em] = 5^2 \\[1em] = 25

Hence, 54 × 53 ÷ 55 = 52 = 25.

Question 2(vii)

Evaluate :

54 ÷ 53 × 55

Answer

Solving,

54÷53×55=54×5553=5953=593=56=15625\Rightarrow 5^4 \div 5^3 \times 5^5 \\[1em] = \dfrac{5^4 \times 5^5}{5^3} \\[1em] = \dfrac{5^9}{5^3} \\[1em] = 5^{9 - 3} \\[1em] = 5^6 \\[1em] = 15625

Hence, 54 ÷ 53 × 55 = 56 = 15625.

Question 2(viii)

Evaluate :

44 ÷ 43 × 40

Answer

Solving,

44÷43×40=44×4043=4443=443=41=4\Rightarrow 4^4 \div 4^3 \times 4^0 \\[1em] = \dfrac{4^4 \times 4^0}{4^3} \\[1em] = \dfrac{4^4}{4^3} \\[1em] = 4^{4 - 3} \\[1em] = 4^1 \\[1em] = 4

Hence, 44 ÷ 43 × 40 = 4.

Question 2(ix)

Evaluate :

(35 × 47 × 58)0

Answer

Solving,

As any non-zero base raised to the power zero is equal to 1,

(35×47×58)0=1\Rightarrow (3^5 \times 4^7 \times 5^8)^0 \\[1em] = 1

Hence, (35 × 47 × 58)0 = 1.

Question 3(i)

Simplify, giving answers with positive index :

2b6 · b3 · 5b4

Answer

Solving,

⇒ 2b6 · b3 · 5b4

⇒ (2 × 5) × b6 + 3 + 4

⇒ 10b13.

Hence, 2b6 · b3 · 5b4 = 10b13.

Question 3(ii)

Simplify, giving answers with positive index :

x2y3 · 6x5y · 9x3y4

Answer

Solving,

⇒ x2y3 · 6x5y · 9x3y4

⇒ (1 × 6 × 9) × x2 + 5 + 3 × y3 + 1 + 4

⇒ 54x10y8.

Hence, x2y3 · 6x5y · 9x3y4 = 54x10y8.

Question 3(iii)

Simplify, giving answers with positive index :

(-a5) (a2)

Answer

Solving,

⇒ (-a5) (a2)

⇒ (-1) (a5) (a2)

⇒ -a5 + 2

⇒ -a7.

Hence, (-a5) (a2) = -a7.

Question 3(iv)

Simplify, giving answers with positive index :

(-y2) (-y3)

Answer

Solving,

⇒ (-y2) (-y3)

⇒ (-1) × (-1) × y2 + 3

⇒ y5.

Hence, (-y2) (-y3) = y5.

Question 3(v)

Simplify, giving answers with positive index :

(-3)2 (3)3

Answer

Solving,

⇒ (-3)2 (3)3

⇒ (-1)2 (3)2 (3)3

⇒ 32 × 33

⇒ 32 + 3

⇒ 35.

Hence, (-3)2 (3)3 = 35.

Question 3(vi)

Simplify, giving answers with positive index :

(-4x) (-5x2)

Answer

Solving,

⇒ (-4x) (-5x2)

⇒ (-4) × (-5) × x1 + 2

⇒ 20x3.

Hence, (-4x) (-5x2) = 20x3.

Question 3(vii)

Simplify, giving answers with positive index :

(5a2b) (2ab2) (a3b)

Answer

Solving,

⇒ (5a2b) (2ab2) (a3b)

⇒ (5 × 2 × 1) × a2 + 1 + 3 × b1 + 2 + 1

⇒ 10a6b4.

Hence, (5a2b) (2ab2) (a3b) = 10a6b4.

Question 3(viii)

Simplify, giving answers with positive index :

x2a + 7 · x2a - 8

Answer

Solving,

⇒ x2a + 7 · x2a - 8

⇒ x(2a + 7) + (2a - 8)

⇒ x4a - 1.

Hence, x2a + 7 · x2a - 8 = x4a - 1.

Question 3(ix)

Simplify, giving answers with positive index :

3y · 32 · 3-4

Answer

Solving,

⇒ 3y · 32 · 3-4

⇒ 3y + 2 + (-4)

⇒ 3y - 2.

Hence, 3y · 32 · 3-4 = 3y - 2.

Question 3(x)

Simplify, giving answers with positive index :

24a · 23a · 2-a

Answer

Solving,

⇒ 24a · 23a · 2-a

⇒ 24a + 3a + (-a)

⇒ 26a.

Hence, 24a · 23a · 2-a = 26a.

Question 3(xi)

Simplify, giving answers with positive index :

4x2y2 ÷ 9x3y3

Answer

Solving,

4x2y2÷9x3y3=4x2y29x3y3=49×x23×y23=49×x1×y1=49xy\Rightarrow 4x^2y^2 ÷ 9x^3y^3\\[1em] = \dfrac{4x^2y^2}{9x^3y^3}\\[1em] = \dfrac{4}{9} \times x^{2-3} \times y^{2-3}\\[1em] = \dfrac{4}{9} \times x^{-1} \times y^{-1}\\[1em] = \dfrac{4}{9xy}

Hence, 4x2y2÷9x3y3=49xy4x^2y^2 ÷ 9x^3y^3 = \dfrac{4}{9xy}.

Question 3(xii)

Simplify, giving answers with positive index :

(102)3 (x8)12

Answer

Solving,

⇒ (102)3 (x8)12

⇒ 102 × 3 × x8 × 12

⇒ 106x96.

Hence, (102)3 (x8)12 = 106x96.

Question 3(xiii)

Simplify, giving answers with positive index :

(a10)10 (16)10

Answer

Solving,

⇒ (a10)10 (16)10

⇒ a10 × 10 × 16 × 10

⇒ a100 × 1

⇒ a100.

Hence, (a10)10 (16)10 = a100.

Question 3(xiv)

Simplify, giving answers with positive index :

(n2)2 (-n2)3

Answer

Solving,

⇒ (n2)2 (-n2)3

⇒ n2 × 2 × (-1)3 × n2 × 3

⇒ n4 × (-1) × n6

⇒ -n4 + 6

⇒ -n10.

Hence, (n2)2 (-n2)3 = -n10.

Question 3(xv)

Simplify, giving answers with positive index :

  • (3ab)2 (-5a2bc4)2

Answer

Solving,

⇒ - (3ab)2 (-5a2bc4)2

⇒ - [32a2b2] × [(-5)2a2 × 2b2c4 × 2]

⇒ - [9a2b2] × [25a4b2c8]

⇒ - (9 × 25) × a2 + 4 × b2 + 2 × c8

⇒ -225a6b4c8.

Hence, - (3ab)2 (-5a2bc4)2 = -225a6b4c8.

Question 3(xvi)

Simplify, giving answers with positive index :

(-2)2 × (0)3 × (3)3

Answer

Solving,

As 0 multiplied with any number gives 0,

⇒ (-2)2 × (0)3 × (3)3

⇒ 4 × 0 × 27

⇒ 0.

Hence, (-2)2 × (0)3 × (3)3 = 0.

Question 3(xvii)

Simplify, giving answers with positive index :

(2a3)4 (4a2)2

Answer

Solving,

⇒ (2a3)4 (4a2)2

⇒ [24a3 × 4] × [42a2 × 2]

⇒ [16a12] × [16a4]

⇒ (16 × 16) × a12 + 4

⇒ 256a16.

Hence, (2a3)4 (4a2)2 = 256a16.

Question 3(xviii)

Simplify, giving answers with positive index :

(4x2y3)3 ÷ (3x2y3)3

Answer

Solving,

(4x2y3)3÷(3x2y3)3=(4x2y3)3(3x2y3)3=(4x2y33x2y3)3=(43)3=6427\Rightarrow (4x^2y^3)^3 ÷ (3x^2y^3)^3\\[1em] = \dfrac{(4x^2y^3)^3}{(3x^2y^3)^3}\\[1em] = \left(\dfrac{4x^2y^3}{3x^2y^3}\right)^3\\[1em] = \left(\dfrac{4}{3}\right)^3\\[1em] = \dfrac{64}{27}

Hence, (4x2y3)3÷(3x2y3)3=6427(4x^2y^3)^3 ÷ (3x^2y^3)^3 = \dfrac{64}{27}.

Question 3(xix)

Simplify, giving answers with positive index :

(12x)3×(6x)2\left(\dfrac{1}{2x}\right)^3 \times (6x)^2

Answer

Solving,

(12x)3×(6x)2=123x3×62x2=18x3×36x2=36x28x3=368×x23=92×x1=92x\Rightarrow \left(\dfrac{1}{2x}\right)^3 \times (6x)^2\\[1em] = \dfrac{1}{2^3x^3} \times 6^2x^2\\[1em] = \dfrac{1}{8x^3} \times 36x^2\\[1em] = \dfrac{36x^2}{8x^3}\\[1em] = \dfrac{36}{8} \times x^{2-3}\\[1em] = \dfrac{9}{2} \times x^{-1}\\[1em] = \dfrac{9}{2x}

Hence, (12x)3×(6x)2=92x\left(\dfrac{1}{2x}\right)^3 \times (6x)^2 = \dfrac{9}{2x}.

Question 3(xx)

Simplify, giving answers with positive index :

(14ab2c)2÷(32a2bc2)4\left(\dfrac{1}{4ab^2c}\right)^2 \div \left(\dfrac{3}{2a^2bc^2}\right)^4

Answer

Solving,

(14ab2c)2÷(32a2bc2)4=142a2b4c2÷3424a8b4c8=116a2b4c2÷8116a8b4c8=116a2b4c2×16a8b4c881=16a8b4c816×81×a2b4c2=181×a82×b44×c82=181×a6×b0×c6=a6c681\Rightarrow \left(\dfrac{1}{4ab^2c}\right)^2 ÷ \left(\dfrac{3}{2a^2bc^2}\right)^4\\[1em] = \dfrac{1}{4^2a^2b^4c^2} ÷ \dfrac{3^4}{2^4a^8b^4c^8}\\[1em] = \dfrac{1}{16a^2b^4c^2} ÷ \dfrac{81}{16a^8b^4c^8}\\[1em] = \dfrac{1}{16a^2b^4c^2} \times \dfrac{16a^8b^4c^8}{81}\\[1em] = \dfrac{16a^8b^4c^8}{16 \times 81 \times a^2b^4c^2}\\[1em] = \dfrac{1}{81} \times a^{8-2} \times b^{4-4} \times c^{8-2}\\[1em] = \dfrac{1}{81} \times a^6 \times b^0 \times c^6\\[1em] = \dfrac{a^6c^6}{81}

Hence, (14ab2c)2÷(32a2bc2)4=a6c681\left(\dfrac{1}{4ab^2c}\right)^2 ÷ \left(\dfrac{3}{2a^2bc^2}\right)^4 = \dfrac{a^6c^6}{81}.

Question 3(xxi)

Simplify, giving answers with positive index :

(5x7)3(10x2)2(2x6)7\dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7}

Answer

Solving,

(5x7)3(10x2)2(2x6)7=53x21102x427x42=125x21100x4128x42=125×100128×x21+4x42=12500128×x2542=312532×x17=312532x17\Rightarrow \dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7}\\[1em] = \dfrac{5^3x^{21} \cdot 10^2x^4}{2^7x^{42}}\\[1em] = \dfrac{125x^{21} \cdot 100x^4}{128x^{42}}\\[1em] = \dfrac{125 \times 100}{128} \times \dfrac{x^{21 + 4}}{x^{42}}\\[1em] = \dfrac{12500}{128} \times x^{25 - 42}\\[1em] = \dfrac{3125}{32} \times x^{-17}\\[1em] = \dfrac{3125}{32x^{17}}

Hence, (5x7)3(10x2)2(2x6)7=312532x17\dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7} = \dfrac{3125}{32x^{17}}.

Question 3(xxii)

Simplify, giving answers with positive index :

(7p2q9r5)2(4pqr)3(14p6q10r4)2\dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2}

Answer

Solving,

(7p2q9r5)2(4pqr)3(14p6q10r4)2=72p4q18r10×43p3q3r3142p12q20r8=49p4q18r10×64p3q3r3196p12q20r8=49×64196×p4+3q18+3r10+3p12q20r8=3136196×p712×q2120×r138=16×p5×q1×r5=16qr5p5\Rightarrow \dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2}\\[1em] = \dfrac{7^2p^4q^{18}r^{10} \times 4^3p^3q^3r^3}{14^2p^{12}q^{20}r^8}\\[1em] = \dfrac{49p^4q^{18}r^{10} \times 64p^3q^3r^3}{196p^{12}q^{20}r^8}\\[1em] = \dfrac{49 \times 64}{196} \times \dfrac{p^{4 + 3}q^{18 + 3}r^{10 + 3}}{p^{12}q^{20}r^8}\\[1em] = \dfrac{3136}{196} \times p^{7 - 12} \times q^{21 - 20} \times r^{13 - 8}\\[1em] = 16 \times p^{-5} \times q^1 \times r^5\\[1em] = \dfrac{16qr^5}{p^5}

Hence, (7p2q9r5)2(4pqr)3(14p6q10r4)2=16qr5p5\dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2} = \dfrac{16qr^5}{p^5}.

Question 4(i)

Simplify and express the answer in the positive exponent form :

(3)3×266×23\dfrac{(-3)^3 \times 2^6}{6 \times 2^3}

Answer

Solving,

(3)3×266×23=(1)3×33×262×3×23=33×263×24=331×264=32×22=(22×32)\Rightarrow \dfrac{(-3)^3 \times 2^6}{6 \times 2^3}\\[1em] = \dfrac{(-1)^3 \times 3^3 \times 2^6}{2 \times 3 \times 2^3}\\[1em] = \dfrac{-3^3 \times 2^6}{3 \times 2^4}\\[1em] = -3^{3-1} \times 2^{6-4}\\[1em] = -3^2 \times 2^2\\[1em] = -(2^2 \times 3^2)

Hence, (3)3×266×23=(22×32)\dfrac{(-3)^3 \times 2^6}{6 \times 2^3} = -(2^2 \times 3^2).

Question 4(ii)

Simplify and express the answer in the positive exponent form :

(23)5×5443×52\dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2}

Answer

Solving,

(23)5×5443×52=23×5×54(22)3×52=215×5426×52=2156×542=29×52\Rightarrow \dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2}\\[1em] = \dfrac{2^{3 \times 5} \times 5^4}{(2^2)^3 \times 5^2}\\[1em] = \dfrac{2^{15} \times 5^4}{2^6 \times 5^2}\\[1em] = 2^{15-6} \times 5^{4-2}\\[1em] = 2^9 \times 5^2

Hence, (23)5×5443×52=29×52\dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2} = 2^9 \times 5^2.

Question 4(iii)

Simplify and express the answer in the positive exponent form :

36×(6)2×36123×35\dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5}

Answer

Solving,

36×(6)2×36123×35=(22×32)×(22×32)×36(22×3)3×35=24×31026×33×35=24×31026×38=246×3108=22×32=3222=(32)2\Rightarrow \dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5}\\[1em] = \dfrac{(2^2 \times 3^2) \times (2^2 \times 3^2) \times 3^6}{(2^2 \times 3)^3 \times 3^5}\\[1em] = \dfrac{2^4 \times 3^{10}}{2^6 \times 3^3 \times 3^5}\\[1em] = \dfrac{2^4 \times 3^{10}}{2^6 \times 3^8}\\[1em] = 2^{4-6} \times 3^{10-8}\\[1em] = 2^{-2} \times 3^2\\[1em] = \dfrac{3^2}{2^2}\\[1em] = \left(\dfrac{3}{2}\right)^2

Hence, 36×(6)2×36123×35=(32)2\dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5} = \left(\dfrac{3}{2}\right)^2.

Question 4(iv)

Simplify and express the answer in the positive exponent form :

1282187-\dfrac{128}{2187}

Answer

Solving,

1282187=2×2×2×2×2×2×23×3×3×3×3×3×3=2737=(23)7=(23)7\Rightarrow -\dfrac{128}{2187}\\[1em] = -\dfrac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}\\[1em] = -\dfrac{2^7}{3^7}\\[1em] = -\left(\dfrac{2}{3}\right)^7\\[1em] = \left(-\dfrac{2}{3}\right)^7

Hence, 1282187=(23)7-\dfrac{128}{2187} = \left(-\dfrac{2}{3}\right)^7.

Question 4(v)

Simplify and express the answer in the positive exponent form :

a7×b7×c5×d4a3×b5×c3×d8\dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8}

Answer

Solving,

a7×b7×c5×d4a3×b5×c3×d8=a73×b7(5)×c5(3)×d48=a10×b2×c8×d4=c8a10×b2×d4\Rightarrow \dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8}\\[1em] = a^{-7-3} \times b^{-7-(-5)} \times c^{5-(-3)} \times d^{4-8}\\[1em] = a^{-10} \times b^{-2} \times c^8 \times d^{-4}\\[1em] = \dfrac{c^8}{a^{10} \times b^2 \times d^4}

Hence, a7×b7×c5×d4a3×b5×c3×d8=c8a10b2d4\dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8} = \dfrac{c^8}{a^{10}b^2d^4}.

Question 4(vi)

Simplify and express the answer in the positive exponent form :

(a3b-5)-2

Answer

Solving,

(a3b5)2=a3×(2)×b5×(2)=a6×b10=b10a6\Rightarrow (a^3b^{-5})^{-2} \\[1em] = a^{3 \times (-2)} \times b^{-5 \times (-2)} \\[1em] = a^{-6} \times b^{10} \\[1em] = \dfrac{b^{10}}{a^6}

Hence, (a3b5)2=b10a6(a^3b^{-5})^{-2} = \dfrac{b^{10}}{a^6}.

Question 5(i)

Evaluate :

6-2 ÷ (4-2 × 3-2)

Answer

Solving,

62÷(42×32)=162÷(142×132)=136÷(116×19)=136÷1144=136×144=4\Rightarrow 6^{-2} ÷ (4^{-2} \times 3^{-2})\\[1em] = \dfrac{1}{6^2} ÷ \left(\dfrac{1}{4^2} \times \dfrac{1}{3^2}\right)\\[1em] = \dfrac{1}{36} ÷ \left(\dfrac{1}{16} \times \dfrac{1}{9}\right)\\[1em] = \dfrac{1}{36} ÷ \dfrac{1}{144}\\[1em] = \dfrac{1}{36} \times 144\\[1em] = 4

Hence, 6-2 ÷ (4-2 × 3-2) = 4.

Question 5(ii)

Evaluate :

[(56)2×94]÷[(32)2×125216]\left[\left(\dfrac{5}{6}\right)^2 \times \dfrac{9}{4}\right] \div \left[\left(-\dfrac{3}{2}\right)^2 \times \dfrac{125}{216}\right]

Answer

Solving,

[(56)2×94]÷[(32)2×125216]=[2536×94]÷[94×125216]=2516÷12596=2516×96125=25×9616×125=65=115\Rightarrow \left[\left(\dfrac{5}{6}\right)^2 \times \dfrac{9}{4}\right] \div \left[\left(-\dfrac{3}{2}\right)^2 \times \dfrac{125}{216}\right]\\[1em] = \left[\dfrac{25}{36} \times \dfrac{9}{4}\right] \div \left[\dfrac{9}{4} \times \dfrac{125}{216}\right]\\[1em] = \dfrac{25}{16} \div \dfrac{125}{96}\\[1em] = \dfrac{25}{16} \times \dfrac{96}{125}\\[1em] = \dfrac{25 \times 96}{16 \times 125}\\[1em] = \dfrac{6}{5}\\[1em] = 1\dfrac{1}{5}

Hence, the value is 1151\dfrac{1}{5}.

Question 5(iii)

Evaluate :

53 × 32 + (17)0 × 73

Answer

Solving,

53×32+(17)0×73=(125×9)+(1×343)=1125+343=1468\Rightarrow 5^3 \times 3^2 + (17)^0 \times 7^3 \\[1em] = (125 \times 9) + (1 \times 343) \\[1em] = 1125 + 343 \\[1em] = 1468

Hence, 53 × 32 + (17)0 × 73 = 1468.

Question 5(iv)

Evaluate :

25 × 150 + (-3)3 - (27)2\left(\dfrac{2}{7}\right)^{-2}

Answer

Solving,

25×150+(3)3(27)2=(32×1)+(27)(72)2=3227494=5494=204494=20494=294\Rightarrow 2^5 \times 15^0 + (-3)^3 - \left(\dfrac{2}{7}\right)^{-2}\\[1em] = (32 \times 1) + (-27) - \left(\dfrac{7}{2}\right)^2\\[1em] = 32 - 27 - \dfrac{49}{4}\\[1em] = 5 - \dfrac{49}{4}\\[1em] = \dfrac{20}{4} - \dfrac{49}{4}\\[1em] = \dfrac{20 - 49}{4}\\[1em] = -\dfrac{29}{4}

Hence, the value is 294-\dfrac{29}{4}.

Question 5(v)

Evaluate :

(22)0 + 2-4 ÷ 2-6 + (12)3\left(\dfrac{1}{2}\right)^{-3}

Answer

Solving,

(22)0+24÷26+(12)3=1+24(6)+23=1+22+8=1+4+8=13\Rightarrow (2^2)^0 + 2^{-4} \div 2^{-6} + \left(\dfrac{1}{2}\right)^{-3} \\[1em] = 1 + 2^{-4 - (-6)} + 2^3 \\[1em] = 1 + 2^2 + 8 \\[1em] = 1 + 4 + 8 \\[1em] = 13

Hence, the value is 13.

Question 5(vi)

Evaluate :

5n × 25n-1 ÷ (5n-1 × 25n-1)

Answer

Solving,

5n×25n1÷(5n1×25n1)=5n×25n15n1×25n1=5n5n1=5n(n1)=51=5\Rightarrow 5^n \times 25^{n-1} ÷ (5^{n-1} \times 25^{n-1})\\[1em] = \dfrac{5^n \times 25^{n-1}}{5^{n-1} \times 25^{n-1}}\\[1em] = \dfrac{5^n}{5^{n-1}}\\[1em] = 5^{n-(n-1)}\\[1em] = 5^1\\[1em] = 5

Hence, 5n × 25n-1 ÷ (5n-1 × 25n-1) = 5.

Question 6(i)

If m = - 2 and n = 2; find the value of :

m2 + n2 - 2mn

Answer

Solving,

m2+n22mn=(2)2+(2)22×(2)×(2)=4+4(8)=8+8=16\Rightarrow m^2 + n^2 - 2mn \\[1em] = (-2)^2 + (2)^2 - 2 \times (-2) \times (2) \\[1em] = 4 + 4 - (-8) \\[1em] = 8 + 8 \\[1em] = 16

Hence, m2 + n2 - 2mn = 16.

Question 6(ii)

If m = - 2 and n = 2; find the value of :

mn + nm

Answer

Solving,

mn+nm=(2)2+(2)2=4+122=4+14=164+14=174=414\Rightarrow m^n + n^m\\[1em] = (-2)^2 + (2)^{-2}\\[1em] = 4 + \dfrac{1}{2^2}\\[1em] = 4 + \dfrac{1}{4}\\[1em] = \dfrac{16}{4} + \dfrac{1}{4}\\[1em] = \dfrac{17}{4}\\[1em] = 4\dfrac{1}{4}

Hence, mn+nm=414m^n + n^m = 4\dfrac{1}{4}.

Question 6(iii)

If m = - 2 and n = 2; find the value of :

6m-3 + 4n2

Answer

Solving,

6m3+4n2=6×(2)3+4×(2)2=6(2)3+4×4=68+16=34+16=34+644=614=1514\Rightarrow 6m^{-3} + 4n^2\\[1em] = 6 \times (-2)^{-3} + 4 \times (2)^2\\[1em] = \dfrac{6}{(-2)^3} + 4 \times 4\\[1em] = \dfrac{6}{-8} + 16\\[1em] = -\dfrac{3}{4} + 16\\[1em] = -\dfrac{3}{4} + \dfrac{64}{4}\\[1em] = \dfrac{61}{4}\\[1em] = 15\dfrac{1}{4}

Hence, 6m3+4n2=15146m^{-3} + 4n^2 = 15\dfrac{1}{4}.

Question 6(iv)

If m = - 2 and n = 2; find the value of :

2n3 - 3m

Answer

Solving,

2n33m=2×(2)33×(2)=2×8(6)=16+6=22\Rightarrow 2n^3 - 3m \\[1em] = 2 \times (2)^3 - 3 \times (-2) \\[1em] = 2 \times 8 - (-6) \\[1em] = 16 + 6 \\[1em] = 22

Hence, 2n3 - 3m = 22.

Question 7

State true or false :

(i) 8 × 815 = 6416

(ii) 168 ÷ 42 = 46

(iii) 270 = 549030

(iv) (-1)n = 1, if n is an even whole number

(v) (-1)n = -1, if n is an odd or even whole number

(vi) (-3)-3 = +9

(vii) 4-4 = -16

Answer

(i) False. 8 × 815 = 81 + 15 = 816, whereas 6416 = (82)16 = 832. Since 816 ≠ 832, the statement is false.

(ii) False. 168 ÷ 42 = (42)8 ÷ 42 = 416 ÷ 42 = 414, which is not equal to 46.

(iii) True. Any non-zero base raised to the power zero is equal to 1, so 270 = 1 = 549030.

(iv) True. When n is an even whole number, (-1)n = 1.

(v) False. Here the expression is read as (-1)n. For an even whole number n, (-1)n = 1 and not -1; it equals -1 only when n is odd. So the value is not -1 for every odd or even whole number n.

(vi) False. (3)3=1(3)3=127=127(-3)^{-3} = \dfrac{1}{(-3)^3} = \dfrac{1}{-27} = -\dfrac{1}{27}, which is not +9.

(vii) False. 44=144=12564^{-4} = \dfrac{1}{4^4} = \dfrac{1}{256}, which is not -16.

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