NCERT solutions for class 8 Maths Chapter 12 Exponents And Powers

Solution for Exercise 12.1

1.Evaluate:
(i) \(3^{-2}\)

(ii) \((-4)^{-2}\)

(iii) \((\frac { 1 }{ 2 })^{-5}\)


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Answer :

(i) We have:
\(3^{-2}=\frac{1}{3^2}=\frac{1}9\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]

(ii)We have:
\((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]

(iii)We have:
\((\frac { 1 }{ 2 })^{-5}=2^5=32\quad\)[∵\((\frac{a}b)^{-n}=(\frac{b}a)^n\)]

2.Simplify and express the result in power notation with a positive exponent. (i)\((-4)^5÷(-4)^8\)
(ii)\((\frac{1}{2^3})^2\)

(iii)\((-3)^4\times (\frac{5}3)^4\)

(iv)\((3^{-7}÷3^{-10})\times 3^{-5}\)

(v)\(2^{-3}\times(-7)^{-3}\)


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Answer :

(i)We have:\((-4)^5÷(-4)^8\)
\(=(-4)^{5-8}=(-4)^{-3}=\frac{1}{(-4)^3}\quad\)[∵\(a^{m}÷a^{n}=a^{m-n}\)]

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

(ii)We have:\((\frac{1}{2^3})^2\)

\((\frac{1}{2^3})^2=\frac{(1)^2}{(2^3)^2}=\frac{1}{2^6}=(\frac{1}2)^6\)

(iii)We have:\((-3)^4\times (\frac{5}3)^4\)

\(=(-3)^4\times (\frac{5}3)^4=(-3)^4\times \frac{(5)^4}{(3)^4}\)

\(=\frac{(3)^4\times(5)^4}{(3)^4}=(5)^4\)

(iv)We have:\((3^{-7}÷3^{-10})\times 3^{-5}\)

\(=(3^{-7}÷3^{-10})\times 3^{-5}=3^{-7-(-10)}\times 3^{-5}\)

\(=3^{-7+10}\times 3^{-5}=3^3\times 3^{-5}=3^{(3-5)}\)

\(=3^{-2}=\frac{1}{3^2}=(\frac{1}3)^2\)

(v)We have:\(2^{-3}\times(-7)^{-3}\)

\(=2^{-3}\times(-7)^{-3}=[2\times(-7)]^{-3}=(-14)^{-3}\)

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

3.Find the value of
(i)\(3^0+4^{-1}\times 2^2\)

(ii)\(2^{-1}\times4^{-1}÷ 2^{-2}\)

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

(iv)\((3^{-1}+4^{-1}+5^{-1})^0\)

(v)\([(\frac{-2}3)^{-2}]^2\)


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Answer :

(i)We have:\(3^0+4^{-1}\times 2^2\)

\(=3^0+4^{-1}\times 2^2=(1+\frac{1}4)\times4\)

\(=(\frac{4+1}4)\times4=\frac{5}4\times 4=5\)

(ii)We have:\(2^{-1}\times4^{-1}÷ 2^{-2}\)

\(=2^{-1}\times4^{-1}÷ 2^{-2}=(\frac{1}2\times\frac{1}4)÷\frac{1}{2^2}\)

\(=\frac{1}8÷\frac{1}4=\frac{1}8\times\frac{4}1\)

\(=\frac{1}2\)

(iii)We have:\((\frac{1}2)^{-2}+(\frac{1}3)^{-2}+(\frac{1}4)^{-2}\)

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

\(=4+9+16=29\)

(iv)We have:\((3^{-1}+4^{-1}+5^{-1})^0\)

\(=(3^{-1}+4^{-1}+5^{-1})^0=1\quad\)[∵\(a^0=1\)]

(v)We have:\([(\frac{-2}3)^{-2}]^2\)

\(=[(\frac{-2}3)^{-2}]^2=[(-\frac{3}2)^2]^2\)

\(=(\frac{9}4)^2=\frac{81}{16}\)

4.Evaluate
(i) \(\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } } \)

(ii) \((5^{-1} \times 2^{-1}) \times 6^{-1}\)


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Answer :

(i)We have:\(\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } } \)

\(=\frac{8^{-1}\times 5^3}{2^{-4}}=\frac{1}8\times 5^3\times 2^4\)

\(=\frac{1}8\times125\times 16=125\times 2=250\)

(ii)We have:\((5^{-1} \times 2^{-1}) \times 6^{-1}\)

\(=(\frac{1}5\times\frac{1}2)\times\frac{1}6\)

\(=\frac{1}{10}\times\frac{1}6=\frac{1}{60}\)

5.Find the value of m for which \(5^m ÷ 5^{-3} = 5^5\).


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Answer :

We have given: \(5^m ÷ 5^{-3} = 5^5\).

\(\Rightarrow 5^{m-(-3)} = 5^5\quad\) [∵\( a^m ÷ a^n = a^{m-n}\)]

\(\Rightarrow 5^{m+3} = 5^5\)

Now, Comparing the powers of equal bases, we have m + 3 = 5

\(\Rightarrow m = 5 – 3 = 2\), i.e., m = 2

6.Evaluate:
(i)\([(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}\)

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


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Answer :

(i)We have:\([(\frac{1}3)^{-1}-(\frac{1}4)^{-1}]^{-1}\)

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

(ii)We have: \((\frac{5}8)^{-7}\times (\frac{8}5)^{-4}\)
\(=(\frac{8}5)^7\times(\frac{5}8)^4=\frac{8^7}{5^7}\times\frac{5^4}{8^4}\)

\(=\frac{8^{7-4}}{5^{7-4}}=\frac{8^3}{5^3}=\frac{512}{125}\)

7.Simplify :
(i)\(\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)\)

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


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Answer :

(i)We have:\(\frac{25\times t^{-4}}{5^{-3}\times10\times t^{-8}}(t\neq 0)\)

\(=\frac{25\times 5^3}{10}\times t^{(-4+8)}=\frac{5\times 5^3}2\times t^4\)

\(=\frac{625}2t^4\)
(ii)We have:\(\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}\)

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

\(=\frac{5^{7}\times (3\times 2)^{5}\times 5^3}{3^{5}\times (2\times5)^{5}}=\frac{5^{7}\times 3^{5}\times 2^5\times 5^3}{3^{5}\times 2^5\times5^{5}}=5^{7+3-5}=5^5\)

Solution for Exercise 12.2

1.Express the following numbers in standard form:
(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000


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Answer :

(i)We have:0.0000000000085
\(=\frac{85}{10000000000000}=\frac{8.5\times 10}{10^{13}}\)

\(=8.5\times 10^{(1-13)}=8.5\times 10^{-12}\)

(ii)We have:0.00000000000942
\(=\frac{942}{100000000000000}=\frac{9.42\times 10^2}{10^{14}}\)

\(=9.42\times 10^{(2-14)}=9.42\times 10^{-12}\)

(iii)We have:6020000000000000=\(6.02\times1000000000000000\)

\(=6.02\times 10^{15}\)

(iv)We have:0.00000000837

\(=\frac{837}{100000000000}=\frac{8.37\times 100}{10^{11}}\)

\(=8.37\times 10^{(2-11)}=8.37\times 10^{-9}\)

(v)We have: 31860000000

\(= 3.186 \times 10000000000\)

\(= 3.186 \times 10^10\)

Hence, the required standard from = \(3.186 × 10^10\)

2.Express the following numbers in usual form.
(i) \(3.02 × 10^{-6}\)

(ii) \(4.5 × 10^4\)

(iii) \(3 × 10^{-8}\)

(iv) \(1.0001 × 10^9\)

(v) \(5.8 × 10^{12}\)

(vi) \(3.61492 \times 10^6\)


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Answer :

(i)We have:\(3.02 × 10^{-6}\)

\(=\frac{302}{100}\times\frac{1}{10^6}=\frac{302}{100000000}\)

\(=302\times 10^{-8}=0.00000302\)

Thus, we have \(3.02 × 10^{-6}=0.00000302\)

(ii)We have:\(4.5 × 10^4\)

\(=\frac{45}{10}\times10^4=45\times 10^3\)

\(=45000\)

Thus, we have \(4.5 × 10^4=45000\)

(iii)We have:\(3 × 10^{-8}\)

\(=\frac{3}{10^8}=\frac{3}{100000000}\)

\(=0.00000003\)

Thus, we have \(3 × 10^{-8}=0.00000003\)

(iv)We have: \(1.0001 × 10^9\)

\(=\frac{10001}{10000}\times10^9=\frac{10001}{10^4}\times10^9\)

\(=10001\times 10^5=1000100000\)

Thus, we have\(1.0001 × 10^9=1000100000\)

(v)We have: \(5.8 × 10^{12}\)

\(=\frac{58}{10}\times10^{12}=58\times 10^{11}\)

\(=5800000000000\)

Thus, we have \(5.8 × 10^{12}=5800000000000\)

(vi)We have: \(3.61492 \times 10^6\)

\(=\frac{361492}{100000}\times10^6=\frac{361492}{10^5}\times10^6\)

\(=361492\times 10=3614920\)

Thus, we have \(3.61492 \times 10^6=3614920\)

3.Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to \(\frac { 1 }{ 1000000 }\) m.

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm.


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Answer :

(i)We have 1 micron=\(\frac { 1 }{ 1000000 }\)

\(=\frac{1}{10^6}m=10^{-6}\)

(ii)We know charge of an electron=0.000,000,000,000,000,000,16

\(\frac{16}{1,000,000,000,000,000,000,00}=\frac{1.6\times 10}{1,000,000,000,000,000,000,00}\)

\(=\frac{1.6\times 10}{10^{20}}=1.6\times 10^{1-20}=1.6\times 10^{-19}\)

(iii)We have size of bacteria=0.0000005m

\(=\frac{5}{10000000}m=\frac{0.5\times 10}{10^7}m\)

\(=0.5\times 10^{1-7}m=0.5\times 10^{-6}m\)

\(=5\times 10^{-7}m\)

(iv)We have size of plant cell=0.00001275m

\(=\frac{1275}{100000000}m=\frac{1.275\times 10^3}{10^8}m\)

\(=1.275\times 10^{3-8}m=1.275\times 10^{-5}m\)

\(=1.275\times 10^{-5}m\)

(v)We have thickness of paper=0.07mm

\(=\frac{7}{100}mm=\frac{0.7\times 10}{10^2}mm\)

\(=0.7\times 10^{1-2}mm=0.7\times 10^{-1}mm\)

\(=7\times 10^{-2}mm\)

4.In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?


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Answer :

We have the thickness of books = \(5 \times 20 = 100 mm\)

Also, the thickness of 5 paper sheets = \(5 \times 0.016 mm = 0.080 mm\).

Total thickness of the stack = \(100 mm + 0.080 mm = 100.080 mm = 1.0008 \times 10^2 mm\)



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