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Class 8 NCERT Solutions- Chapter 12 Exponents and Powers – Exercise 12.1

  • Last Updated : 10 Mar, 2021

Question 1. Evaluate:

Solution:

(i) 3–2 

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3-2\frac{1}{(3^2)} = \frac{1}{9}                 (Property used: a-n\frac{1}{a^n} )



(ii) (– 4)– 2 

(-4)-2\frac{1}{(-4)^2} = \frac{1}{16}         (Property used: a-n\frac{1}{a^n} )

(iii) (\frac{1}{2} ) -5 

(\frac{1}{2} )-5  = (2)5 = 32           (Property used: (\frac{b}{a})^{-n} = \frac{a^n}{b^n} )

Question 2. Simplify and express the result in power notation with a positive exponent.

Solution:

(i) (-4)5 ÷ (-4)8

= (-4)5-8 = (-4) -3                      (Property used: am ÷ an= am-n)

\frac{1}{(-4)^3}  



\mathbf{(\frac{1}{(-4)})^3}

(ii) (\frac{1}{2^3})^2

\frac{(1)^2}{(2^3)^2}                             (Property used: (am)n = am×n) 

\frac{1}{2^6}

\mathbf{\frac{1}{2^6}}

(iii) (-3)4 × (\frac{5}{3} )4

= ((3)4 × \frac{5^4}{3^4}                                  (Property used: (a/b)n = an/ bn  & (-a)n = an if a is positive number and n is even)

= 54

(iv) (3-7 ÷ 3-10) × 3-5

= 3 (-7-(-10)) × 3-5                                     (Property used: am ÷ an= am-n)



= 3 (-7+10) × 3-5

= 33 × 3-5

= 3 (3+(-5))                                                   (Property used: am × an = a m + n)

= 3-2                                                             (Property used: a-m  =\frac{1}{a^m} )

\frac{1}{3^2}

\mathbf{\frac{1}{3^2}}

(v) 2-3 × (-7)-3

= (2 × (-7))-3                                           (Property used: am × bm = (a×b)m)

= (-14)-3                                                     (Property used: a-m  =\frac{1}{a^m} )

\frac{1}{(-14)^3}



=\mathbf{\frac{1}{(-14)^3}}

Question 3. Find the value of

Solution:

(i) (30 + 4-1) × 22

= (1 + (\frac{1}{4} )) × 4                     (a0 = 1  a 0)

= (\frac{5}{4} ) × 4

= 5

(ii) (2-1 × 4-1) ÷ 2-2

= (2 × 4)-1 ÷ \frac{1}{2^2}                                (Property used: am × bm = (a×b)m)

= (8)-1 ÷ \frac{1}{4}

= (\frac{1}{8} ) ÷ \frac{1}{4}



=(\frac{1}{8} ) × 4

= (\frac{1}{2} )

(iii) (1/2)-2 + (1/3)-2 + (1/4)-2

= 22 + 32 + 42                                        (Property used: (\frac{1}{a})^{-m}   =am)

= 4 + 9 + 16

= 29

(iv) (3-1 + 4-1 + 5-1)0

= (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} )0                             (a0 = 1 (a 0)

= 0 

(v) {(\frac{-2}{3} )-2}2



= (\frac{-2}{3} ) -2×2                                      (Property used: (am)n = am×n

= (\frac{-2}{3} )-4    = (\frac{-3}{2} )4                     (Property used: (b/a)-n = an/bn)

\frac{3^4}{2^4}

\frac{81}{16}

Question 4. Evaluate 

Solution:

(i) (8-1 × 53) / 2-4

= (\frac{1}{8}  × 125) / (2-4)                      (Property used: (b/a)-n = an/bn)

= (\frac{1}{8} ) × 125 × 24

= 250

(ii) (5-1 × 2-1) × 6-1



= (5 × 2)-1 × 6-1                           (Property used: am × bm = (a×b)m)

= 10-1 × 6-1

= (10 × 6)-1                                     (Property used: am × bm = (a×b)m)

= 60-1

\frac{1}{60}

Question 5. Find the value of m for which 5m ÷ 5– 3 = 5

Solution:

5m-(– 3) = 55                       (Property used: am ÷ an= am-n)

5m+3 = 5

m+3 = 5

m = 5-3

m = 2

Question 6. Evaluate 

Solution:

(i) {(\frac{1}{3} )-1 – (\frac{1}{4} )-1}-1

= (31 – 41) -1                          (Property used: (1/a)-m  = am)

= (-1)-1

= (1/(-1))1

= (-1)

(ii) (\frac{5}{8} )-7 × (\frac{8}{5} )-4

= (\frac{5}{8} )7 × (\frac{8}{5} )-4                       (Property used: (b/a)-n = (a/b)n)

= (\frac{8}{5} ) 7+ (-4)                               (Property used: am × an = a m + n)



= (\frac{8}{5} )3  = 83/53

\frac{512}{125}

Question 7. Simplify

Solution:

(i) \frac{(25 \times t^{-4})}{(5^{-3} \times 10 \times t^{-8})}  (t ≠ 0)

\frac{(5^2 \times t^{-4})}{(5^{-3} \times 10 \times t^{-8})}          (Property used: am ÷ an= am-n ) (25 = 52)

=\frac{(5^{2-(-3)} \times t^{-4-(-8)})}{ 10}

\frac{(5^{5} \times t^{4})}{ 10}

\frac{(625 \times t^{4})}{ 2}

(ii) \frac{(3^{-5} \times 10^{-5} \times 125)}{(5^{-7} \times 6^{-5})}

\frac{(3^{-5} \times (2\times 5)^{-5} \times 125)}{(5^{-7} \times (2 \times 3)^{-5})}

\frac{(3^{-5} \times 2^{-5}\times 5^{-5} \times 125)}{(5^{-7} \times 2^{-5} \times 3^{-5})}                   (Property used: (a×b)m = am × bm)

= (3-5-(-5) × 2-5-(-5) × 5 (-5)+3+7)                                      (Property used: am ÷ an= am-n )

= (30 × 20 × 55)                                                                (a0 = 1 (a ≠ 0)                   

= 55




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