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Chapter 2

Exponents and Powers — Exercise 2(A)

Class - 8 RS Aggarwal Mathematics Solutions

Exercise 2A

Question 1(i)

Simplify:

a6×a8a^6 \times a^8

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m + n}

Solving,

⇒a6×a8⇒a6+8⇒a14\Rightarrow a^6 \times a^8\\[1em] \Rightarrow a^{6+8}\\[1em] \Rightarrow a^{14}

Hence, a6×a8=a14\bm{a^6 \times a^8 = a^{14}}.

Question 1(ii)

Simplify:

x5×x−3x^5 \times x^{-3}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n}

Solving,

⇒x5×x−3⇒x5+(−3)⇒x2\Rightarrow x^5 \times x^{-3}\\[1em] \Rightarrow x^{5+(-3)}\\[1em] \Rightarrow x^{2}

Hence, x5×x−3=x2\bm{x^5 \times x^{-3} = x^{2}}.

Question 1(iii)

Simplify:

z9×z3×z−6z^9 \times z^3 \times z^{-6}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n}

Solving,

⇒z9×z3×z−6⇒z9+3+(−6)⇒z6\Rightarrow z^9 \times z^3 \times z^{-6}\\[1em] \Rightarrow z^{9+3+(-6)}\\[1em] \Rightarrow z^{6}

Hence, z9×z3×z−6=z6\bm{z^9 \times z^3 \times z^{-6} = z^{6}}.

Question 1(iv)

Simplify:

a2b3×a5b2a^2b^3 \times a^5b^2

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n}

Solving,

⇒a2b3×a5b2⇒a2+5×b3+2⇒a7b5\Rightarrow a^2b^3 \times a^5b^2\\[1em] \Rightarrow a^{2+5} \times b^{3+2}\\[1em] \Rightarrow a^{7}b^{5}

Hence, a2b3×a5b2=a7b5\bm{a^2b^3 \times a^5b^2 = a^{7}b^{5}}.

Question 1(v)

Simplify:

5x7×3x45x^7 \times 3x^4

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n}

Solving,

⇒5x7×3x4⇒(5×3)×x7+4⇒15x11\Rightarrow 5x^7 \times 3x^4\\[1em] \Rightarrow (5 \times 3) \times x^{7+4}\\[1em] \Rightarrow 15x^{11}

Hence, 5x7×3x4=15x11\bm{5x^7 \times 3x^4 = 15x^{11}}.

Question 1(vi)

Simplify:

p3q4×p5q−5p^3q^4 \times p^5q^{-5}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒p3q4×p5q−5⇒p3+5×q4+(−5)⇒p8q−1⇒p8q\Rightarrow p^3q^4 \times p^5q^{-5}\\[1em] \Rightarrow p^{3+5} \times q^{4+(-5)}\\[1em] \Rightarrow p^{8}q^{-1}\\[1em] \Rightarrow \dfrac{p^8}{q}

Hence, p3q4×p5q−5=p8q\bm{p^3q^4 \times p^5q^{-5} = \dfrac{p^8}{q}}.

Question 1(vii)

Simplify:

x7y−5×x−5y3x^7y^{-5} \times x^{-5}y^3

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒x7y−5×x−5y3⇒x7+(−5)×y−5+3⇒x2y−2⇒x2y2\Rightarrow x^7y^{-5} \times x^{-5}y^3\\[1em] \Rightarrow x^{7+(-5)} \times y^{-5+3}\\[1em] \Rightarrow x^{2}y^{-2}\\[1em] \Rightarrow \dfrac{x^2}{y^2}

Hence, x7y−5×x−5y3=x2y2\bm{x^7y^{-5} \times x^{-5}y^3 = \dfrac{x^2}{y^2}}.

Question 1(viii)

Simplify:

x−2y5×x0y−7x^{-2}y^5 \times x^0y^{-7}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+n,a0=1a^m \times a^n = a^{m+n}, a^0 = 1 and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒x−2y5×x0y−7⇒x−2+0×y5+(−7)⇒x−2y−2⇒1x2y2\Rightarrow x^{-2}y^5 \times x^0y^{-7}\\[1em] \Rightarrow x^{-2+0} \times y^{5+(-7)}\\[1em] \Rightarrow x^{-2}y^{-2}\\[1em] \Rightarrow \dfrac{1}{x^2y^2}

Hence, x−2y5×x0y−7=1x2y2\bm{x^{-2}y^5 \times x^0y^{-7} = \dfrac{1}{x^2y^2}}.

Question 1(ix)

Simplify:

x6y4z−2×x−3y−5z−1×x2z4x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+na^m \times a^n = a^{m+n} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒x6y4z−2×x−3y−5z−1×x2z4⇒x6+(−3)+2×y4+(−5)×z−2+(−1)+4⇒x5y−1z1⇒x5zy\Rightarrow x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4\\[1em] \Rightarrow x^{6+(-3)+2} \times y^{4+(-5)} \times z^{-2+(-1)+4}\\[1em] \Rightarrow x^{5}y^{-1}z^{1}\\[1em] \Rightarrow \dfrac{x^5z}{y}

Hence, x6y4z−2×x−3y−5z−1×x2z4=x5zy\bm{x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4 = \dfrac{x^5z}{y}}.

Question 2(i)

Simplify:

x12x7\dfrac{x^{12}}{x^7}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒x12x7⇒x12−7⇒x5\Rightarrow \dfrac{x^{12}}{x^7}\\[1em] \Rightarrow x^{12-7}\\[1em] \Rightarrow x^{5}

Hence, x12x7=x5\bm{\dfrac{x^{12}}{x^7} = x^{5}}.

Question 2(ii)

Simplify:

z6z−3\dfrac{z^6}{z^{-3}}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒z6z−3⇒z6−(−3)⇒z9\Rightarrow \dfrac{z^6}{z^{-3}}\\[1em] \Rightarrow z^{6-(-3)}\\[1em] \Rightarrow z^{9}

Hence, z6z−3=z9\bm{\dfrac{z^6}{z^{-3}} = z^{9}}.

Question 2(iii)

Simplify:

m5n3m2n−4\dfrac{m^5n^3}{m^2n^{-4}}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒m5n3m2n−4⇒m5−2×n3−(−4)⇒m3n7\Rightarrow \dfrac{m^5n^3}{m^2n^{-4}}\\[1em] \Rightarrow m^{5-2} \times n^{3-(-4)}\\[1em] \Rightarrow m^{3}n^{7}

Hence, m5n3m2n−4=m3n7\bm{\dfrac{m^5n^3}{m^2n^{-4}} = m^{3}n^{7}}.

Question 2(iv)

Simplify:

18x9÷6x718x^9 \div 6x^7

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒18x9÷6x7⇒18x96x7⇒186×x9−7⇒3x2\Rightarrow 18x^9 \div 6x^7\\[1em] \Rightarrow \dfrac{18x^9}{6x^7}\\[1em] \Rightarrow \dfrac{18}{6} \times x^{9-7}\\[1em] \Rightarrow 3x^{2}

Hence, 18x9÷6x7=3x2\bm{18x^9 \div 6x^7 = 3x^{2}}.

Question 2(v)

Simplify:

7a12÷56a157a^{12} \div 56a^{15}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒7a12÷56a15⇒7a1256a15⇒756×a12−15⇒18×a−3⇒18a3\Rightarrow 7a^{12} \div 56a^{15}\\[1em] \Rightarrow \dfrac{7a^{12}}{56a^{15}}\\[1em] \Rightarrow \dfrac{7}{56} \times a^{12-15}\\[1em] \Rightarrow \dfrac{1}{8} \times a^{-3}\\[1em] \Rightarrow \dfrac{1}{8a^3}

Hence, 7a12÷56a15=18a3\bm{7a^{12} \div 56a^{15} = \dfrac{1}{8a^3}}.

Question 2(vi)

Simplify:

a13b7÷a5b−3a^{13}b^7 \div a^5b^{-3}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒a13b7÷a5b−3⇒a13b7a5b−3⇒a13−5×b7−(−3)⇒a8b10\Rightarrow a^{13}b^7 \div a^5b^{-3}\\[1em] \Rightarrow \dfrac{a^{13}b^7}{a^5b^{-3}}\\[1em] \Rightarrow a^{13-5} \times b^{7-(-3)}\\[1em] \Rightarrow a^{8}b^{10}

Hence, a13b7÷a5b−3=a8b10\bm{a^{13}b^7 \div a^5b^{-3} = a^{8}b^{10}}.

Question 2(vii)

Simplify:

7x14÷21x−107x^{14} \div 21x^{-10}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n}

Solving,

⇒7x14÷21x−10⇒7x1421x−10⇒721×x14−(−10)⇒13×x24⇒x243\Rightarrow 7x^{14} \div 21x^{-10}\\[1em] \Rightarrow \dfrac{7x^{14}}{21x^{-10}}\\[1em] \Rightarrow \dfrac{7}{21} \times x^{14-(-10)}\\[1em] \Rightarrow \dfrac{1}{3} \times x^{24}\\[1em] \Rightarrow \dfrac{x^{24}}{3}

Hence, 7x14÷21x−10=x243\bm{7x^{14} \div 21x^{-10} = \dfrac{x^{24}}{3}}.

Question 2(viii)

Simplify:

p11÷p11p^{11} \div p^{11}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n} and a0=1a^0 = 1

Solving,

⇒p11÷p11⇒p11p11⇒p11−11⇒p0⇒1\Rightarrow p^{11} \div p^{11}\\[1em] \Rightarrow \dfrac{p^{11}}{p^{11}}\\[1em] \Rightarrow p^{11-11}\\[1em] \Rightarrow p^{0}\\[1em] \Rightarrow 1

Hence, p11÷p11=1\bm{p^{11} \div p^{11} = 1}.

Question 2(ix)

Simplify:

a7b5c4a−2b3c6\dfrac{a^7b^5c^4}{a^{-2}b^3c^6}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n\dfrac{a^m}{a^n} = a^{m-n} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒a7b5c4a−2b3c6⇒a7−(−2)×b5−3×c4−6⇒a9b2c−2⇒a9b2c2\Rightarrow \dfrac{a^7b^5c^4}{a^{-2}b^3c^6}\\[1em] \Rightarrow a^{7-(-2)} \times b^{5-3} \times c^{4-6}\\[1em] \Rightarrow a^{9}b^{2}c^{-2}\\[1em] \Rightarrow \dfrac{a^9b^2}{c^2}

Hence, a7b5c4a−2b3c6=a9b2c2\bm{\dfrac{a^7b^5c^4}{a^{-2}b^3c^6} = \dfrac{a^9b^2}{c^2}}.

Question 3(i)

Simplify:

(a3)2(a^3)^2

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn}

Solving,

⇒(a3)2⇒a3×2⇒a6\Rightarrow (a^3)^2\\[1em] \Rightarrow a^{3 \times 2}\\[1em] \Rightarrow a^{6}

Hence, (a3)2=a6\bm{(a^3)^2 = a^{6}}.

Question 3(ii)

Simplify:

(2x2y)4(2x^2y)^4

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm(ab)^m = a^mb^m and (am)n=amn(a^m)^n = a^{mn}

Solving,

⇒(2x2y)4⇒24×x2×4×y4⇒16x8y4\Rightarrow (2x^2y)^4\\[1em] \Rightarrow 2^4 \times x^{2 \times 4} \times y^4\\[1em] \Rightarrow 16x^{8}y^{4}

Hence, (2x2y)4=16x8y4\bm{(2x^2y)^4 = 16x^{8}y^{4}}.

Question 3(iii)

Simplify:

(m2n−3)4(m^2n^{-3})^4

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(m2n−3)4⇒m2×4×n−3×4⇒m8n−12⇒m8n12\Rightarrow (m^2n^{-3})^4\\[1em] \Rightarrow m^{2 \times 4} \times n^{-3 \times 4}\\[1em] \Rightarrow m^{8}n^{-12}\\[1em] \Rightarrow \dfrac{m^8}{n^{12}}

Hence, (m2n−3)4=m8n12\bm{(m^2n^{-3})^4 = \dfrac{m^8}{n^{12}}}.

Question 3(iv)

Simplify:

(x2y−3)−2(x^2y^{-3})^{-2}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(x2y−3)−2⇒x2×(−2)×y−3×(−2)⇒x−4y6⇒y6x4\Rightarrow (x^2y^{-3})^{-2}\\[1em] \Rightarrow x^{2 \times (-2)} \times y^{-3 \times (-2)}\\[1em] \Rightarrow x^{-4}y^{6}\\[1em] \Rightarrow \dfrac{y^6}{x^4}

Hence, (x2y−3)−2=y6x4\bm{(x^2y^{-3})^{-2} = \dfrac{y^6}{x^4}}.

Question 3(v)

Simplify:

(3x3y−3)2(3x^3y^{-3})^2

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(3x3y−3)2⇒32×x3×2×y−3×2⇒9x6y−6⇒9x6y6\Rightarrow (3x^3y^{-3})^2\\[1em] \Rightarrow 3^2 \times x^{3 \times 2} \times y^{-3 \times 2}\\[1em] \Rightarrow 9x^{6}y^{-6}\\[1em] \Rightarrow \dfrac{9x^6}{y^6}

Hence, (3x3y−3)2=9x6y6\bm{(3x^3y^{-3})^2 = \dfrac{9x^6}{y^6}}.

Question 3(vi)

Simplify:

(5m4n−3)3(5m^4n^{-3})^3

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(5m4n−3)3⇒53×m4×3×n−3×3⇒125m12n−9⇒125m12n9\Rightarrow (5m^4n^{-3})^3\\[1em] \Rightarrow 5^3 \times m^{4 \times 3} \times n^{-3 \times 3}\\[1em] \Rightarrow 125m^{12}n^{-9}\\[1em] \Rightarrow \dfrac{125m^{12}}{n^9}

Hence, (5m4n−3)3=125m12n9\bm{(5m^4n^{-3})^3 = \dfrac{125m^{12}}{n^9}}.

Question 4(i)

Evaluate:

(36)1/2

Answer

We know that,

For any non-zero rational number (aa)

amn=(am)1na^{\dfrac{m}{n}} = (a^m)^{\dfrac{1}{n}} and (am)n=amn(a^m)^n = a^{mn}

We know that 36 = 62

Solving,

⇒(36)12⇒(62)12⇒62×12⇒61⇒6\Rightarrow (36)^{\dfrac{1}{2}}\\[1em] \Rightarrow (6^2)^{\dfrac{1}{2}}\\[1em] \Rightarrow 6^{2 \times \dfrac{1}{2}}\\[1em] \Rightarrow 6^1\\[1em] \Rightarrow 6

Hence, (36)12=6\mathbf{(36)^{\dfrac{1}{2}} = 6}.

Question 4(ii)

Evaluate:

(27)2/3

Answer

We know that,

For any non-zero rational number (aa)

amn=(am)1na^{\dfrac{m}{n}} = (a^m)^{\dfrac{1}{n}} and (am)n=amn(a^m)^n = a^{mn}

We know that 27 = 33

Solving,

⇒(27)23⇒(33)23⇒33×23⇒32⇒9\Rightarrow (27)^{\dfrac{2}{3}}\\[1em] \Rightarrow (3^3)^{\dfrac{2}{3}}\\[1em] \Rightarrow 3^{3 \times \dfrac{2}{3}}\\[1em] \Rightarrow 3^2\\[1em] \Rightarrow 9

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

Question 4(iii)

Evaluate:

(16)-3/4

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 16 = 24

Solving,

⇒(16)−34⇒(24)−34⇒24×−34⇒2−3⇒123⇒18\Rightarrow (16)^{\dfrac{-3}{4}}\\[1em] \Rightarrow (2^4)^{\dfrac{-3}{4}}\\[1em] \Rightarrow 2^{4 \times \dfrac{-3}{4}}\\[1em] \Rightarrow 2^{-3}\\[1em] \Rightarrow \dfrac{1}{2^3}\\[1em] \Rightarrow \dfrac{1}{8}

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

Question 4(iv)

Evaluate:

(64)-1/3

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 64 = 43

Solving,

⇒(64)−13⇒(43)−13⇒43×−13⇒4−1⇒14\Rightarrow (64)^{\dfrac{-1}{3}}\\[1em] \Rightarrow (4^3)^{\dfrac{-1}{3}}\\[1em] \Rightarrow 4^{3 \times \dfrac{-1}{3}}\\[1em] \Rightarrow 4^{-1}\\[1em] \Rightarrow \dfrac{1}{4}

Hence, (64)−13=14\mathbf{(64)^{\dfrac{-1}{3}} = \dfrac{1}{4}}.

Question 4(v)

Evaluate:

(81)-1/4

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 81 = 34

Solving,

⇒(81)−14⇒(34)−14⇒34×−14⇒3−1⇒13\Rightarrow (81)^{\dfrac{-1}{4}}\\[1em] \Rightarrow (3^4)^{\dfrac{-1}{4}}\\[1em] \Rightarrow 3^{4 \times \dfrac{-1}{4}}\\[1em] \Rightarrow 3^{-1}\\[1em] \Rightarrow \dfrac{1}{3}

Hence, (81)−14=13\mathbf{(81)^{\dfrac{-1}{4}} = \dfrac{1}{3}}.

Question 4(vi)

Evaluate:

(32)-4/5

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 32 = 25

Solving,

⇒(32)−45⇒(25)−45⇒25×−45⇒2−4⇒124⇒116\Rightarrow (32)^{\dfrac{-4}{5}}\\[1em] \Rightarrow (2^5)^{\dfrac{-4}{5}}\\[1em] \Rightarrow 2^{5 \times \dfrac{-4}{5}}\\[1em] \Rightarrow 2^{-4}\\[1em] \Rightarrow \dfrac{1}{2^4}\\[1em] \Rightarrow \dfrac{1}{16}

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

Question 5(i)

Simplify:

(25a2)12(25a^2)^{\dfrac{1}{2}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm(ab)^m = a^mb^m and (am)n=amn(a^m)^n = a^{mn}

We know that 25 = 52

Solving,

⇒(25a2)12⇒(52)12×a2×12⇒51×a1⇒5a\Rightarrow (25a^2)^{\dfrac{1}{2}}\\[1em] \Rightarrow (5^2)^{\dfrac{1}{2}} \times a^{2 \times \dfrac{1}{2}}\\[1em] \Rightarrow 5^1 \times a^1\\[1em] \Rightarrow 5a

Hence, (25a2)12=5a\bm{(25a^2)^{\dfrac{1}{2}} = 5a}.

Question 5(ii)

Simplify:

(27x−3)13(27x^{-3})^{\dfrac{1}{3}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 27 = 33

Solving,

⇒(27x−3)13⇒(33)13×x−3×13⇒31×x−1⇒3x\Rightarrow (27x^{-3})^{\dfrac{1}{3}}\\[1em] \Rightarrow (3^3)^{\dfrac{1}{3}} \times x^{-3 \times \dfrac{1}{3}}\\[1em] \Rightarrow 3^1 \times x^{-1}\\[1em] \Rightarrow \dfrac{3}{x}

Hence, (27x−3)13=3x\bm{(27x^{-3})^{\dfrac{1}{3}} = \dfrac{3}{x}}.

Question 5(iii)

Simplify:

(64m−6n3)23(64m^{-6}n^3)^{\dfrac{2}{3}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 64 = 43

Solving,

⇒(64m−6n3)23⇒(43)23×m−6×23×n3×23⇒42×m−4×n2⇒16m−4n2⇒16n2m4\Rightarrow (64m^{-6}n^3)^{\dfrac{2}{3}}\\[1em] \Rightarrow (4^3)^{\dfrac{2}{3}} \times m^{-6 \times \dfrac{2}{3}} \times n^{3 \times \dfrac{2}{3}}\\[1em] \Rightarrow 4^2 \times m^{-4} \times n^2\\[1em] \Rightarrow 16m^{-4}n^2\\[1em] \Rightarrow \dfrac{16n^2}{m^4}

Hence, (64m−6n3)23=16n2m4\bm{(64m^{-6}n^3)^{\dfrac{2}{3}} = \dfrac{16n^2}{m^4}}.

Question 5(iv)

Simplify:

(81a4b8c−4)14(81a^4b^8c^{-4})^{\dfrac{1}{4}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 81 = 34

Solving,

⇒(81a4b8c−4)14⇒(34)14×a4×14×b8×14×c−4×14⇒31×a1×b2×c−1⇒3ab2c\Rightarrow (81a^4b^8c^{-4})^{\dfrac{1}{4}}\\[1em] \Rightarrow (3^4)^{\dfrac{1}{4}} \times a^{4 \times \dfrac{1}{4}} \times b^{8 \times \dfrac{1}{4}} \times c^{-4 \times \dfrac{1}{4}}\\[1em] \Rightarrow 3^1 \times a^1 \times b^2 \times c^{-1}\\[1em] \Rightarrow \dfrac{3ab^2}{c}

Hence, (81a4b8c−4)14=3ab2c\bm{(81a^4b^8c^{-4})^{\dfrac{1}{4}} = \dfrac{3ab^2}{c}}.

Question 5(v)

Simplify:

(3x−3y3)−2(3x^{-3}y^3)^{-2}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(3x−3y3)−2⇒3−2×x−3×(−2)×y3×(−2)⇒132×x6×y−6⇒x69y6\Rightarrow (3x^{-3}y^3)^{-2}\\[1em] \Rightarrow 3^{-2} \times x^{-3 \times (-2)} \times y^{3 \times (-2)}\\[1em] \Rightarrow \dfrac{1}{3^2} \times x^6 \times y^{-6}\\[1em] \Rightarrow \dfrac{x^6}{9y^6}

Hence, (3x−3y3)−2=x69y6\bm{(3x^{-3}y^3)^{-2} = \dfrac{x^6}{9y^6}}.

Question 5(vi)

Simplify:

(6ab2c−3)−1(6ab^2c^{-3})^{-1}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(6ab2c−3)−1⇒6−1×a−1×b2×(−1)×c−3×(−1)⇒16×a−1×b−2×c3⇒c36ab2\Rightarrow (6ab^2c^{-3})^{-1}\\[1em] \Rightarrow 6^{-1} \times a^{-1} \times b^{2 \times (-1)} \times c^{-3 \times (-1)}\\[1em] \Rightarrow \dfrac{1}{6} \times a^{-1} \times b^{-2} \times c^3\\[1em] \Rightarrow \dfrac{c^3}{6ab^2}

Hence, (6ab2c−3)−1=c36ab2\bm{(6ab^2c^{-3})^{-1} = \dfrac{c^3}{6ab^2}}.

Question 5(vii)

Simplify:

(−3a3/4b−1/4)4\left(-3a^{3/4}b^{-1/4}\right)^4

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(−3a3/4b−1/4)4⇒(−3)4×a(3/4)×4×b(−1/4)×4⇒81×a3×b−1⇒81a3b\Rightarrow \left(-3a^{3/4}b^{-1/4}\right)^4\\[1em] \Rightarrow (-3)^4 \times a^{(3/4) \times 4} \times b^{(-1/4) \times 4}\\[1em] \Rightarrow 81 \times a^3 \times b^{-1}\\[1em] \Rightarrow \dfrac{81a^3}{b}

Hence, (−3a3/4b−1/4)4=81a3b\bm{\left(-3a^{3/4}b^{-1/4}\right)^4 = \dfrac{81a^3}{b}}.

Question 5(viii)

Simplify:

(32a10b−5)15(32a^{10}b^{-5})^{\dfrac{1}{5}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm,(am)n=amn(ab)^m = a^mb^m, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

We know that 32 = 25

Solving,

⇒(32a10b−5)15⇒(25)15×a10×15×b−5×15⇒21×a2×b−1⇒2a2b\Rightarrow (32a^{10}b^{-5})^{\dfrac{1}{5}}\\[1em] \Rightarrow (2^5)^{\dfrac{1}{5}} \times a^{10 \times \dfrac{1}{5}} \times b^{-5 \times \dfrac{1}{5}}\\[1em] \Rightarrow 2^1 \times a^2 \times b^{-1}\\[1em] \Rightarrow \dfrac{2a^2}{b}

Hence, (32a10b−5)15=2a2b\bm{(32a^{10}b^{-5})^{\dfrac{1}{5}} = \dfrac{2a^2}{b}}.

Question 5(ix)

Simplify:

x18y−12z33\sqrt[3]{x^{18}y^{-12}z^3}

Answer

We know that,

For any non-zero rational number (aa)

an=a1n,(am)n=amn\sqrt[n]{a} = a^{\dfrac{1}{n}}, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒x18y−12z33⇒(x18y−12z3)13⇒x18×13×y−12×13×z3×13⇒x6×y−4×z1⇒x6zy4\Rightarrow \sqrt[3]{x^{18}y^{-12}z^3}\\[1em] \Rightarrow \left(x^{18}y^{-12}z^3\right)^{\dfrac{1}{3}}\\[1em] \Rightarrow x^{18 \times \dfrac{1}{3}} \times y^{-12 \times \dfrac{1}{3}} \times z^{3 \times \dfrac{1}{3}}\\[1em] \Rightarrow x^6 \times y^{-4} \times z^1\\[1em] \Rightarrow \dfrac{x^6z}{y^4}

Hence, x18y−12z33=x6zy4\bm{\sqrt[3]{x^{18}y^{-12}z^3} = \dfrac{x^6z}{y^4}}.

Question 6(i)

Show that:

xm+n×xn+l×xl+m(xm×xn×xl)2=1\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{\left(x^m \times x^n \times x^l\right)^2} = 1

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+n,aman=am−n,(am)n=amna^m \times a^n = a^{m+n}, \dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn} and a0=1a^0 = 1

Solving L.H.S.,

⇒xm+n×xn+l×xl+m(xm×xn×xl)2⇒xm+n+n+l+l+m(xm+n+l)2⇒x2m+2n+2lx2(m+n+l)⇒x2m+2n+2lx2m+2n+2l⇒x(2m+2n+2l)−(2m+2n+2l)⇒x0⇒1\Rightarrow \dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{\left(x^m \times x^n \times x^l\right)^2}\\[1em] \Rightarrow \dfrac{x^{m+n+n+l+l+m}}{\left(x^{m+n+l}\right)^2}\\[1em] \Rightarrow \dfrac{x^{2m+2n+2l}}{x^{2(m+n+l)}}\\[1em] \Rightarrow \dfrac{x^{2m+2n+2l}}{x^{2m+2n+2l}}\\[1em] \Rightarrow x^{(2m+2n+2l)-(2m+2n+2l)}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, xm+n×xn+l×xl+m(xm×xn×xl)2=1\bm{\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{\left(x^m \times x^n \times x^l\right)^2} = 1}.

Question 6(ii)

Show that:

xp−q×xq−r×xr−p=1\sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}} = 1

Answer

We know that,

For any non-zero rational number (aa)

a=a12,(am)n=amn,am×an=am+n\sqrt{a} = a^{\dfrac{1}{2}}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n} and a0=1a^0 = 1

Solving L.H.S.,

⇒xp−q×xq−r×xr−p⇒(xp−q)12×(xq−r)12×(xr−p)12⇒xp−q2×xq−r2×xr−p2⇒xp−q2+q−r2+r−p2⇒xp−q+q−r+r−p2⇒x02⇒x0⇒1\Rightarrow \sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}}\\[1em] \Rightarrow \left(x^{p-q}\right)^{\dfrac{1}{2}} \times \left(x^{q-r}\right)^{\dfrac{1}{2}} \times \left(x^{r-p}\right)^{\dfrac{1}{2}}\\[1em] \Rightarrow x^{\dfrac{p-q}{2}} \times x^{\dfrac{q-r}{2}} \times x^{\dfrac{r-p}{2}}\\[1em] \Rightarrow x^{\dfrac{p-q}{2}+\dfrac{q-r}{2}+\dfrac{r-p}{2}}\\[1em] \Rightarrow x^{\dfrac{p-q+q-r+r-p}{2}}\\[1em] \Rightarrow x^{\dfrac{0}{2}}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, xp−q×xq−r×xr−p=1\bm{\sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}} = 1}.

Question 7

Show that: (xpxq)r×(xqxr)p×(xrxp)q=1\left(\dfrac{x^p}{x^q}\right)^r \times \left(\dfrac{x^q}{x^r}\right)^p \times \left(\dfrac{x^r}{x^p}\right)^q = 1

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n} and a0=1a^0 = 1

Solving L.H.S.,

⇒(xpxq)r×(xqxr)p×(xrxp)q⇒(xp−q)r×(xq−r)p×(xr−p)q⇒xpr−qr×xpq−pr×xqr−pq⇒x(pr−qr)+(pq−pr)+(qr−pq)⇒xpr−qr+pq−pr+qr−pq⇒x0⇒1\Rightarrow \left(\dfrac{x^p}{x^q}\right)^r \times \left(\dfrac{x^q}{x^r}\right)^p \times \left(\dfrac{x^r}{x^p}\right)^q\\[1em] \Rightarrow \left(x^{p-q}\right)^r \times \left(x^{q-r}\right)^p \times \left(x^{r-p}\right)^q\\[1em] \Rightarrow x^{pr-qr} \times x^{pq-pr} \times x^{qr-pq}\\[1em] \Rightarrow x^{(pr-qr)+(pq-pr)+(qr-pq)}\\[1em] \Rightarrow x^{pr - qr + pq - pr + qr - pq}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, (xpxq)r×(xqxr)p×(xrxp)q=1\bm{\left(\dfrac{x^p}{x^q}\right)^r \times \left(\dfrac{x^q}{x^r}\right)^p \times \left(\dfrac{x^r}{x^p}\right)^q = 1}.

Question 8(i)

Show that:

(xa+b)a−b×(xb+c)b−c×(xc+a)c−a=1\left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a} = 1

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn,am×an=am+n,a0=1(a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1 and (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2

Solving L.H.S.,

⇒(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)⇒xa2−b2×xb2−c2×xc2−a2⇒x(a2−b2)+(b2−c2)+(c2−a2)⇒xa2−b2+b2−c2+c2−a2⇒x0⇒1\Rightarrow \left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a}\\[1em] \Rightarrow x^{(a+b)(a-b)} \times x^{(b+c)(b-c)} \times x^{(c+a)(c-a)}\\[1em] \Rightarrow x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}\\[1em] \Rightarrow x^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}\\[1em] \Rightarrow x^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, (xa+b)a−b×(xb+c)b−c×(xc+a)c−a=1\bm{\left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a} = 1}.

Question 8(ii)

Show that:

(xax−b)a−b×(xbx−c)b−c×(xcx−a)c−a=1\left(\dfrac{x^a}{x^{-b}}\right)^{a-b} \times \left(\dfrac{x^b}{x^{-c}}\right)^{b-c} \times \left(\dfrac{x^c}{x^{-a}}\right)^{c-a} = 1

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n,a0=1\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1 and (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2

Solving L.H.S.,

⇒(xax−b)a−b×(xbx−c)b−c×(xcx−a)c−a⇒(xa−(−b))a−b×(xb−(−c))b−c×(xc−(−a))c−a⇒(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)⇒xa2−b2×xb2−c2×xc2−a2⇒xa2−b2+b2−c2+c2−a2⇒x0⇒1\Rightarrow \left(\dfrac{x^a}{x^{-b}}\right)^{a-b} \times \left(\dfrac{x^b}{x^{-c}}\right)^{b-c} \times \left(\dfrac{x^c}{x^{-a}}\right)^{c-a}\\[1em] \Rightarrow \left(x^{a-(-b)}\right)^{a-b} \times \left(x^{b-(-c)}\right)^{b-c} \times \left(x^{c-(-a)}\right)^{c-a}\\[1em] \Rightarrow \left(x^{a+b}\right)^{a-b} \times \left(x^{b+c}\right)^{b-c} \times \left(x^{c+a}\right)^{c-a}\\[1em] \Rightarrow x^{(a+b)(a-b)} \times x^{(b+c)(b-c)} \times x^{(c+a)(c-a)}\\[1em] \Rightarrow x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}\\[1em] \Rightarrow x^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, (xax−b)a−b×(xbx−c)b−c×(xcx−a)c−a=1\bm{\left(\dfrac{x^a}{x^{-b}}\right)^{a-b} \times \left(\dfrac{x^b}{x^{-c}}\right)^{b-c} \times \left(\dfrac{x^c}{x^{-a}}\right)^{c-a} = 1}.

Question 8(iii)

Show that:

(xa+bxc)a−b×(xb+cxa)b−c×(xc+axb)c−a=1\left(\dfrac{x^{a+b}}{x^c}\right)^{a-b} \times \left(\dfrac{x^{b+c}}{x^a}\right)^{b-c} \times \left(\dfrac{x^{c+a}}{x^b}\right)^{c-a} = 1

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n,a0=1\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1 and (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2

Solving L.H.S.,

⇒(xa+bxc)a−b×(xb+cxa)b−c×(xc+axb)c−a⇒(xa+b−c)a−b×(xb+c−a)b−c×(xc+a−b)c−a⇒x(a+b)(a−b)−c(a−b)×x(b+c)(b−c)−a(b−c)×x(c+a)(c−a)−b(c−a)⇒xa2−b2−ca+cb×xb2−c2−ab+ac×xc2−a2−bc+ab\Rightarrow \left(\dfrac{x^{a+b}}{x^c}\right)^{a-b} \times \left(\dfrac{x^{b+c}}{x^a}\right)^{b-c} \times \left(\dfrac{x^{c+a}}{x^b}\right)^{c-a}\\[1em] \Rightarrow \left(x^{a+b-c}\right)^{a-b} \times \left(x^{b+c-a}\right)^{b-c} \times \left(x^{c+a-b}\right)^{c-a}\\[1em] \Rightarrow x^{(a+b)(a-b)-c(a-b)} \times x^{(b+c)(b-c)-a(b-c)} \times x^{(c+a)(c-a)-b(c-a)}\\[1em] \Rightarrow x^{a^2-b^2-ca+cb} \times x^{b^2-c^2-ab+ac} \times x^{c^2-a^2-bc+ab}

Adding the powers,

⇒x(a2−b2−ca+cb)+(b2−c2−ab+ac)+(c2−a2−bc+ab)⇒x(a2−b2+b2−c2+c2−a2)+(−ca+ac)+(cb−bc)+(−ab+ab)⇒x0⇒1\Rightarrow x^{(a^2-b^2-ca+cb)+(b^2-c^2-ab+ac)+(c^2-a^2-bc+ab)}\\[1em] \Rightarrow x^{(a^2-b^2+b^2-c^2+c^2-a^2)+(-ca+ac)+(cb-bc)+(-ab+ab)}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, (xa+bxc)a−b×(xb+cxa)b−c×(xc+axb)c−a=1\bm{\left(\dfrac{x^{a+b}}{x^c}\right)^{a-b} \times \left(\dfrac{x^{b+c}}{x^a}\right)^{b-c} \times \left(\dfrac{x^{c+a}}{x^b}\right)^{c-a} = 1}.

Question 8(iv)

Show that:

(xa2xb2)1a+b×(xb2xc2)1b+c×(xc2xa2)1c+a=1\left(\dfrac{x^{a^2}}{x^{b^2}}\right)^{\dfrac{1}{a+b}} \times \left(\dfrac{x^{b^2}}{x^{c^2}}\right)^{\dfrac{1}{b+c}} \times \left(\dfrac{x^{c^2}}{x^{a^2}}\right)^{\dfrac{1}{c+a}} = 1

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n,a0=1\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1 and a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)

Solving L.H.S.,

⇒(xa2xb2)1a+b×(xb2xc2)1b+c×(xc2xa2)1c+a⇒(xa2−b2)1a+b×(xb2−c2)1b+c×(xc2−a2)1c+a⇒x(a−b)(a+b)a+b×x(b−c)(b+c)b+c×x(c−a)(c+a)c+a⇒xa−b×xb−c×xc−a⇒x(a−b)+(b−c)+(c−a)⇒xa−b+b−c+c−a⇒x0⇒1\Rightarrow \left(\dfrac{x^{a^2}}{x^{b^2}}\right)^{\dfrac{1}{a+b}} \times \left(\dfrac{x^{b^2}}{x^{c^2}}\right)^{\dfrac{1}{b+c}} \times \left(\dfrac{x^{c^2}}{x^{a^2}}\right)^{\dfrac{1}{c+a}}\\[1em] \Rightarrow \left(x^{a^2-b^2}\right)^{\dfrac{1}{a+b}} \times \left(x^{b^2-c^2}\right)^{\dfrac{1}{b+c}} \times \left(x^{c^2-a^2}\right)^{\dfrac{1}{c+a}}\\[1em] \Rightarrow x^{\dfrac{(a-b)(a+b)}{a+b}} \times x^{\dfrac{(b-c)(b+c)}{b+c}} \times x^{\dfrac{(c-a)(c+a)}{c+a}}\\[1em] \Rightarrow x^{a-b} \times x^{b-c} \times x^{c-a}\\[1em] \Rightarrow x^{(a-b)+(b-c)+(c-a)}\\[1em] \Rightarrow x^{a-b+b-c+c-a}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Hence, (xa2xb2)1a+b×(xb2xc2)1b+c×(xc2xa2)1c+a=1\bm{\left(\dfrac{x^{a^2}}{x^{b^2}}\right)^{\dfrac{1}{a+b}} \times \left(\dfrac{x^{b^2}}{x^{c^2}}\right)^{\dfrac{1}{b+c}} \times \left(\dfrac{x^{c^2}}{x^{a^2}}\right)^{\dfrac{1}{c+a}} = 1}.

Question 9

Show that: (xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+ca+a2=1\left(\dfrac{x^a}{x^b}\right)^{a^2+ab+b^2} \times \left(\dfrac{x^b}{x^c}\right)^{b^2+bc+c^2} \times \left(\dfrac{x^c}{x^a}\right)^{c^2+ca+a^2} = 1

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n,a0=1\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}, a^0 = 1 and (a−b)(a2+ab+b2)=a3−b3(a-b)(a^2+ab+b^2) = a^3 - b^3

Solving L.H.S.,

⇒(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⇒x(a3−b3)+(b3−c3)+(c3−a3)⇒xa3−b3+b3−c3+c3−a3⇒x0⇒1\Rightarrow \left(\dfrac{x^a}{x^b}\right)^{a^2+ab+b^2} \times \left(\dfrac{x^b}{x^c}\right)^{b^2+bc+c^2} \times \left(\dfrac{x^c}{x^a}\right)^{c^2+ca+a^2}\\[1em] \Rightarrow \left(x^{a-b}\right)^{a^2+ab+b^2} \times \left(x^{b-c}\right)^{b^2+bc+c^2} \times \left(x^{c-a}\right)^{c^2+ca+a^2}\\[1em] \Rightarrow 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] \Rightarrow x^{a^3-b^3} \times x^{b^3-c^3} \times x^{c^3-a^3}\\[1em] \Rightarrow x^{(a^3-b^3)+(b^3-c^3)+(c^3-a^3)}\\[1em] \Rightarrow x^{a^3-b^3+b^3-c^3+c^3-a^3}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

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

Question 10(i)

Evaluate:

(xaxb)1ab×(xbxc)1bc×(xcxa)1ca\left(\dfrac{x^a}{x^b}\right)^{\dfrac{1}{ab}} \times \left(\dfrac{x^b}{x^c}\right)^{\dfrac{1}{bc}} \times \left(\dfrac{x^c}{x^a}\right)^{\dfrac{1}{ca}}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,(am)n=amn,am×an=am+n\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, a^m \times a^n = a^{m+n} and a0=1a^0 = 1

Solving,

⇒(xaxb)1ab×(xbxc)1bc×(xcxa)1ca⇒(xa−b)1ab×(xb−c)1bc×(xc−a)1ca⇒xa−bab×xb−cbc×xc−aca⇒x(1b−1a)×x(1c−1b)×x(1a−1c)⇒x1b−1a+1c−1b+1a−1c⇒x0⇒1\Rightarrow \left(\dfrac{x^a}{x^b}\right)^{\dfrac{1}{ab}} \times \left(\dfrac{x^b}{x^c}\right)^{\dfrac{1}{bc}} \times \left(\dfrac{x^c}{x^a}\right)^{\dfrac{1}{ca}}\\[1em] \Rightarrow \left(x^{a-b}\right)^{\dfrac{1}{ab}} \times \left(x^{b-c}\right)^{\dfrac{1}{bc}} \times \left(x^{c-a}\right)^{\dfrac{1}{ca}}\\[1em] \Rightarrow x^{\dfrac{a-b}{ab}} \times x^{\dfrac{b-c}{bc}} \times x^{\dfrac{c-a}{ca}}\\[1em] \Rightarrow x^{\left(\dfrac{1}{b}-\dfrac{1}{a}\right)} \times x^{\left(\dfrac{1}{c}-\dfrac{1}{b}\right)} \times x^{\left(\dfrac{1}{a}-\dfrac{1}{c}\right)}\\[1em] \Rightarrow x^{\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{c}}\\[1em] \Rightarrow x^0\\[1em] \Rightarrow 1

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

Question 10(ii)

Evaluate:

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

Answer

We know that,

For any non-zero rational number (aa)

am−n=amana^{m-n} = \dfrac{a^m}{a^n}

Solving,

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

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

Question 11

Simplify: (ab)y−z⋅(bc)z−x⋅(ca)x−y(ab)^{y-z} \cdot (bc)^{z-x} \cdot (ca)^{x-y}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(ab)m=ambm(ab)^m = a^mb^m and am×an=am+na^m \times a^n = a^{m+n}

Solving,

⇒(ab)y−z⋅(bc)z−x⋅(ca)x−y⇒ay−zby−z⋅bz−xcz−x⋅cx−yax−y\Rightarrow (ab)^{y-z} \cdot (bc)^{z-x} \cdot (ca)^{x-y}\\[1em] \Rightarrow a^{y-z}b^{y-z} \cdot b^{z-x}c^{z-x} \cdot c^{x-y}a^{x-y}

Collecting the powers of a,ba, b and cc separately,

⇒a(y−z)+(x−y)⋅b(y−z)+(z−x)⋅c(z−x)+(x−y)⇒ay−z+x−y⋅by−z+z−x⋅cz−x+x−y⇒ax−z⋅by−x⋅cz−y\Rightarrow a^{(y-z)+(x-y)} \cdot b^{(y-z)+(z-x)} \cdot c^{(z-x)+(x-y)}\\[1em] \Rightarrow a^{y-z+x-y} \cdot b^{y-z+z-x} \cdot c^{z-x+x-y}\\[1em] \Rightarrow a^{x-z} \cdot b^{y-x} \cdot c^{z-y}

Hence, (ab)y−z⋅(bc)z−x⋅(ca)x−y=ax−zby−xcz−y\bm{(ab)^{y-z} \cdot (bc)^{z-x} \cdot (ca)^{x-y} = a^{x-z}b^{y-x}c^{z-y}}.

Question 12

Simplify: x2n+3⋅x(2n+1)(n+2)(x3)2n+1⋅xn(2n+1)\dfrac{x^{2n+3} \cdot x^{(2n+1)(n+2)}}{\left(x^3\right)^{2n+1} \cdot x^{n(2n+1)}}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+n,aman=am−na^m \times a^n = a^{m+n}, \dfrac{a^m}{a^n} = a^{m-n} and (am)n=amn(a^m)^n = a^{mn}

Solving the numerator,

⇒x2n+3⋅x(2n+1)(n+2)⇒x(2n+3)+(2n2+4n+n+2)⇒x2n+3+2n2+5n+2⇒x2n2+7n+5\Rightarrow x^{2n+3} \cdot x^{(2n+1)(n+2)}\\[1em] \Rightarrow x^{(2n+3)+(2n^2+4n+n+2)}\\[1em] \Rightarrow x^{2n+3+2n^2+5n+2}\\[1em] \Rightarrow x^{2n^2+7n+5}

Solving the denominator,

⇒(x3)2n+1⋅xn(2n+1)⇒x3(2n+1)⋅x2n2+n⇒x(6n+3)+(2n2+n)⇒x2n2+7n+3\Rightarrow \left(x^3\right)^{2n+1} \cdot x^{n(2n+1)}\\[1em] \Rightarrow x^{3(2n+1)} \cdot x^{2n^2+n}\\[1em] \Rightarrow x^{(6n+3)+(2n^2+n)}\\[1em] \Rightarrow x^{2n^2+7n+3}

So,

⇒x2n2+7n+5x2n2+7n+3⇒x(2n2+7n+5)−(2n2+7n+3)⇒x2\Rightarrow \dfrac{x^{2n^2+7n+5}}{x^{2n^2+7n+3}}\\[1em] \Rightarrow x^{(2n^2+7n+5)-(2n^2+7n+3)}\\[1em] \Rightarrow x^{2}

Hence, x2n+3⋅x(2n+1)(n+2)(x3)2n+1⋅xn(2n+1)=x2\bm{\dfrac{x^{2n+3} \cdot x^{(2n+1)(n+2)}}{\left(x^3\right)^{2n+1} \cdot x^{n(2n+1)}} = x^2}.

Question 13

Simplify: a7+2n⋅(a2)3n+2(a4)2n+3\dfrac{a^{7+2n} \cdot \left(a^2\right)^{3n+2}}{\left(a^4\right)^{2n+3}}

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+n,aman=am−n,(am)n=amna^m \times a^n = a^{m+n}, \dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn} and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒a7+2n⋅(a2)3n+2(a4)2n+3⇒a7+2n⋅a2(3n+2)a4(2n+3)⇒a7+2n⋅a6n+4a8n+12⇒a(7+2n)+(6n+4)a8n+12⇒a8n+11a8n+12⇒a(8n+11)−(8n+12)⇒a−1⇒1a\Rightarrow \dfrac{a^{7+2n} \cdot \left(a^2\right)^{3n+2}}{\left(a^4\right)^{2n+3}}\\[1em] \Rightarrow \dfrac{a^{7+2n} \cdot a^{2(3n+2)}}{a^{4(2n+3)}}\\[1em] \Rightarrow \dfrac{a^{7+2n} \cdot a^{6n+4}}{a^{8n+12}}\\[1em] \Rightarrow \dfrac{a^{(7+2n)+(6n+4)}}{a^{8n+12}}\\[1em] \Rightarrow \dfrac{a^{8n+11}}{a^{8n+12}}\\[1em] \Rightarrow a^{(8n+11)-(8n+12)}\\[1em] \Rightarrow a^{-1}\\[1em] \Rightarrow \dfrac{1}{a}

Hence, a7+2n⋅(a2)3n+2(a4)2n+3=1a\bm{\dfrac{a^{7+2n} \cdot \left(a^2\right)^{3n+2}}{\left(a^4\right)^{2n+3}} = \dfrac{1}{a}}.

Question 14(i)

Evaluate:

(16625)1/4\left(\dfrac{16}{625}\right)^{1/4}

Answer

We know that,

For any non-zero rational number (aa)

(ab)m=ambm\Big(\dfrac{a}{b}\Big)^m = \dfrac{a^m}{b^m} and (am)n=amn(a^m)^n = a^{mn}

We know that 16 = 24 and 625 = 54

Solving,

⇒(16625)14⇒(2454)14⇒(25)4×14⇒(25)1⇒25\Rightarrow \Big(\dfrac{16}{625}\Big)^{\dfrac{1}{4}}\\[1em] \Rightarrow \Big(\dfrac{2^4}{5^4}\Big)^{\dfrac{1}{4}}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{4 \times \dfrac{1}{4}}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{1}\\[1em] \Rightarrow \dfrac{2}{5}

Hence, (16625)14=25\mathbf{\Big(\dfrac{16}{625}\Big)^{\dfrac{1}{4}} = \dfrac{2}{5}}.

Question 14(ii)

Evaluate:

(8116)−1/4\left(\dfrac{81}{16}\right)^{-1/4}

Answer

We know that,

For any non-zero rational number (aa)

(ab)−m=(ba)m\Big(\dfrac{a}{b}\Big)^{-m} = \Big(\dfrac{b}{a}\Big)^{m} and (am)n=amn(a^m)^n = a^{mn}

We know that 81 = 34 and 16 = 24

Solving,

⇒(8116)−14⇒(1681)14⇒(2434)14⇒(23)4×14⇒23\Rightarrow \Big(\dfrac{81}{16}\Big)^{\dfrac{-1}{4}}\\[1em] \Rightarrow \Big(\dfrac{16}{81}\Big)^{\dfrac{1}{4}}\\[1em] \Rightarrow \Big(\dfrac{2^4}{3^4}\Big)^{\dfrac{1}{4}}\\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{4 \times \dfrac{1}{4}}\\[1em] \Rightarrow \dfrac{2}{3}

Hence, (8116)−14=23\mathbf{\Big(\dfrac{81}{16}\Big)^{\dfrac{-1}{4}} = \dfrac{2}{3}}.

Question 14(iii)

Evaluate:

(64)2/3+1253+30+12−5+(27)−2/3×(259)−1/2(64)^{2/3} + \sqrt[3]{125} + 3^0 + \dfrac{1}{2^{-5}} + (27)^{-2/3} \times \left(\dfrac{25}{9}\right)^{-1/2}

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn,a0=1,1a−m=am(a^m)^n = a^{mn}, a^0 = 1, \dfrac{1}{a^{-m}} = a^m and (ab)−m=(ba)m\Big(\dfrac{a}{b}\Big)^{-m} = \Big(\dfrac{b}{a}\Big)^{m}

We know that 64 = 43, 125 = 53, 27 = 33, 25 = 52 and 9 = 32

Solving each term separately,

⇒(64)23=(43)23=42=16⇒1253=(53)13=5⇒30=1⇒12−5=25=32⇒(27)−23=(33)−23=3−2=19⇒(259)−12=(925)12=(3252)12=35\Rightarrow (64)^{\dfrac{2}{3}} = (4^3)^{\dfrac{2}{3}} = 4^2 = 16\\[1em] \Rightarrow \sqrt[3]{125} = (5^3)^{\dfrac{1}{3}} = 5\\[1em] \Rightarrow 3^0 = 1\\[1em] \Rightarrow \dfrac{1}{2^{-5}} = 2^5 = 32\\[1em] \Rightarrow (27)^{\dfrac{-2}{3}} = (3^3)^{\dfrac{-2}{3}} = 3^{-2} = \dfrac{1}{9}\\[1em] \Rightarrow \Big(\dfrac{25}{9}\Big)^{\dfrac{-1}{2}} = \Big(\dfrac{9}{25}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{3^2}{5^2}\Big)^{\dfrac{1}{2}} = \dfrac{3}{5}

Substituting the values,

⇒16+5+1+32+19×35⇒54+345⇒54+115⇒810+115⇒81115⇒54115\Rightarrow 16 + 5 + 1 + 32 + \dfrac{1}{9} \times \dfrac{3}{5}\\[1em] \Rightarrow 54 + \dfrac{3}{45}\\[1em] \Rightarrow 54 + \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{810 + 1}{15}\\[1em] \Rightarrow \dfrac{811}{15}\\[1em] \Rightarrow 54\dfrac{1}{15}

Hence, the value of the given expression is 54115\mathbf{54\dfrac{1}{15}}.

Question 14(iv)

Evaluate:

(81)−1×3−5×39×(64)5/6×(33)6(81)^{-1} \times 3^{-5} \times 3^9 \times (64)^{5/6} \times \left(\sqrt[3]{3}\right)^6

Answer

We know that,

For any non-zero rational number (aa)

am×an=am+n,(am)n=amn,a−n=1ana^m \times a^n = a^{m+n}, (a^m)^n = a^{mn}, a^{-n} = \dfrac{1}{a^n} and a0=1a^0 = 1

We know that 81 = 34, 64 = 26 and 33=313\sqrt[3]{3} = 3^{\dfrac{1}{3}}

Solving,

⇒(81)−1×3−5×39×(64)56×(33)6⇒(34)−1×3−5×39×(26)56×(313)6⇒3−4×3−5×39×25×32⇒3−4−5+9+2×32⇒32×32⇒9×32⇒288\Rightarrow (81)^{-1} \times 3^{-5} \times 3^9 \times (64)^{\dfrac{5}{6}} \times \left(\sqrt[3]{3}\right)^6\\[1em] \Rightarrow (3^4)^{-1} \times 3^{-5} \times 3^9 \times (2^6)^{\dfrac{5}{6}} \times \left(3^{\dfrac{1}{3}}\right)^6\\[1em] \Rightarrow 3^{-4} \times 3^{-5} \times 3^9 \times 2^5 \times 3^2\\[1em] \Rightarrow 3^{-4-5+9+2} \times 32\\[1em] \Rightarrow 3^{2} \times 32\\[1em] \Rightarrow 9 \times 32\\[1em] \Rightarrow 288

Hence, the value of the given expression is 288.

Question 14(v)

Evaluate:

y3x×yx\sqrt{\dfrac{y^3}{x}} \times \sqrt{\dfrac{y}{x}}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

a×b=ab\sqrt{a} \times \sqrt{b} = \sqrt{ab}

Solving,

⇒y3x×yx⇒y3x×yx⇒y4x2⇒y4x2⇒y2x\Rightarrow \sqrt{\dfrac{y^3}{x}} \times \sqrt{\dfrac{y}{x}}\\[1em] \Rightarrow \sqrt{\dfrac{y^3}{x} \times \dfrac{y}{x}}\\[1em] \Rightarrow \sqrt{\dfrac{y^4}{x^2}}\\[1em] \Rightarrow \dfrac{\sqrt{y^4}}{\sqrt{x^2}}\\[1em] \Rightarrow \dfrac{y^2}{x}

Hence, y3x×yx=y2x\bm{\sqrt{\dfrac{y^3}{x}} \times \sqrt{\dfrac{y}{x}} = \dfrac{y^2}{x}}.

Question 15(i)

Find the value of xx when:

(−311)x+5÷(−311)−2x+3=(−311)2x−5×[(−311)−2](x+4)\left(\dfrac{-3}{11}\right)^{x+5} \div \left(\dfrac{-3}{11}\right)^{-2x+3} = \left(\dfrac{-3}{11}\right)^{2x-5} \times \left[\left(\dfrac{-3}{11}\right)^{-2}\right]^{(x+4)}

Answer

We know that,

For any non-zero rational number (aa)

aman=am−n,am×an=am+n\dfrac{a^m}{a^n} = a^{m-n}, a^m \times a^n = a^{m+n} and (am)n=amn(a^m)^n = a^{mn}

Solving L.H.S.,

⇒(−311)x+5÷(−311)−2x+3⇒(−311)(x+5)−(−2x+3)⇒(−311)x+5+2x−3⇒(−311)3x+2\Rightarrow \Big(\dfrac{-3}{11}\Big)^{x+5} \div \Big(\dfrac{-3}{11}\Big)^{-2x+3}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{(x+5)-(-2x+3)}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{x+5+2x-3}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{3x+2}

Solving R.H.S.,

⇒(−311)2x−5×[(−311)−2](x+4)⇒(−311)2x−5×(−311)−2(x+4)⇒(−311)(2x−5)+(−2x−8)⇒(−311)−13\Rightarrow \Big(\dfrac{-3}{11}\Big)^{2x-5} \times \Big[\Big(\dfrac{-3}{11}\Big)^{-2}\Big]^{(x+4)}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{2x-5} \times \Big(\dfrac{-3}{11}\Big)^{-2(x+4)}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{(2x-5)+(-2x-8)}\\[1em] \Rightarrow \Big(\dfrac{-3}{11}\Big)^{-13}

So,

⇒(−311)3x+2=(−311)−13\Rightarrow \Big(\dfrac{-3}{11}\Big)^{3x+2} = \Big(\dfrac{-3}{11}\Big)^{-13}

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

⇒3x+2=−13⇒3x=−13−2⇒3x=−15⇒x=−153⇒x=−5\Rightarrow 3x + 2 = -13\\[1em] \Rightarrow 3x = -13 - 2\\[1em] \Rightarrow 3x = -15\\[1em] \Rightarrow x = \dfrac{-15}{3}\\[1em] \Rightarrow x = -5

Hence, the value of x is -5.

Question 15(ii)

Find the value of xx when:

[{(25)2r}4]x+2=[{(25)−2}(x−1)]−3\left[\bigg\lbrace \left(\dfrac{2}{5}\right)^2 r \bigg\rbrace ^4\right]^{x+2} = \left[\bigg\lbrace\left(\dfrac{2}{5}\right)^{-2}\bigg\rbrace^{(x-1)}\right]^{-3}

Answer

We know that,

For any non-zero rational number (aa)

(am)n=amn(a^m)^n = a^{mn}

Solving L.H.S.,

⇒[{(25)2}4]x+2⇒[(25)2×4]x+2⇒(25)8(x+2)⇒(25)8x+16\Rightarrow \Big[\Big\lbrace\Big(\dfrac{2}{5}\Big)^2\Big\rbrace^4\Big]^{x+2}\\[1em] \Rightarrow \Big[\Big(\dfrac{2}{5}\Big)^{2 \times 4}\Big]^{x+2}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{8(x+2)}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{8x+16}

Solving R.H.S.,

⇒[{(25)−2}(x−1)]−3⇒[(25)−2(x−1)]−3⇒(25)(2−2x)×(−3)⇒(25)−6+6x\Rightarrow \Big[\Big\lbrace\Big(\dfrac{2}{5}\Big)^{-2}\Big\rbrace^{(x-1)}\Big]^{-3}\\[1em] \Rightarrow \Big[\Big(\dfrac{2}{5}\Big)^{-2(x-1)}\Big]^{-3}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{(2-2x) \times (-3)}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{-6+6x}

So,

⇒(25)8x+16=(25)−6+6x\Rightarrow \Big(\dfrac{2}{5}\Big)^{8x+16} = \Big(\dfrac{2}{5}\Big)^{-6+6x}

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

⇒8x+16=−6+6x⇒8x−6x=−6−16⇒2x=−22⇒x=−222⇒x=−11\Rightarrow 8x + 16 = -6 + 6x\\[1em] \Rightarrow 8x - 6x = -6 - 16\\[1em] \Rightarrow 2x = -22\\[1em] \Rightarrow x = \dfrac{-22}{2}\\[1em] \Rightarrow x = -11

Hence, the value of x\bm{x} is -11.

Question 16(i)

Simplify:

[{2p−1q2r}3]−2\left[\bigg\lbrace2p^{-1}q^2r\bigg\rbrace^3\right]^{-2}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

(am)n=amn,(ab)m=ambm(a^m)^n = a^{mn}, (ab)^m = a^mb^m and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒[{2p−1q2r}3]−2⇒(2p−1q2r)3×(−2)⇒(2p−1q2r)−6⇒2−6×p−1×(−6)×q2×(−6)×r−6⇒126×p6×q−12×r−6⇒p664q12r6\Rightarrow \Big[\Big\lbrace 2p^{-1}q^2r \Big\rbrace^3\Big]^{-2}\\[1em] \Rightarrow \Big(2p^{-1}q^2r\Big)^{3 \times (-2)}\\[1em] \Rightarrow \Big(2p^{-1}q^2r\Big)^{-6}\\[1em] \Rightarrow 2^{-6} \times p^{-1 \times (-6)} \times q^{2 \times (-6)} \times r^{-6}\\[1em] \Rightarrow \dfrac{1}{2^6} \times p^{6} \times q^{-12} \times r^{-6}\\[1em] \Rightarrow \dfrac{p^6}{64q^{12}r^6}

Hence, [{2p−1q2r}3]−2=p664q12r6\bm{\Big[\Big\lbrace 2p^{-1}q^2r \Big\rbrace^3\Big]^{-2} = \dfrac{p^6}{64q^{12}r^6}}.

Question 16(ii)

Simplify:

(3p2qr−22p−1q3)2÷(2p3r)−1\left(\dfrac{3p^2qr^{-2}}{2p^{-1}q^3}\right)^2 \div \left(2p^3r\right)^{-1}

Answer

We know that,

For any non-zero rational numbers (aa and bb)

aman=am−n,(am)n=amn,(ab)m=ambm\dfrac{a^m}{a^n} = a^{m-n}, (a^m)^n = a^{mn}, (ab)^m = a^mb^m and a−n=1ana^{-n} = \dfrac{1}{a^n}

Solving,

⇒(3p2qr−22p−1q3)2÷(2p3r)−1⇒(32p2×2q2r−2×222p−1×2q3×2)÷2−1p−3r−1⇒(9p4q2r−44p−2q6)×12−1p−3r−1⇒(94p4+2q2−6r−4)×2p3r⇒94p6q−4r−4×2p3r⇒9×24p6+3q−4r−4+1⇒92p9q−4r−3⇒9p92q4r3\Rightarrow \Big(\dfrac{3p^2qr^{-2}}{2p^{-1}q^3}\Big)^2 \div \Big(2p^3r\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{3^2p^{2 \times 2}q^2r^{-2 \times 2}}{2^2p^{-1 \times 2}q^{3 \times 2}}\Big) \div 2^{-1}p^{-3}r^{-1}\\[1em] \Rightarrow \Big(\dfrac{9p^4q^2r^{-4}}{4p^{-2}q^6}\Big) \times \dfrac{1}{2^{-1}p^{-3}r^{-1}}\\[1em] \Rightarrow \Big(\dfrac{9}{4}p^{4+2}q^{2-6}r^{-4}\Big) \times 2p^3r\\[1em] \Rightarrow \dfrac{9}{4}p^6q^{-4}r^{-4} \times 2p^3r\\[1em] \Rightarrow \dfrac{9 \times 2}{4}p^{6+3}q^{-4}r^{-4+1}\\[1em] \Rightarrow \dfrac{9}{2}p^9q^{-4}r^{-3}\\[1em] \Rightarrow \dfrac{9p^9}{2q^4r^3}

Hence, (3p2qr−22p−1q3)2÷(2p3r)−1=9p92q4r3\bm{\Big(\dfrac{3p^2qr^{-2}}{2p^{-1}q^3}\Big)^2 \div \Big(2p^3r\Big)^{-1} = \dfrac{9p^9}{2q^4r^3}}.

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