# NCERT Solutions For Class 8 Maths Chapter 12

## NCERT Solutions Class 8 Maths Exponents and Powers

NCERT Solutions for class 8 Maths chapter 12 Exponents and Powers is crucial for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 12 Exponents and Powers. Student can download the NCERT Solution for class 8 Maths Chapter 12 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 8 maths chapter 12.

### NCERT Solutions Class 8 Maths Chapter 12 Exercises

Exercise-12.1

Q.1. (i)43$4^{-3}$

(ii)(5)5$(-5)^{-5}$

(iii)(12)4$(\frac{1}{2})^{-4}$

Solution:

(i)42$4^{-2}$

(4)2=142=116$(4)^{-2}=\frac{1}{4^{2}}=\frac{1}{16}$

(ii)(5)2$(-5)^{-2}$

(5)2=1(5)2=125$(-5)^{-2}=\frac{1}{(-5)^{2}}=\frac{1}{25}$

(iii)(12)4$(\frac{1}{2})^{-4}$

(12)2=(21)2=(2)2=4$(\frac{1}{2})^{-2}=(\frac{2}{1})^{2}=(2)^{2}=4$

Formula: am=1am$a^{-m}=\frac{1}{a^{m}}$

Q.2. (i)(5)4÷(5)7$(-5)^{4}\div (-5)^{7}$

(ii)(143)2$(\frac{1}{4^{3}})^{2}$

(iii)(4)3×(54)3$(-4)^{3}\times (\frac{5}{4})^{3}$

(iv)(27÷210)×25$(2^{-7}\div 2^{-10})\times 2^{-5}$

(v)33×(7)3$3^{-3}\times (-7)^{-3}$

Solution:

(i)(5)4÷(5)7$(-5)^{4}\div (-5)^{7}$

(5)4÷(5)7=(5)47=(5)3=1(5)3$(-5)^{4}\div (-5)^{7}=(-5)^{4-7}\\ \\ =(-5)^{-3}\\ \\ =\frac{1}{(-5)^{3}}$

(ii)(143)2$(\frac{1}{4^{3}})^{2}$

(143)2=12(43)2=143×2=146$(\frac{1}{4^{3}})^{2}=\frac{1^{2}}{(4^{3})^{2}}\\ \\ =\frac{1}{4^{3\times 2}}\\ \\ =\frac{1}{4^{6}}$

(iii)(4)3×(54)3$(-4)^{3}\times (\frac{5}{4})^{3}$

(4)2×(54)2=(4)2×5242=[(1)2×42]×5242=422×52=40×52=52$(-4)^{2}\times (\frac{5}{4})^{2}=(-4)^{2}\times \frac{5^{2}}{4^{2}}\\ \\ =[(-1)^{2}\times 4^{2}]\times \frac{5^{2}}{4^{2}}\\ \\ =4^{2-2}\times 5^{2}=4^{0}\times 5^{2}=5^{2}$

(iv)(27÷210)×25$(2^{-7}\div 2^{-10})\times 2^{-5}$

(27÷210)×25=27(10)×25=27+10×25=23×25=235=22=122$(2^{-7}\div 2^{-10})\times 2^{-5}=2^{-7-(-10)}\times 2^{-5}\\ \\ =2^{-7+10}\times 2^{-5}\\ \\ =2^{3}\times 2^{-5}=2^{3-5}\\ \\ =2^{-2}=\frac{1}{2^{2}}$

(v)33×(7)3$3^{-3}\times (-7)^{-3}$

33×(7)3=133×1(7)3=1[3×(7)]3=1(21)3$3^{-3}\times (-7)^{-3}=\frac{1}{3^{3}}\times \frac{1}{(-7)^{3}}\\ \\ =\frac{1}{[3\times (-7)]^{3}}=\frac{1}{(-21)^{3}}$

Formulae:

1.(am)n=am×n$(a^{m})^{n}=a^{m\times n}$

2.(ab)m=ambm$(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$

3.(ab)m=ambm$(ab)^{m}=a^{m} b^{m}$

4.am÷an=amn$a^{m}\div a^{n}=a^{m-n}$

5.am×an=am+n$a^{m}\times a^{n}=a^{m+n}$

Q.3. Find the value of:

(i)(30+41)×22$(3^{0}+4^{-1})\times 2^{2}$

(ii)(21×41)÷22$(2^{-1}\times 4^{-1})\div 2^{-2}$

(iii)(12)2+(13)2+(14)2$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}$

(iv)(31+41+51)0$(3^{-1}+4^{-1}+5^{-1})^{0}$

(v)[(23)2]2$[(\frac{-2}{3})^{-2}]^{2}$

Solution:

(i)(30+41)×22$(3^{0}+4^{-1})\times 2^{2}$

(1+14)×22=(4+14)×22=54×22=5×222=5×20=5×1=5$(1+\frac{1}{4})\times 2^{2}\\ \\ =(\frac{4+1}{4})\times 2^{2}=\frac{5}{4}\times 2^{2}\\ \\ =5\times 2^{2-2}\\ \\ =5\times 2^{0}=5\times 1=5$

(ii)(21×41)÷22$(2^{-1}\times 4^{-1})\div 2^{-2}$

(21×41)÷22=(121×141)÷22=(12×122)÷22=123÷22=23÷22=23(2)=23+2=21=12$(2^{-1}\times 4^{-1})\div 2^{-2}=(\frac{1}{2^{1}}\times \frac{1}{4^{1}})\div 2^{-2}\\ \\ =(\frac{1}{2}\times \frac{1}{2^{2}})\div 2^{-2}=\frac{1}{2^{3}}\div 2^{-2}\\ \\ =2^{-3}\div 2^{-2}=2^{-3-(-2)}=2^{-3+2}=2^{-1}\\ \\ =\frac{1}{2}$

(iii)(12)2+(13)2+(14)2$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}$

(12)2+(13)2+(14)2=(21)2+(31)2+(41)2=21×(2)+31×(2)+41×(2)=22+32+42=4+9+16=29$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}=(2^{-1})^{2}+(3^{-1})^{2}+(4^{-1})^{2}\\ \\ =2^{-1\times (-2)}+3^{-1\times (-2)}+4^{-1\times (-2)}\\ \\ =2^{2}+3^{2}+4^{2}=4+9+16=29$

(iv)(31+41+51)0$(3^{-1}+4^{-1}+5^{-1})^{0}$

(31+41+51)0=(13+14+15)0=(20+15+1260)0=(4760)0=1$(3^{-1}+4^{-1}+5^{-1})^{0}=(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{0}\\ \\ =(\frac{20+15+12}{60})^{0}=(\frac{47}{60})^{0}= 1$

(v)[(23)2]2$[(\frac{-2}{3})^{-2}]^{2}$

[(23)2]2=(23)2×2=(23)4=(32)4=8116$[(\frac{-2}{3})^{-2}]^{2}=(\frac{-2}{3})^{-2\times 2}\\ \\ =(\frac{-2}{3})^{-4}=(\frac{-3}{2})^{4}=\frac{81}{16}$

Q.4. Evaluate:

(i)81×5324$\frac{8^{-1}\times 5^{3}}{2^{-4}}$

(ii)(51×21)×61$(5^{-1}\times 2^{-1})\times 6^{-1}$

Solution:

(i)81×5324$\frac{8^{-1}\times 5^{3}}{2^{-4}}$

81×5324=(23)1×5324=23×5324=23(4)×53=23+4×53=2×125=250$\frac{8^{-1}\times 5^{3}}{2^{-4}}=\frac{(2^{3})^{-1}\times 5^{3}}{2^{-4}}=\frac{2^{-3}\times 5^{3}}{2^{-4}}\\ \\ =2^{-3(-4)}\times 5^{3}=2^{-3+4}\times 5^{3}\\ \\ =2\times 125=250$

(ii)(51×21)×61$(5^{-1}\times 2^{-1})\times 6^{-1}$

(51×21)×61=(15×12)×16=110×16=160$(5^{-1}\times 2^{-1})\times 6^{-1}=(\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}\\ \\ =\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$

Q.5. Find the value of m for which5m÷53=55$5^{m}\div 5^{-3}=5^{5}$.

Solution:

5m÷53=555m(3)=555m+3=55$5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}$ 5m÷53=555m(3)=555m+3=55m+3=5m=53m=2$5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}\\ \\ \Rightarrow m+3=5\\ \\ \Rightarrow m=5-3\\ \\ \Rightarrow m=2$

Q.6. Evaluate:

(i) [(13)1(14)1]1$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}$

(ii)(58)7×(85)4$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}$

Solution:

(i) [(13)1(14)1]1$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}$

[(13)1(14)1]1=[(31)1(41)1]( becauseam=1m)=[34]=1$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}=[(\frac{3}{1})^{1}-(\frac{4}{1})^{1}]\, (\ because a^{-m}=\frac{1}{m})\\ \\ =[3-4]=-1$

(ii)(58)7×(85)4$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}$

(58)7×(85)4=5787×8454=57(4)×84(7)=57+4×84+7=53×83=8353=512125$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}\\ \\ =5^{-7(-4)}\times 8^{-4(-7)}\\ \\ =5^{-7+4}\times 8^{-4+7}=5^{-3}\times 8^{3}=\frac{8^{3}}{5^{3}}\\ \\ =\frac{512}{125}$

Q.7. Simplify:

(i)25×t453×10×t8(t0)$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)$

(ii)35×105×12557×65$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$

Solution:

(i)25×t453×10×t8(t0)$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)$

25×t453×10×t8=52×t453×5×2×t8=52(3)1×t4(8)2=52+31×t4+82=54×t42=6252t4$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{5^{2}\times t^{-4}}{5^{-3}\times 5\times2\times t^{-8}}\\ \\ =\frac{5^{2-(-3)-1}\times t^{-4(-8)}}{2}\\ \\ =\frac{5^{2+3-1}\times t^{-4+8}}{2}=\frac{5^{4}\times t^{4}}{2}\\ \\ =\frac{625}{2}t^{4}$

(ii)35×105×12557×65$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$

35×105×12557×65=35×(2×5)5×5357×(2×3)5=35×25×55×5357×25×35=35×25×55+357×25×35=35×25×5257×25×35=35(5)×25(5)×52(7)=35+5×25+5×52+7=30×20×55=1×1×3125=3125$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times (2\times 5)^{-5}\times 5^{3}}{5^{-7}\times (2\times 3)^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5}\times 5^{3}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5+3}}{5^{-7}\times 2^{-5}\times 3^{-5}}=\frac{3^{-5}\times 2^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =3^{-5-(-5)}\times 2^{-5-(-5)}\times 5^{-2-(-7)}\\ \\ =3^{-5+5}\times 2^{-5+5}\times 5^{-2+7}=3^{0}\times 2^{0}\times 5^{5}=1\times 1\times 3125\\ \\ =3125$

Exercise-12.2

Q.1. Converts the numbers into standard form:

(i)0.0000000000085

(ii)0.00000000000942

(iii)6020000000000000

(iv)0.00000000837

(v)31860000000

Solution:

(i)0.0000000000085

0.0000000000085=0.0000000000085×10121012=8.5×1012$0.0000000000085=0.0000000000085\times \frac{10^{12}}{10^{12}}=8.5\times 10^{-12}$

(ii)0.00000000000942

0.00000000000942=0.00000000000942×10121012=9.42×1012$0.00000000000942 =0.00000000000942 \times \frac{10^{12}}{10^{12}}=9.42\times 10^{-12}$

(iii)6020000000000000

6020000000000000=6020000000000000×10151015=6.02×1015$6020000000000000 =6020000000000000 \times \frac{10^{15}}{10^{15}}=6.02\times 10^{15}$

(iv)0.00000000837

0.00000000837=0.00000000837×109109=8.37×109$0.00000000837 =0.00000000837 \times \frac{10^{9}}{10^{9}}=8.37\times 10^{-9}$

(v)31860000000

31860000000=31860000000×10101010=3.186×1010$31860000000 =31860000000 \times \frac{10^{10}}{10^{10}}=3.186\times 10^{10}$

Q.2. Convert the following numbers in usual form:

(i)3.02×106$3.02\times 10^{-6}$

(ii)4.5×104$4.5\times 10^{4}$

(iii)3×108$3\times 10^{-8}$

(iv)1.0001×109$1.0001\times 10^{9}$

(v) 5.8×1012$5.8\times 10^{12}$

(vi)3.61492×106$3.61492\times 10^{6}$

Solution:

(i)3.02×106$3.02\times 10^{-6}$

3.02×106=3.02106=0.00000302$3.02\times 10^{-6}=\frac{3.02}{10^{6}}=0.00000302$

(ii)4.5×104$4.5\times 10^{4}$

4.5×104=4.5×10000=45000$4.5\times 10^{4}=4.5\times 10000=45000$

(iii) 3×108$3\times 10^{-8}$

3×108=3108=0.00000003$3\times 10^{-8}=\frac{3}{10^{8}}=0.00000003$

(iv) 1.0001×109$1.0001 \times 10^{9}$

1.0001×109=1000100000$1.0001 \times 10^{9}=1000100000$

(v) 5.8×1012$5.8\times 10^{12}$

5.8×1012=5.8×1000000000000=5800000000000$5.8\times 10^{12}=5.8\times 1000000000000=5800000000000$

(vi) 3.61492×106$3.61492\times 10^{6}$

3.61492×106=3.61492×1000000=3614920$3.61492\times 10^{6}=3.61492 \times 1000000 =3614920$

Q.3. Conert the given numbers in the following statements in standard form:

(i) 1 micron is equal to 11000000$\frac{1}{1000000}$ m.

(ii)Charge of an electron is 0.000,000,000,000,000,000,28 coulomb.

(iii)Size of bacteria is 0.0000005 m.

(iv)Size of a plant cell is 0.00001275 m.

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

Solution:

(i)1 micron =11000000=1106=1×106m$\frac{1}{1000000}=\frac{1}{10^{6}}=1\times 10^{-6}\: m$

(ii) Charge of an electron is 0.00000000000000000028 coulombs.

=0.00000000000000000028×10191019=2.8×1019coulomb$= 0.00000000000000000028\times \frac{10^{19}}{10^{19}}=2.8\times 10^{-19}\: coulomb$

(iii)Size of bacteria = 0.0000005 =510000000=5107=5×107m$\frac{5}{10000000}=\frac{5}{10^{7}}=5\times 10^{-7\: m}$

(iv)Size of a plant cell is 0.00001275 m =0.00001275×105105=1.275×105m$0.00001275\times \frac{10^{5}}{10^{5}}=1.275\times 10^{-5}\: m$

(v) Thickness of a thick paper = 0.07 mm =7100mm=7102=7×102mm$\frac{7}{100}\: mm=\frac{7}{10^{2}}=7\times 10^{-2}\: mm$

Q.4. In a stack there are 6 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Solution:

Thickness of one book = 20 mm

Thickness of 6 books =20×6=120mm$20\times 6=120\: mm$

Thickness of one paper = 0.016 mm

Thickness of 5 papers =0.016×5=0.08mm$0.016\times 5=0.08\: mm$

Total thickness of a stack =120+0.08

=120.08 mm=120.08×102102=1.2008×102mm$120.08\times \frac{10^{2}}{10^{2}}=1.2008\times 10^{2}\: mm$