Explore topic-wise InterviewSolutions in .

(i) (4^-1 × 3^-1)²  ( a^-n = 1/aⁿ)

= (1/4 × 1/3)²

= (1/12)²

= (1×1 / 12×12)

= 1/144

(ii) (5^-1 ÷ 6^-1)³

= ((1/5) / (1/6))³ (a^-n = 1/aⁿ)

= ((1/5) × 6)³ (1/a ÷ 1/b = 1/a × b/1)

= (6/5)³

= (6×6×6) / (5×5×5)

= 216/125

(iii) (2^-1 + 3^-1)^-1

(1/2 + 1/3)^-1 (a^-n = 1/aⁿ)

LCM of 2 and 3 is 6

= ((1×3 + 1×2)/6)^-1

= (5/6)^-1

= 6/5

(i) \((\frac{3}{2})^{-1}\times (\frac{3}{2})^{-1}\times (\frac{3}{2})^{-1}\times (\frac{3}{2})^{-1}\)

⇒ \((\frac{3}{2})^{-4}=(\frac{2}{3})^{4}\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{n}=a\times a............n\, times\)]

(ii) \((\frac{2}{5})^{-2}\times (\frac{2}{5})^{-2}\times (\frac{2}{5})^{-2}\)

⇒   \((\frac{2}{5})^{-6}=(\frac{5}{2})^{6}\)[Using  \(a^{-n}=\frac{1}{a^{n}}\);\(a^{n}=a\times a............n\, times\)]

Let the number be = x

So, 

(-15)^-1 ÷ x = (-5)^-1

1/-15 × 1/x  = 1/-5

1/x = (1×-15)/-5

1/x = 3

x = 1/3

Let the number \((-15)^{-1}\) should be divided by x

According to the question:

\((-15)^{-1}\)\(\div x\)\(=(-5)^{-1}\)

\(\frac{1}{-15}\div x=-\frac{1}{5}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(-\frac{1}{15}\times \frac{1}{x}=-\frac{1}{5}\)[Using \(p\div q=p\times\frac{1}{q}\)]

\(\frac{1}{x}=\frac{(-1)\times (-15)}{5}\)

\(\frac{1}{x}=3\)

\(x=\frac{1}{3}\)

Therefore \((-15)^{-1}\) should be divided by \(\frac{1}{3}\)

Let the number be = x

So, 

(1/2)^-1 × x = (-4/7)^-1

1/(1/2) × x = 1/(-4/7)  (1/aⁿ = a^-n)

x = (-7/4) / (2/1)

= (-7/4) × (1/2) (1/a ÷ 1/b = 1/a × b/1)

= -7/8

(i) \((\frac{1}{4})^{-4}\times (\frac{1}{4})^{-8}=(\frac{1}{4})^{-4x}\)

\((\frac{1}{4})^{-4}\times (\frac{1}{4})^{-8}=(\frac{1}{4})^{-4x}\)

⇒ \((\frac{1}{4})^{-4-8}\)= \((\frac{1}{4})^{-4x}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)]

Equating coefficients when bases are equal.

-4-8 = -4x

-12 = -4x

-12 = -4x

x = 3

(ii) \((\frac{-1}{2})^{19}\div (\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x+1}\)

\((\frac{-1}{2})^{19}\div (\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x+1}\)

⇒ \((\frac{1}{2})^{-19-8}\)= \((\frac{1}{2})^{-2x+1}\)[Using \(a^{n}\div a^{m}=a^{m-n}\)]

Equating coefficients when bases are equal.

-19-8 = -2x+1

-27 = -2x+1

-27-1 = -2x

-28 = -2x

x = 14

(iii) \((\frac{3}{2})^{-3}\times (\frac{3}{2})^{5}=(\frac{3}{2})^{2x+1}\)

\((\frac{3}{2})^{-3}\times (\frac{3}{2})^{5}=(\frac{3}{2})^{2x+1}\)

 ⇒ \((\frac{3}{2})^{-3+5}\)= \((\frac{3}{2})^{2x+1}\) [Using \(a^{n}\times a^{m}=a^{m+n}\)]

Equating coefficients when bases are equal.

-3+5 = 2x+1

2-1 = 2x

1 = 2x

x = \(\frac{1}{2}\)

(iv) \((\frac{2}{5})^{-3}\times (\frac{2}{5})^{15}=(\frac{2}{5})^{2+3x}\)

\((\frac{2}{5})^{-3}\times (\frac{2}{5})^{15}=(\frac{2}{5})^{2+3x}\)

⇒ \((\frac{2}{5})^{-3+15}=(\frac{2}{5})^{2+3x}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)]

Equating coefficients when bases are equal.

-3+15 = 2+3x

12-2 = 3x

10 = 3x

x = \(\frac{10}{3}\)

(v) \((\frac{5}{4})^{-x}\div (\frac{5}{4})^{-4}=(\frac{5}{4})^{5}\)

\((\frac{5}{4})^{-x}\div (\frac{5}{4})^{-4}=(\frac{5}{4})^{5}\)[Using \(a^{n}\div a^{m}=a^{m-n}\)]

Equating coefficients when bases are equal.

-x+4 = 5

-x = 5 - 4

-x = 1

x = -1

(vi) \((\frac{8}{3})^{2x+1}\times (\frac{8}{3})^{5}=(\frac{8}{3})^{x+2}\)

\((\frac{8}{3})^{2x+1}\times (\frac{8}{3})^{5}=(\frac{8}{3})^{x+2}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)]

Equating coefficients when bases are equal.

2x+1+5 = x+2

2x+6 = x+2

2x-x = 2-6

x = -4

Let the number should \(5^{-1}\) be multiplied by x

According to the question:

\(x\times 5^{-1}=(-7)^{-1}\)

\(x\times \frac{1}{5}=\frac{1}{-7}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(x=\frac{1}{-7}\div \frac{5}{1}=\frac{5}{-7}\)

\(x=\frac{1}{-7}\times \frac{5}{1}=\frac{5}{-7}\)= \(-\frac{5}{7}\)[Using and \(\frac{1}{a}\div \frac{1}{b}=\frac{1}{a}\times \frac{b}{1}\)]

Therefore \(5^{-1}\) should be multiplied by \(\frac{-5}{7}\)

Let the number be = x

So,

5^-1 × x = (-7)^-1

1/5 × x = 1/-7 (1/aⁿ = a^-n)

x = (-1/7) / (1/5)

= (-1/7) × (5/1) (1/a ÷ 1/b = 1/a × b/1)

= -5/7

(i) (1/4)^-4 × (1/4)^-8 = (1/4)^-4x

(1/4)^-4-8 = (1/4)^-4x   (aⁿ × a^m = a^n+m)

(1/4)^-12 = (1/4)^-4x

When the bases are same we can directly equate the coefficients.

-12 = -4x

x = -12/-4

= 3

(ii) (-1/2)^-19 ÷ (-1/2)⁸ = (-1/2)^-2x+1

(1/2)^-19-8 = (1/2)^-2x+1  (aⁿ ÷ a^m = a^n-m)

(1/2)^-27 = (1/2)^-2x+1

When the bases are same we can directly equate the coefficients

-27 = -2x+1

-2x = -27-1

x = -28/-2

= 14

(iii) (3/2)^-3 × (3/2)⁵ = (3/2)^2x+1

(3/2)^-3+5 = (3/2)^2x+1  (aⁿ × a^m = a^n+m)

(3/2)² = (3/2)^2x+1

When the bases are same we can directly equate the coefficients

2 = 2x+1

2x = 2-1

x = 1/2

Let the number should \((\frac{5}{3})^{-2}\) be multiplied by x

According to the question:

\(x\times (\frac{5}{3})^{-2}\)= \((\frac{7}{3})^{-1}\)

 \(x\times (\frac{3}{5})^{2}\)= \(\frac{3}{7}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(x= \frac{3}{7}\div (\frac{3}{5})^{2}\)

\(x=\frac{3}{7}\times \frac{25}{7}=\frac{25}{21}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)]

Therefore

\((\frac{1}{2})^{-1}\) should be multiplied by \(\frac{25}{21}\)

Let the number \((\frac{1}{2})^{-1}\) should be multiplied by x

According to the question:

\(x\times (\frac{1}{2})^{-1}=(-\frac{4}{7})^{-1}\)

\(x\times 2=-\frac{7}{4}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(x=-\frac{7}{4}\div 2\)

\(x=-\frac{7}{4}\times \frac{1}{2}\)= \(\frac{-7}{8}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)]

Therefore

\((\frac{1}{2})^{-1}\) should be multiplied by \(-\frac{7}{8}\)

Let the number \((\frac{1}{2})^{-1}\) should be multiplied by x

According to the question:

\(x\times (\frac{1}{2})^{-1}=(-\frac{4}{7})^{-1}\)

\(x\times 2=-\frac{7}{4}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\)]

x = \(-\frac{7}{4}\times \frac{1}{2}=-\frac{7}{8}\)

Therefore \((\frac{1}{2})^{-1}\) should be multiplied by \(-\frac{7}{8}\)

According to question,

let whose number will be x,

so,

x × (-10/3) = -7/3

x = (-7/3 ) ÷ ( -10/3)

x = -7/3 × -3/10

 x= 7/10

5^2x ÷ 5^-3 = 5⁵

5^2x+3 = 5⁵ (aⁿ ÷ a^m = a^n-m)

When the bases are same we can directly equate the coefficients as below,

2x+3 = 5

2x = 5-3

2x = 2

x = 1

x = (4/5)^-2 ÷ (1/4)²

= (5/4)² ÷ (1/4)²  (1/aⁿ = a^-n)

= (5/4)² × (4/1)²  (1/a ÷ 1/b = 1/a × b/1)

= 25/16 × 16

= 25

x^-1 = 1/25

Let the number \(5^{-1}\) should be multiplied by \(x\)

According to the question:

\(x\times 5^{-1}=(-7)^{-1}\)

\(x\times \frac{1}{5}=\frac{1}{-7}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(x=\frac{1}{-7}\times \frac{5}{1}=\frac{5}{-7}\)

Therefore \(5^{-1}\) should be multiplied by \(\frac{-5}{7}\)

Let the number \((-15)^{-1}\) should be divided by x

According to the question:

\((-15)^{-1}\)÷ x= \((-5)^{-1}\)

\(\frac{1}{-15}÷x=-\frac{1}{5}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

\(\frac{1}{-15}\times \frac{1}{x}=-\frac{1}{5}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)]

\(\frac{1}{x}=\frac{(-1)\times (-15)}{5}\)

\(\frac{1}{x}=3\)

\(x=\frac{1}{3}\)

Therefore

\((-15)^{-1}\) should be divided by \(\frac{1}{3}\)

x= (3/2)² × (2/3)^-4

= (3/2)² × (3/2)⁴  (1/aⁿ = a^-n)

= (3/2)²⁺⁴  (aⁿ × a^m = a^n+m)

= (3/2)6

x^-2 = ((3/2)⁶)^-2

= (3/2)^-12

= (2/3)¹²

(i) (3² + 2²) × (1/2)³

= (9 + 4) × 1/8 

= 13/8

(ii) (3² – 2²) × (2/3)^-3

= (9-4) × (3/2)³

= 5 × (27/8)

= 135/8

(iii) ((1/3)^-3 – (1/2)^-3) ÷ (1/4)^-3

= (3³ – 2³) ÷ 4³  (1/a^-n = aⁿ)

= (27-8) ÷ 64

= 19 × 1/64  (1/a ÷ 1/b = 1/a × b/1)

= 19/64

\(5^{2x}\div 5^{-3}=5^{5}\)

⇒ \((5)^{2x-3}=5^{5}\)[Using \(a^{n}\div a^{m}=a^{m-n}\)]

Equating coefficients when bases are equal.

2x+3 = 5

2x = 5-3

2x = 2

x = 1

(i) 5^-2

= 1/5² ( a^-n = 1/aⁿ)

= 1/25

(ii) (-3)^-2

= (1/-3)² (a^-n = 1/aⁿ)

= 1/9

(iii) (1/3)^-4

= 3⁴ (1/a^-n = aⁿ)

= 81 

(iv) (-1/2)^-1

= -2¹ (1/a^-n = aⁿ)

= -2  

Let the number be x

So, 

5^-1 × x = (-7)^-1

1/5 × x = 1/-7

x = (-1/7) / (1/5)

= (-1/7) × (5/1)

= -5/7

Let the number be = x

So,

(1/2)-1 × x = (-4/7)-1

1/(1/2) × x = 1/(-4/7)

x = (-7/4) / (2/1)

= (-7/4) × (1/2)

= -7/8

(i) (3/2)^-1 × (3/2)^-1 × (3/2)^-1 × (3/2)^-1

= (3/2)^-4  (a^-n = 1/aⁿ, aⁿ = a×a…n times)

(ii) (2/5)^-2 × (2/5)^-2 × (2/5)^-2

= (2/5)^-6 (a^-n = 1/aⁿ, aⁿ = a×a…n times)

\(x = (\frac{3}{2})^{2}\times (\frac{2}{3})^{-4}\)

\(x = (\frac{3}{2})^{2}\times (\frac{3}{2})^{4}\)[Using \(a^{m}\times a^{n} = a^{m+n}\)]

\(x = (\frac{3}{2})^{6}\)

\(x^{-2} = (\frac{2}{3})^{12}\)

(i) ((1/3)^-3 – (1/2)^-3) ÷ (1/4)^-3

= (3³ – 2³) ÷ 4³ (1/aⁿ = a^-n)

= (27-8) ÷ 64

= 19 ÷ 64

= 19 × 1/64 (1/a ÷ 1/b = 1/a × b/1)

= 19/64

(ii) (3² – 2²) × (2/3)^-3

= (9 – 4) × (3/2)³ (1/aⁿ = a^-n)

= 5 × (27/8)

= 135/8

(iii) ((1/2)^-1 × (-4)^-1)^-1

= (2¹ × (1/-4))^-1 (1/aⁿ = a^-n)

= (1/-2)^-1 (1/aⁿ = a^-n)

= -2¹

= -2

(iv) (((-1/4)²)^-2)^-1

= ((-1/16)^-2)^-1 (1/aⁿ = a^-n)

= ((-16)²)^-1 (1/aⁿ = a^-n)

= (256)^-1  (1/aⁿ = a^-n)

= 1/256

(i) \(\{(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}\}\div (\frac{1}{4})^{-3}\)

⇒ \(\{(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}\}\div (\frac{1}{4})^{-3}\)

⇒ \((3^{3}-2^{3})\div 4^{3}\)[Using \(\cfrac{1}{a^{n}}\)= \(a^{-n}\)] 

⇒ (27-8) ÷ \(4^{3}\) 

⇒ 19 ÷ \(4^{3}\)

⇒ \(19\times \frac{1}{64}\)= \(\frac{19}{64}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)] 

(ii) \((3^{2}-2^{2})\times (\frac{2}{3})^{-3}\)

⇒ \((3^{2}-2^{2})\times (\frac{2}{3})^{-3}\)

⇒ \((9-4)\times (\frac{3}{2})^{-3}\)[Using  \(\cfrac{1}{a^{n}}\)= \(a^{-n}\)] 

⇒ \((5)\times (\frac{3}{2})^{3}\)

⇒ \((5)\times (\frac{27}{8})\)[Using \(a^{n}=a\times a ...... ........n\,times\)] 

⇒ \(\frac{135}{8}\)

(iii) \(\{(\frac{1}{2})^{-1}\times (-4)^{-1}\}^{-1}\)

⇒ \(\{(\frac{1}{2})^{-1}\times (-4)^{-1}\}^{-1}\)

⇒ \(\{(2)\times (\frac{1}{-4})\}^{-1}\)[Using  \(\cfrac{1}{a^{n}}\)= \(a^{-n}\)] 

⇒ \(\{-\frac{1}{2}\}^{-1}\)

⇒ -2

(iv) \([\{(\frac{-1}{4})^{2}\}^{-2}]^{-1}\)

⇒ \([\{(\frac{-1}{4})^{2}\}^{-2}]^{-1}\)

⇒ \(\{-\frac{1}{4}\}^{4}\)[Using (aⁿ)ᵐ = aᵐⁿ] 

⇒ \(\frac{1}{256}\)[Using \(a^{n}=a\times a ...... ........n\,times\)] 

(v) \(\{(\frac{2}{3})^{2}\}^{3}\times (\frac{1}{3})^{-4}\times 3^{-1}\times 6^{-1}\)

\((\frac{2}{3})^{6}\times (\frac{1}{3})^{-4}\times \frac{1}{3}\times \frac{1}{2\times 3}\)

\((\frac{2}{3})^{6}\times (\frac{1}{3})^{-4}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{3}\)

\(\cfrac{2^{6-1}}{3^{6-4+1+1}}\)      \(a^{m}\times a^{n}=a^{m+n}\)

\(\cfrac{2^{5}}{3^{4}}\)= \(\cfrac{32}{81}\)

(i) \((3^{2}+2^{2})\times (\frac{1}{2})^{3}\)

⇒ \((9+4)\times \frac{1}{2^{3}}\)[Using \(a^{n}=a\times a.............n\, times\)]

⇒ \(13\times \frac{1}{8}=\frac{13}{8}\)

(ii) \((3^{2}+2^{2})\times (\frac{2}{3})^{-3}\)

⇒ \((9+4)\times (\frac{3}{2})^{3}\)[Using \(a^{-n}=\frac{1}{a^{n}}\) \(a^{n}=a\times a.............n\, times\)]

⇒ \(13\times \frac{27}{8}=\frac{351}{8}\)

(iii) \([(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}]\div (\frac{1}{4})^{-3}\)

⇒ \([(3)^{3}-(2)^{3}]\div (4)^{3}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

⇒ [27-8]\(\div 4^{3}\)

⇒ \(19\times \frac{1}{4^{2}}\)[Using \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)]

⇒ \(19\times \frac{1}{4^{2}}\)= \(\frac{19}{64}\)[Using \(a^{n}=a\times a.............n\, times\)]

(iv) \((2^{2}+3^{2}-4^{2})\div (\frac{3}{2})^{2}\)

⇒  (4+9-16)\(\div (\frac{3}{2})^{2}\)[Using \(a^{n}=a\times a.............n\, times\)]

⇒ \((-3)\div \frac{9}{4}\)

⇒ \(-3\times \frac{4}{9}=-\frac{4}{3}\)[Using  \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)]

(i) (4^-1 × 3^-1)²

= (1/4 × 1/3)² (a^-n = 1/aⁿ)

= (1/12)²

= 1/144

(ii) (5^-1 ÷ 6^-1)³

= (1/5 ÷ 1/6)³ (a^-n = 1/aⁿ)

= (1/5 × 6)³ (1/a ÷ 1/b = 1/a × b/1)

= (6/5)³

= 216/125

(iii) (2^-1 + 3^-1)^-1

= (1/2 + 1/3)^-1 (a^-n = 1/aⁿ)

LCM of 2 and 3 is 6

= ((3+2)/6)^-1

= (5/6)-1 (1/a^-n = aⁿ)

=  6/5

(iv) (3^-1 × 4^-1)^-1 × 5^-1

= (1/3 × 1/4)^-1 × 1/5 (a^-n = 1/aⁿ)

= (1/12)^-1 × 1/5 (1/a^-n = aⁿ)

= 12 × 1/5

= 12/5

i) (3/4)^-2

= (4/3)²  (a/b)^-n = (b/a)ⁿ)

(ii) (5/4)^-3

= (4/3)³  ((a/b)^-n = (b/a)ⁿ)

(iii) 4³ × 4^-9

= (4)^3-9  (aⁿ × a^m = a^n+m)

= 4^-6   (1/aⁿ = a^-n)

= (1/4)⁶

(iv) ((4/3)^-3)^-4

= (4/3)¹²  ((aⁿ)^m = a^nm)

(v) ((3/2)⁴)^-2

= (3/2)^-8 ((aⁿ)^m = a^nm)

= (2/3)⁸ (1/aⁿ = a^-n)

(i) (1/4)³

(4)^-3  (1/aⁿ = a^-n)

(ii) 3⁵ 

= (1/3)^-5  (1/aⁿ = a^-n)

(iii) (3/5)⁴

(5/3)^-4 ((a/b)^-n = (b/a)ⁿ)

(iv) ((3/2)⁴)^-3

= (3/2)^-12 ((aⁿ)^m = a^nm)

(v) ((7/3)⁴)^-3

= (7/3)^-12 ((aⁿ)^m = a^nm)

(i) \((\frac{3}{4})^{-2}\)

⇒ \((\frac{3}{4})^{-2}\)= \((\frac{4}{3})^{2}\)[Using \((\frac{a}{b})^{-n}\)= \((\frac{b}{a})^{n}\)]

(ii) \((\frac{5}{4})^{-3}\)

⇒ \((\frac{5}{4})^{-3}\)= \((\frac{4}{5})^{a}\)[Using \((\frac{a}{b})^{-n}\)= \((\frac{b}{a})^{n}\) ] 

(iii) \(4^{3}\times 4^{-9}\)

⇒  \(4^{3}\times 4^{-9}\)= \(4^{3-9}\)= \(4^{-6}\)[Using (aⁿ× aᵐ = aᵐ⁺ⁿ ] 

⇒ \(4^{-6}\)= \((\frac{1}{4})^{6}\)[Using \(\frac{1}{a^{n}}\)= \(a^{-n}\)] 

(iv) \(\{(\frac{4}{3})^{-3}\}^{-4}\)

⇒ \(\{(\frac{4}{3})^{-3}\}^{-4}\)= \((\frac{4}{3})^{12}\)[Using (aⁿ)ᵐ = aᵐⁿ ] 

(v) \(\{(\frac{3}{2})^{4}\}^{-2}\)

⇒ \(\{(\frac{3}{2})^{4}\}^{-2}\)= \((\frac{3}{2})^{-8}\)= \((\frac{2}{3})^{8}\)[Using (aⁿ)ᵐ = aᵐⁿ and \(\frac{1}{a^{n}}\)= \(a^{-n}\)]

As (we know that a^-n = 1/an)

(i) 2^-3 = 1/2³ = 1/2×2×2 

= 1/8 

As (we know that a^-n = 1/an)

(ii) (-4)^-2 = 1/-4² = 1/-4×-4 

= 1/16 

As (we know that 1/a^-n = aⁿ)

(iii) 1/(3)^-2 = 3² 

= 3×3 

= 9 

As (we know that a^-n = 1/an)

(iv) (1/2)^-5 = 25 / 15 

= 2×2×2×2×2 

= 32 

As (we know that a^-n = 1/an)

(v) (2/3)^-2 = 3² / 2² 

= 3×3 / 2×2 

= 9/4

(i) \(2^{-3}\) = \(\frac{1}{2^{8}}\) = \(\frac{1}{8}\)[Using \(a^{-n} = \frac{1}{a^{n}}\)]

(ii) \((-4)^{-2}\) = \(\frac{1}{-4^{2}}\) = \(\frac{1}{(-4)\times (-4)}\) =\(\frac{1}{16}\)[Using \(a^{-n} = \frac{1}{a^{n}}\) ]

(iii) \(\frac{1}{3^{-2}}\) = \(\frac{3\times 3}{1}\) = \(\frac{9}{1}\)[Using \(\frac{1}{a^{-n}}\) = \(a^{n}\)]

(iv) \((\frac{1}{2})^{-5}\) = \(\frac{2^{5}}{1^{5}}\) = \(\frac{2^{5}}{1}\) = \(\frac{32}{1}\)[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\)]

(v) \((\frac{2}{3})^{-2}\) = \(\frac{3^{2}}{2^{2}}\) = \(\frac{9}{4}\)[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\);\(a^{2}\) = \(a\times a\)]

(i) 6^-1

= 1/6¹ ( a^-n = 1/aⁿ)

= 1/6

(ii) (-7)^-1

= 1/-7¹ ( a^-n = 1/aⁿ)

= -1/7 

(iii) (1/4)^-1

= 4¹ (1/a^-n = aⁿ)

= 4

(iv) (-4)^-1 × (-3/2)^-1

= 1/-4¹ × (2/-3)¹ (a^-n = 1/aⁿ, 1/a^-n = aⁿ)

= 1/-2 × -1/3

= 1/6

(i) \((\frac{1}{4})^{3}\)

⇒ \((\frac{1}{4})^{3}\)= \((4)^{-3}\)[Using \(\frac{1}{a^{n}}\)= \(a^{-n}\)] 

(ii) \(3^{5}\)

⇒ \(3^{5}\)= \((\frac{1}{3})^{-5}\)= \((4)^{-3}\)[Using  \(\frac{1}{a^{n}}\)= \(a^{-n}\)] 

(iii) \((\frac{3}{5})^{4}\)

⇒ \((\frac{3}{5})^{4}\)= \((\frac{5}{3})^{-4}\)[Using \((\frac{a}{b})^{-n}\)= \((\frac{b}{n})^{n}\)] 

(iv) \(\{(\frac{3}{2})^{4}\}^{-3}\)

⇒ \(\{(\frac{3}{2})^{4}\}^{-3}\) = \((\frac{3}{2})^{-12}\)[Using (aⁿ)ᵐ = aᵐⁿ] 

(v) \(\{(\frac{7}{3})^{4}\}^{-3}\)

⇒ \(\{(\frac{7}{3})^{4}\}^{-3}\)= \((\frac{7}{3})^{-12}\)[Using (aⁿ)ᵐ = aᵐⁿ]

(i) \(5^{-2}\)

⇒ \(\frac{1}{5^{8}}\)=\(\frac{1}{25}\)[Using \(a^{-n}=\frac{1}{a^{n}}\) ;\(a^{n}=a\times a........n\, times\)]

(ii) \((-3)^{-2}\)

⇒ \((\frac{1}{-3})^{2}\)=\(\frac{1}{-3}\times \frac{1}{-3}\)=\(\frac{1}{9}\)[Using   \(a^{-n}=\frac{1}{a^{n}}\) ;\(a^{n}=a\times a........n\, times\)]

(iii) \((\frac{1}{3})^{-4}\)

⇒ \((\frac{3}{1})^{4}\)=\(3^{4}\)=81[Using   \(a^{-n}=\frac{1}{a^{n}}\) ;\(a^{n}=a\times a........n\, times\)]

(iv) \((\frac{-1}{2})^{-1}\)

⇒ \(\frac{-2}{1}\)= -2[Using   \(a^{-n}=\frac{1}{a^{n}}\) ;\(a^{n}=a\times a........n\, times\)]

x = \((\frac{4}{5})^{-2}\div (\frac{1}{4})^{2}\)

On using, [Using \(a^{-n}=\frac{1}{a^{n}}\)]

we get,

 x = \((\frac{5}{4})^{2}\div (\frac{1}{4})^{2}\)

[Using and \(\frac{1}{a}\div \frac{1}{b}\)=\(\frac{1}{a}\times \frac{b}{1}\)] 

x = \((\frac{5}{4})^{2}\times (\frac{4}{1})^{2}\)

x = \(\frac{5}{4}\times \frac{5}{4}\times 4\times 4\)

x = 25

\(x^{-1} = \frac{1}{25}\)

(i) \((\frac{1}{2})^{-1}+(\frac{1}{3})^{-1}+(\frac{1}{4})^{-1}\)

⇒ (2)+(3)+(4)[Using \(a^{-n}=\frac{1}{a^{n}}\)]

⇒ 2+3+4 = 9

(ii) \((\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}\)

⇒ \((2^{2})+(3^{2})+(4^{2})\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{2}=a\times a\)]

⇒ 4+9+16 = 29

(iii) \((2^{-1}\times 4^{-1})÷2^{2}\)

⇒ \((\frac{1}{2}\times \frac{1}{4})÷(\frac{1}{2^{2}})\)[Using \(a^{-n}=\frac{1}{a^{n}}\)and \(\frac{1}{a}\div \frac{1}{b}\) = \(\frac{1}{a}\times \frac{b}{1}\)]

⇒ \((\frac{1\times 1}{2\times 4})\times 2^{2}\)[Using \(\frac{1}{a}\times \frac{b}{1}\)= \(\frac{1}{ab}\);\(a^{2}=a\times a\)] 

⇒ \(\frac{1}{8}\times 4\) = \(\frac{1}{2}\)

(iv) \((5^{-1}\times 2^{-1})÷6^{-1}\)

⇒ \((\frac{1}{5}\times \frac{1}{2})÷(\frac{1}{6})\)[Using  \(a^{-n}=\frac{1}{a^{n}}\)and \(\frac{1}{a}\div \frac{1}{b}\) = \(\frac{1}{a}\times \frac{b}{1}\)]

⇒ \((\frac{1\times 1}{5\times 2})\times (\frac{6}{1})\)[Using  \(\frac{1}{a}\times \frac{b}{1}\)= \(\frac{1}{ab}\)] 

⇒   \(\frac{1}{10}\times 6\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)

(i) (1/2)^-1 + (1/3)^-1 + (1/4)^-1

= 2¹ + 3¹ + 4¹   (1/a^-n = aⁿ)

= 2+3+4

= 9

(ii) (1/2)^-2 + (1/3)^-2 + (1/4)^-2

22 + 32 + 42  (1/a^-n = aⁿ)

= 2×2 + 3×3 + 4×4

= 4+9+16

= 29

(iii) (2^-1 × 4^-1) ÷ 2^-2

= (1/2¹ × 1/4¹) / (1/2²)   (a^-n = 1/aⁿ)

= (1/2 × 1/4) × 4/1   (1/a ÷ 1/b = 1/a × b/1)

= 1/2

(i) 6020000000000000

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

In this question total number of digits leaving one digit from left are 15.

Therefore the standard form is: \(6.02\times 10^{15}\)

(ii) 0.00000000000942

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

If the number has all the digits to the right of the decimal then powers will be negative. In this question total number of digits after decimal are 12.

Therefore the standard form is: \(9.42\times 10^{-12}\)

(iii) 0.00000000085

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

If the number has all the digits to the right of the decimal then powers will be negative. In this question total number of digits after decimal are 12.

Therefore the standard form is: \(8.5\times 10^{-10}\)

(iv) \(846\times 10^{-7}\)

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

In this question total number of digits are 2.

Therefore the standard form is: \(8.46\times 10^{9}\)

(v) \(3759\times 10^{-4}\)

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

In this question total number of digits are 3.

Therefore the standard form is: \(8.46\times 10^{-1}\)

(vi) 0.00072984

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit. If the number has all the digits to the right of the decimal then powers will be negative.

In this question total number of digits after decimal are 4.

Therefore the standard form is: \(7.2984\times 10^{-4}\)

(vii) 0.000437\(\times 10^{4}\)

To write in the standard form, count the number of digits leaving one digit from the left. The total number of digits so obtained becomes power of 10. Decimal comes after first left digit.

If the number has all the digits to the right of the decimal then powers will be negative. In this question total number of digits after decimal are 4.

Therefore the standard form is: 4.37

(viii) 4÷100000

To write in the standard form, Count the number of zeros of the divisor. This number of zeros becomes negative power of 10.

Therefore the standard form is: \(4\times 10^{-5}\)

(i) 3^-1 + 4^-1

1/3 + 1/4 (we know that a^-n = 1/aⁿ)

LCM of 3 and 4 is 12

= (1×4 + 1×3)/12

= (4+3)/12

= 7/12

(ii) (3⁰ + 4^-1) × 2²

(1 + 1/4) × 4 (we know that a-n = 1/aⁿ, a⁰ = 1)

LCM of 1 and 4 is 4

= (1×4 + 1×1)/4 × 4

= (4+1)/4 × 4

= 5/4 × 4

= 5

(iii) (3^-1 + 4^-1 + 5^-1)⁰

(We know that a⁰ = 1)

(3^-1 + 4^-1 + 5^-1)⁰ = 1

(iv) ((1/3)^-1 – (1/4)^-1)^-1

(3¹ – 4¹)^-1 (we know that 1/a-n = an, a-n = 1/an)

= (3-4)^-1

= (-1)^-1

= 1/-1 

= -1

(i) \(3^{-1}+4^{-1}\)

⇒ \(\frac{1}{3}+\frac{1}{4}\)= \(\frac{4+2}{12}\)= \(\frac{7}{12}\)(LCM of 3 and 4 is 12)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\)]

(ii) (\(3^{0}+4^{-1}\)) x \(2^{2}\)

⇒ \((1+\frac{1}{4})\times 4\) = \((\frac{4+1}{4})\times 4\) = \(\frac{5\times 4}{4}\) = 5(LCM of 1 and 4 is 4)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\); \(a^{0}\) = 1;\(a^{2}\) = \(a\times a\)]

(iii) \((3^{-1}+4^{-1}+5^{-1})^{0}\)

⇒ \((3^{-1}+4^{-1}+5^{-1})^{0}\) = 1

[Using \(a^{0}\) = 1]

We know that any number to power zero is always equal to 1.

(iv) \(\{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}+(\frac{1}{4})^{-2}\}\)

⇒ (3)-(4)+\((4^{2})\)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\);  \(a^{2}\) = \(a\times a\)]

3-4+16 = 15

(i) \(6^{-1}\)

⇒ \(6^{-1}\)= \(\frac{1}{6}\)[Using \(a^{-n}\)= \(\frac{1}{a^{n}}\)]

 (ii) \((-7)^{-1}\)

⇒ \((-7)^{-1}\)= \(\frac{1}{-7}\)= \(-\frac{1}{7}\)[Using \(a^{-n}\)= \(\frac{1}{a^{n}}\)]

 (iii) \((\frac{1}{4})^{-1}\)

⇒ \((\frac{1}{4})^{-1}\)= 4[Using \(a^{-n}\)= \(\frac{1}{a^{n}}\)]

 (iv) \((-4)^{-1}\times (\frac{-3}{2})^{-1}\)

⇒ \((-4)^{-1}\times (\frac{-3}{2})^{-1}\)=\(\frac{1}{-4}\times \frac{2}{-3}\)=\(\frac{2}{12}\)=\(\frac{1}{6}\)[Using \(a^{-n}\)= \(\frac{1}{a^{n}}\)]

 (v) \((\frac{3}{5})^{-1}\times (\frac{5}{2})^{-1}\)

⇒ \((\frac{3}{5})^{-1}\times (\frac{5}{2})^{-1}\)=  \(\frac{5}{3}\times \frac{2}{5}\)=\(\frac{10}{15}\)=\(\frac{2}{3}\) [Using \(a^{-n}\)= \(\frac{1}{a^{n}}\)]

(i) \(\{4^{-1}\times 3^{-1}\}^{2}\)

⇒ \((\frac{1}{4}\times \frac{1}{3})^{2}\)= \((\frac{1}{12})^{2}\)= \(\frac{1}{144}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\); \(a^{n}\)=\(a\times a...... .......n\, times\)] 

(ii) \(\{5^{-1}\div 6^{-1}\}^{3}\)

⇒ \((\frac{1}{5}\div \frac{1}{6})^{3}\)[Using  \(a^{-n}\)=\(\frac{1}{a^{n}}\) ] 

⇒ \((\frac{1}{5}\times 6)^{3}\)= \(\frac{6^3}{5^3}\)=\(\frac{216}{125}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{1}{b}\)] 

⇒ \(\frac{6^3}{5^3}\)=\(\frac{216}{125}\)[Using  \(a^{n}\)=\(a\times a...... .......n\, times\)] 

(iii) \((2^{-1}+3^{-1})^{-1}\)

⇒ \((\frac{1}{2}+\frac{1}{3})^{-1}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\)] 

⇒ \((\frac{3+2}{6})^{-1}\)= \((\frac{5}{6})^{-1}\)= \(\frac{6}{5}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\)] 

(iv) \(\{3^{-1}\times 4^{-1}\}^{-1}\times 5^{-1}\)

⇒ \((\frac{1}{3}\times \frac{1}{4})^{-1}\times \frac{1}{5}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\)] 

⇒ \((\frac{1}{12})^{-1}\times \frac{1}{5}\)= \(\frac{12}{5}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\)] 

(v) \((4^{-1}-5^{-1})^{-1}\div 3^{-1}\)

⇒ \((\frac{1}{4}-\frac{1}{5})^{-1}\div \frac{1}{3}\)[Using \(a^{-n}\)=\(\frac{1}{a^{n}}\) ] 

⇒ \((\frac{5-4}{20})^{-1}\times 3\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{1}{b}\) ]

⇒ \(\frac{1}{20}\times 3\)= \(\frac{3}{20}\)

(i) \((4^{-1}\times 3^{-1})^{2}\)

⇒ \((\frac{1}{4}\times \frac{1}{3})^{2}\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{2}=a\times a\)] 

⇒ \(\frac{1}{12}\times \frac{1}{12}\) = \(\frac{1}{144}\)

(ii) \((5^{-1}\div 6^{-1})^{3}\)

⇒ \((\frac{1}{5}\times \frac{1}{6})^{3}\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{n}=a\times a..........n\,times\)]

⇒  \((\frac{1}{5}\times \frac{6}{1})^{3}\) = \((\frac{6}{5})^{3}\)=\(\frac{216}{125}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)=\(\frac{1}{a}\times \frac{b}{1}\)] 

(iii) \((2^{-1}\div 4^{-1})\div 2^{-2}\)

⇒   \((\frac{1}{2}\times \frac{1}{4})\)\(\div \frac{1}{2^{2}}\)[Using \(a^{-n}=\frac{1}{a^{n}}\);\(a^{2}=a\times a\)]

⇒ \(\frac{1}{8}\times \frac{4}{1}\) = \(\frac{1}{2}\)[Using and \(\frac{1}{a}\div \frac{1}{b}\)=\(\frac{1}{a}\times \frac{b}{1}\)] 

(iv) \((3^{-1}\times 4^{-1})\times 5^{-1}\)

⇒ \((\frac{1}{3}\times \frac{1}{4})\times \frac{1}{5}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)] 

⇒   \(\frac{1}{12}\times \frac{1}{5}\) = \(\frac{1}{60}\)

(i) \(4.83\times 10^{7}\)

In case of positive power of 10. The usual form of the number is written after multiplying the given numbers and then counts the numbers from the right and put decimal.

Step 1: Multiply the given numbers: 4.83× 10000000 = 4830000000

Step 2: Put decimal after two places from the right: 48300000.00

Step 3: Write the number after in the usual form: 48300000

(ii) \(3.02\times 10^{-6}\)

In case of negative powers, decimal sifts to left equal to the power of 10.

Step 1: Here the power of 10 is negative 6.

Step 2: Therefore decimal will sift six places to the left. i.e: 0.00000302

(iii) \(4.5\times 10^{4}\)

In case of positive power of 10. The usual form of the number is written after multiplying the given numbers and then counts the numbers from the right and put decimal.

Step 1: Multiply the given numbers: 4.5× 10000 = 450000

Step 2: Put decimal after one place from the right: 45000.0

Step 3: Write the number after in the usual form: 45000

(iv) \(3\times 10^{-8}\)

In case of negative powers, decimal sifts to left equal to the power of 10.

Step 1: Here the power of 10 is negative 8.

Step 2: Therefore decimal will sift eight places to the left, and we write zeros before the number to make eight places . i.e: 0.00000003

(v) \(1.0001\times 10^{9}\)

In case of positive power of 10. The usual form of the number is written after multiplying the given numbers and then counts the numbers from the right and put decimal.

Step 1: Multiply the given numbers: 1.0001× 1000000000 = 10001000000000

Step 2: Put decimal after four places from the right: 1000100000.0000

Step 3: Write the number after in the usual form: 1000100000

(vi) \(5.8\times 10^{2}\)

In case of positive power of 10. The usual form of the number is written after multiplying the given numbers and then counts the numbers from the right and put decimal.

Step 1: Multiply the given numbers: 5.8× 100 = 5800

Step 2: Put decimal after one place from the right: 580.0

Step 3: Write the number after in the usual form: 580

(vii) \(3.61492\times 10^{6}\)

In case of positive power of 10. The usual form of the number is written after multiplying the given numbers and then counts the numbers from the right and put decimal.

Step 1: Multiply the given numbers: 3.61492 × 1000000 = 5800

Step 2: Put decimal after five places from the right: 3614920.00000

Step 3: Write the number after in the usual form: 3614920

(viii) \(3.25\times 10^{-7}\)

In case of negative powers, decimal sifts to left equal to the power of 10.

Step 1: Here the power of 10 is negative 8.

Step 2: Therefore decimal will sift seven places to the left, and we write zeros before the number to make seven places . i.e: 0.000000325