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S.chand publication solution of class 8 chapter 2 exercise 2 Exponents

EXERCISE 2


Question 1

Evaluate:

(i) 2-3

Sol :

Using property a^{-n}=\dfrac{1}{a^n}

\dfrac{1}{2^3}=\dfrac{1}{2}\times \dfrac{1}{2} \times \dfrac{1}{2}

\dfrac{1}{2\times 2 \times 2}=\dfrac{1}{8}


(ii) 60

Sol :

Using property: a0=1

60=1


(iii) (-3)-2

Sol :

Using property a^{-n}=\dfrac{1}{a^n}

\dfrac{1}{(-3)^2}=\dfrac{1}{-3}\times \dfrac{1}{-3}

\dfrac{1}{-3\times -3 }=\dfrac{1}{-9}


(iv) 85 . 8-5

Sol :

Using property am×an=a(m+n)

⇒85 . 8-5 or

⇒85 × 8-5 

⇒8(5-5)


(v) \dfrac{1}{2^{-2}}

Sol :

Using property \dfrac{1}{a}=a^{-1}

⇒2(-2)×-1

⇒22

⇒4


(vi) 3(8-1)0

Sol :

Using property: a0=1

⇒3×1

⇒3


(vii) \dfrac{1}{2^{-1}}+\dfrac{1}{3^{-1}}

Sol :

Using property : \dfrac{1}{a^{-1}}=a

⇒2+3

⇒5


(viii) 7x0

Sol :

Using property: a0=1

⇒7×1

⇒7


(ix) a2 ÷ a-2

Sol :

Using property : a^{-n}=\dfrac{1}{a^n}

a^2 \div \dfrac{1}{a^2}

a^2 \times \dfrac{a^2}{1}

⇒a2×a2

Using property : an×am=a(n+m)

⇒a4


(x) \dfrac{y^{-6}}{y^4}

Sol :

Using property : a^{-n}=\dfrac{1}{a}

\dfrac{1}{(y^4)\times (y^6)}

Using property: am×an=a(m+n)

\dfrac{1}{y^{(6+4)}}=\dfrac{1}{y^{(10)}} or

⇒y(-10)


(xi) (x -3)0

Sol :

Using property: (am)n=a(m×n)

⇒x0

⇒1


(xii) 60 + 6 -1 + 3 -1

Sol :

Using property: a0=1

⇒6 -1 + 3 -1

Using property: a^{-1}=\dfrac{1}{a}

\dfrac{1}{6}+\dfrac{1}{3}

L.C.M of 6 and 3 is =2×3=6

26,3
33,3
1,1

\dfrac{1\times 1+1\times 2}{6}=\dfrac{1+2}{6}

or \dfrac{3}{6}=\dfrac{1}{2}


(xiii) \dfrac{1}{a^0 + b^0}

Sol :

Using property: a0=1

\dfrac{1}{1+ 1}=\dfrac{1}{2}


(xiv) (3x2)3

Sol :

Using property: (an)m=a(m×n)

⇒ 3(3)× x(2×3)

⇒ 27× x6

⇒ 27x6


(xv) \left[\left(\dfrac{5}{7}\right)^{-6}\right]^{-0}

Sol :

Using property: (am)n=a(m×n)

\left(\dfrac{5}{7}\right)^{-6\times -0}

\left(\dfrac{5}{7}\right)^0

Using property : a0=1

⇒1


Question 2

Simplify:

(i) x5 . x3

Sol :

Using property : am×an=a(m+n)

⇒x(5+3)


(ii) \dfrac{x^8}{x^3}

Sol :

Using property : \dfrac{1}{a^n}=a^{-n}

⇒x8 × x(-3)

Using property : am×an=a(m+n)

⇒x(8-3)

⇒x5


(iii) (x3)2

Sol :

Using property : am×an=a(m+n)

⇒x(3×2)

⇒x6


(iv) (3x2)3

Sol :

Using property : am×an=a(m+n)

⇒33×(x2)3

⇒27×x6

⇒27x6


(v) (5x-3yz2)-2

Sol :

Using property : a^{-n}=\dfrac{1}{a^n}

\dfrac{1}{5x^{-3}yz^2}


Question 3

Write the option that indicates the correct meaning of the given expression

(i) 4n3 means

(a) 4 times n + n + n

(b) 4n . 4n . 4n

(c) 4 . n . n . n

Sol :

⇒4n3

⇒4.n.n.n or (4×n×n×n)

Option (c) is correct


(ii) 6x3y2 means

(a) 6 . xy . xy . xy . xy . xy

(b) (6 . x. x. x . y)(7. x. x. x . y)

(c) 6 . x . x . x . y .y

Sol :

Option (c) is correct


Question 4

Simplify:

(i) \dfrac{(a^0b^{-2})^5}{2a^{-1}}

Sol :

\dfrac{(a^{0\times 5}b^{-2\times 5})}{2a^{-1}}

\dfrac{(a^{0}b^{-10})}{2a^{-1}}

Using property: \dfrac{1}{a^{-1}}=a

\dfrac{a^0 \times a^1}{2b^{10}}

Using property: a0=1

\dfrac{a^1}{2b^{10}}


(ii) \dfrac{4a^{-3}b^{0}}{2a^{2}b^{-1}}

Sol :

\dfrac{\not{4}a^{-3}b^{0}}{\not{2}a^{2}b^{-1}}

Using property: \dfrac{1}{a}=a^{-1}

⇒2a-3×b0× a-2 × b+1

Using property: an×am=a(m+n)

⇒2a-3-2×b0+1

⇒2a-5×b1

\dfrac{2b}{a^5}


(iii) \dfrac{2x^{2}y^{-2}}{2^{-1}x^{2}y^{2}}

Sol :

Using property : \dfrac{1}{a^{-1}}=a^1

⇒(2x2y-2)(21x-2y-2)

⇒(2×2)x2-2y-2-2

⇒4x0y-4

Using property : a0=1

⇒4y-4

\dfrac{4}{y^4}


(iv) \dfrac{(t^{-4})^3}{(t^{3})^{-4}}

Sol :

Using property: (am)n=a(m×n)

\dfrac{t^{-12}}{t^{-12}}

⇒1


(v) 4k2(4-1k+4k-2)

Sol :

⇒4k2(4-1k)+4k2(4k-2)

using property: \dfrac{1}{a}=a^{-1}

\dfrac{4k^{2}\times k}{4}+\dfrac{4k^{2}\times 4}{k^{2}}

\dfrac{k^{2+1}}{1}+\dfrac{16k^{2}}{k^{2}}

⇒k3+16


(vi) \left(\dfrac{3x^{-2}}{2y^{-1}}\right)^{-2}

Sol :

\dfrac{(3x^{-2})^{-2}}{(2y^{-1})^{-2}}

Using property : (am)n=a(m×n)

\dfrac{3^{-2}x^{4})}{2^{-2}y^{2}}

\dfrac{9^{-1}x^{4})}{4^{-1}y^{2}}

\dfrac{4x^{4})}{9y^{2}}


(vii) \left[\dfrac{3^{-1}}{(-2)^{-2}}\right]^{-2}

Sol :

\dfrac{(3^{-1})^{-2}}{\{(-2)^{-2}\}^{-2}}

Using property : (am)n=a(m×n)

\dfrac{3^{2}}{(-2)^{4}}=\dfrac{9}{16}

 


Question 5

Express each of the following as a power of 2 : 8 , 8x , 16x+3

Sol :

⇒8=2×2×2=23

⇒8x=(2×2×2)x=23x

⇒16x+3=(2×2×2×2)x+3


Question 6

If a = 2m and b = 2{m+1} , show that \dfrac{8a^3}{b^2}=2^{m+1}

Sol :

\dfrac{8a^3}{b^2}

\dfrac{8(2^m)^3}{(2^{m+1})^2}

Using property : (am)n=a(m×n)

\dfrac{2^3\times 2^{3m}}{2^{2m+2}}

Using property : a(m×n)=(am)n

\dfrac{2^3\times 2^{2m+m}}{2^{2m}+2^{2}}

\dfrac{2^3\times 2^{2m}\times 2^{m}}{2^{2m} \times 2^{2}}

\dfrac{2^3\times \not{(2^{2m} )}\times 2^{m}}{\not{(2^{2m}) }\times 2^{2}}

⇒2×2m

Using property : am×an=a(m+n)

⇒2m+1

Hence proved


Question 7

If x = 2k and y = 2k+3 , what is the value of \dfrac{x}{y} ?

Sol :

\dfrac{x}{y}

\dfrac{2^k}{2^{k+3}}

Using property: a(m+n)=am×an

\dfrac{\not{(2^k)}}{\not{(2^{k})}+2^{3}}

\dfrac{1}{8}


Question 8

Write each expression such that there is no variable in the denominator .

(i) \dfrac{3r^{-2}s^{3}}{12r^{-3}s^{7}}

Sol :

⇒(3r-2s3)(r3s-7)/12

Using property: am×an=a(m+n)

\dfrac{3}{12}(r^{-2+3}s^{3-7})

\dfrac{1}{4}(r^{1}s^{-4})

\dfrac{r^{1}s^{4}}{4}


(ii) \dfrac{12x^3y^{-2}z^{4}}{6x^{7}y^{-5}z^{-3}}

Sol :

\dfrac{12}{6}(x^3 y^{-2} z^4)(x^{-7} y^{5} z^{3})

Using property: am×an=a(m+n)

⇒2(x3-7×y-2+5×z4+3)

⇒2(x4×y3×z7)

⇒2x4y3z7


(iii) \dfrac{(5m)^{0}n^{-2}}{4mn^{-3}}

Sol :

\dfrac{5^0}{4}(m^0 \times n^{-2})(m^{-1}n^{3})

Using property: a0=1

\dfrac{1}{4}(1 \times n^{-2})(m^{-1}n^{3})

Using property: am×an=a(m+n)

\dfrac{1}{4}( n^{-2+3})(m^{-1})

\dfrac{nm^{-1}}{4}


(iv) \dfrac{14a^{-4}k}{7^{0}a^{3}k^{6}}

Sol :

\dfrac{14}{7^0}(a^{-4}k)(a^{-3}k^{-8})

Using property: a0=1

⇒14 (a-4k)(a-3k-8)

⇒14 (a-4-3k1-8)

⇒14 (a-7k-7)

 


Question 9

Simplify:

(i) \dfrac{2^{-n}. 8^{2n-1} . 16^{2n}}{4^{3n}}

Sol :

\dfrac{2^{-n}\times 2^{3(2n-1)} \times 2^{4(2n)}}{2^{2(3n)}}

\dfrac{2^{-n}\times 2^{6n-3} \times 2^{8n}}{2^{6n}}

\dfrac{2^{-n}\times \not{(2^{6n})} \times 2^{-3} \times 2^{8n}}{\not{(2^{6n})}}

⇒2-n× 2-3× 28n

Using property : am×an=a(m+n)

⇒2-n-3+8n

⇒2+7n-3


(ii) \dfrac{2^{n+4}-2.2^{n}}{2.2^{n+3}}+2^{-3}

Sol : wrong

Using property : am×an=a(m+n)

\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}+2^{-3}

\dfrac{\not{(2^{n+4})}}{\not{(2^{n+4})}}-\dfrac{2^{n+1}}{2^{n+4}}+2^{-3}

1-\dfrac{2^{n+1}}{2^{(n+1)+(3)}}+2^{-3}

1-\dfrac{\not{(2^{n+1})}}{\not{(2^{(n+1)})}\times 2^{(3)}}+2^{-3}

1-\dfrac{1}{2^3}+\dfrac{1}{2^3}

⇒1


(iii)  \dfrac{(0.6)^0-(0.1)^{-1}}{\left(\dfrac{3}{2^{3}}\right)^{-1}.\left(\dfrac{3}{2}\right)^{3}+\left(-\dfrac{1}{3}\right)^{-1}}

Sol :

Using property: a0=1

\dfrac{1-\left(\dfrac{1}{10}\right)^{-1}}{\left(\dfrac{3}{2^{3}}\right)^{-1} \times \left(\dfrac{3}{2}\right)^{3} + \left(-\dfrac{1}{3}\right)^{-1}}

Using property: a^{-1}=\dfrac{1}{a}

\dfrac{1-\dfrac{10}{1}}{\left(\dfrac{2^{3}}{3}\right) \times \left(\dfrac{3^3}{2^3}\right) - \left(\dfrac{3}{1}\right)}

\dfrac{1-10}{\left(\dfrac{\not{(8)}}{\not{(3)}}\right) \times \left(\dfrac{\not{(27)}}{\not{(8)}}\right) -3}

\dfrac{-9}{9-3}=-\dfrac{9}{6}=-\dfrac{3}{2}

 


Question 10

If \dfrac{9^n \times 3^5 \times (27)^3}{3\times (81)^4}=27 , then find the value of n

Sol :

\dfrac{9^n \times 3^5 \times (27)^3}{3\times (81)^4}

\dfrac{3^{2n} \times 3^5 \times 3^{9}}{3^{1}\times 3^{16}}

Using property: am×an=a(m+n)

\dfrac{3^{2n+5+9}}{3^{16+1}}

\dfrac{3^{2n+14}}{3^{17}}

⇒32n×314×3-17

⇒32n×314-17

⇒32n×3-3

\dfrac{3^{2n}}{3^3}=27  according to question

⇒32n=33×33

⇒32n=36

⇒32n=32×3

⇒n=3


Question 11

Show that :

(i) \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

Sol :

Using property : \dfrac{a^m}{a^n}=a{(m-n)}

⇒{x(a-b)}(a+b) × {x(b-c)}(b+c) × {x(c-a)}(c+a)

Using property : (am)n=am×n

⇒x(a-b)×(a+b) × x(b-c)×(b+c) × x(c-a)×(c+a)

Using property : (a+b)(a-b)=a2-b2

x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}

Using property : am×an=a(m+n)

x^{a^2-b^2+b^2-c^2+c^2-a^2}

⇒x0

Using property : a0=1

⇒1

Hence proved


(ii) \dfrac{x^{a+b}\times x^{b+c}\times x^{c+a}}{(x^a \times x^b \times x^c)^2}=1

Sol :

Using property : am×an=a(m+n)

\dfrac{x^{a+b+b+c+c+a}}{(x^{a+b+c})^2}

\dfrac{x^{2a+2b+2c}}{(x^{a+b+c})^2}

\dfrac{\not{(x^{2a+2b+2c})}}{\not{(x^{2a+2b+2c})}}

⇒1

Hence proved


Question 12

Find x such that

(i) \left(\dfrac{7}{4}\right)^{-3} \times \left(\dfrac{7}{4}\right)^{-5}=\left(\dfrac{7}{4}\right)^{3x-2}

Sol :

Using property : am×an=a(m+n)

\left(\dfrac{7}{4}\right)^{-3-5}=\left(\dfrac{7}{4}\right)^{3x-2}

\left(\dfrac{7}{4}\right)^{-8}=\left(\dfrac{7}{4}\right)^{3x-2}

On comparing both sides

⇒-8=3x-2

Transposing -2

⇒-8+2=3x

Transposing 3

x=\dfrac{-6}{3}=-2


(ii) \left(\dfrac{125}{8}\right)\times \left(\dfrac{125}{8}\right)^x=\left(\dfrac{5}{2}\right)^{18}

Sol :

Using property : am×an=a(m+n)

\left(\dfrac{125}{8}\right)^{x+1}=\left(\dfrac{5}{2}\right)^{18}

\left(\dfrac{5^3}{2^3}\right)^{x+1}=\left(\dfrac{5}{2}\right)^{18}

\left(\dfrac{5}{2}\right)^{3(x+1)}=\left(\dfrac{5}{2}\right)^{18}

Comparing both sides

⇒3(x+1)=18

⇒3x+3=18

⇒3x=18-3

x=\dfrac{15}{3}=5

 


Question 13

Find the value of x such that

(i) 3^{2x-1}=\dfrac{1}{27^{x-3}}

Sol :

3^{2x-1}=\dfrac{1}{3^{3(x-3)}}

32x-1=3-3x+9

On comparing both sides

⇒2x-1=-3x+9

⇒3x+2x=9+1

⇒5x=10

x=\dfrac{10}{5}=2


(ii) \left(\dfrac{p}{q}\right)^{3x+2}=\left(\dfrac{q}{p}\right)^{2-x}

Sol :

\left(\dfrac{p}{q}\right)^{3x+2}=\left(\dfrac{p}{q}\right)^{-2+x}

On comparing both sides

⇒3x+2=-2+x

⇒3x-x=-2-2

⇒2x=-4

x=\dfrac{-4}{2}=-2


 

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