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

## EXERCISE 2

Question 1

Evaluate:

(i) 2-3

Sol :

Using property

(ii) 60

Sol :

Using property: a0=1

60=1

(iii) (-3)-2

Sol :

Using property

(iv) 85 . 8-5

Sol :

Using property am×an=a(m+n)

⇒85 . 8-5 or

⇒85 × 8-5

⇒8(5-5)

(v)

Sol :

Using property

⇒2(-2)×-1

⇒22

⇒4

(vi) 3(8-1)0

Sol :

Using property: a0=1

⇒3×1

⇒3

(vii)

Sol :

Using property :

⇒2+3

⇒5

(viii) 7x0

Sol :

Using property: a0=1

⇒7×1

⇒7

(ix) a2 ÷ a-2

Sol :

Using property :

⇒a2×a2

Using property : an×am=a(n+m)

⇒a4

(x)

Sol :

Using property :

Using property: am×an=a(m+n)

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:

L.C.M of 6 and 3 is =2×3=6

 2 6,3 3 3,3 1,1

or

(xiii)

Sol :

Using property: a0=1

(xiv) (3x2)3

Sol :

Using property: (an)m=a(m×n)

⇒ 3(3)× x(2×3)

⇒ 27× x6

⇒ 27x6

(xv)

Sol :

Using property: (am)n=a(m×n)

Using property : a0=1

⇒1

Question 2

Simplify:

(i) x5 . x3

Sol :

Using property : am×an=a(m+n)

⇒x(5+3)

(ii)

Sol :

Using property :

⇒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 :

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)

Sol :

Using property:

Using property: a0=1

(ii)

Sol :

Using property:

⇒2a-3×b0× a-2 × b+1

Using property: an×am=a(m+n)

⇒2a-3-2×b0+1

⇒2a-5×b1

(iii)

Sol :

Using property :

⇒(2x2y-2)(21x-2y-2)

⇒(2×2)x2-2y-2-2

⇒4x0y-4

Using property : a0=1

⇒4y-4

(iv)

Sol :

Using property: (am)n=a(m×n)

⇒1

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

Sol :

⇒4k2(4-1k)+4k2(4k-2)

using property:

⇒k3+16

(vi)

Sol :

Using property : (am)n=a(m×n)

(vii)

Sol :

Using property : (am)n=a(m×n)

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

Sol :

Using property : (am)n=a(m×n)

Using property : a(m×n)=(am)n

⇒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 ?

Sol :

Using property: a(m+n)=am×an

Question 8

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

(i)

Sol :

⇒(3r-2s3)(r3s-7)/12

Using property: am×an=a(m+n)

(ii)

Sol :

Using property: am×an=a(m+n)

⇒2(x3-7×y-2+5×z4+3)

⇒2(x4×y3×z7)

⇒2x4y3z7

(iii)

Sol :

Using property: a0=1

Using property: am×an=a(m+n)

(iv)

Sol :

Using property: a0=1

⇒14 (a-4k)(a-3k-8)

⇒14 (a-4-3k1-8)

⇒14 (a-7k-7)

#### Question 9

Simplify:

(i)

Sol :

⇒2-n× 2-3× 28n

Using property : am×an=a(m+n)

⇒2-n-3+8n

⇒2+7n-3

(ii)

Sol : wrong

Using property : am×an=a(m+n)

⇒1

(iii)

Sol :

Using property: a0=1

Using property:

#### Question 10

If , then find the value of n

Sol :

Using property: am×an=a(m+n)

⇒32n×314×3-17

⇒32n×314-17

⇒32n×3-3

according to question

⇒32n=33×33

⇒32n=36

⇒32n=32×3

⇒n=3

#### Question 11

Show that :

(i)

Sol :

Using property :

⇒{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

Using property : am×an=a(m+n)

⇒x0

Using property : a0=1

⇒1

Hence proved

(ii)

Sol :

Using property : am×an=a(m+n)

⇒1

Hence proved

#### Question 12

Find x such that

(i)

Sol :

Using property : am×an=a(m+n)

On comparing both sides

⇒-8=3x-2

Transposing -2

⇒-8+2=3x

Transposing 3

(ii)

Sol :

Using property : am×an=a(m+n)

Comparing both sides

⇒3(x+1)=18

⇒3x+3=18

⇒3x=18-3

#### Question 13

Find the value of x such that

(i)

Sol :

32x-1=3-3x+9

On comparing both sides

⇒2x-1=-3x+9

⇒3x+2x=9+1

⇒5x=10

(ii)

Sol :

On comparing both sides

⇒3x+2=-2+x

⇒3x-x=-2-2

⇒2x=-4

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