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 chapter for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. Exponents are applied in various algebra questions, therefore, it is essential to understand the rules of exponents. There are 2 easy “rules of 1” to master. First, any amount raised to the power of “1” equals itself. Secondly, 1 raised to any power is one. Multiplication of 2 powers having identical base can be simplified by adding them. For raising a power to a power we simply need to multiply the exponents. Two powers can be divided into the same base by subtracting the exponents. Any non-zero number raised to the power of 0 = 1. Any non-zero number raised to a negative power is equal to its reciprocal raised to the opposite positive power.

Several times large numbers are tough to understand, read, operate, therefore, superscripts are applied to transform massive numbers into the modest form with the assistant of powers and exponents. Exponents describe the number of times the base is multiplied. whereas, when it comes to power, it comprises of 2 parts, i.e. the exponent and a base number. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 12 Exponents and Powers. We have provided a comprehensive study material for Class 8 Maths Chapter 12 Exponents and Powers NCERT Solutions. The students can download and understand all the exercise-wise NCERT problems on Exponents and Powers from the links provided below.

NCERT Solutions Class 8 Maths Chapter 12 Exercises

Exercise-12.1

Q.1. (i)$4^{-3}$

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

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

Solution:

(i)$4^{-2}$

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

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

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

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

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

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

Q.2. (i)$(-5)^{4}\div (-5)^{7}$

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

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

(iv)$(2^{-7}\div 2^{-10})\times 2^{-5}$

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

Solution:

(i)$(-5)^{4}\div (-5)^{7}$

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

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

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

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

$(-4)^{2}\times (\frac{5}{4})^{2}=(-4)^{2}\times \frac{5^{2}}{4^{2}}\\ \\ =[(-1)^{2}\times 4^{2}]\times \frac{5^{2}}{4^{2}}\\ \\ =4^{2-2}\times 5^{2}=4^{0}\times 5^{2}=5^{2}$

(iv)$(2^{-7}\div 2^{-10})\times 2^{-5}$

$(2^{-7}\div 2^{-10})\times 2^{-5}=2^{-7-(-10)}\times 2^{-5}\\ \\ =2^{-7+10}\times 2^{-5}\\ \\ =2^{3}\times 2^{-5}=2^{3-5}\\ \\ =2^{-2}=\frac{1}{2^{2}}$

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

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

Formulae:

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

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

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

4.$a^{m}\div a^{n}=a^{m-n}$

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

Q.3. Find the value of:

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

(ii)$(2^{-1}\times 4^{-1})\div 2^{-2}$

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

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

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

Solution:

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

$(1+\frac{1}{4})\times 2^{2}\\ \\ =(\frac{4+1}{4})\times 2^{2}=\frac{5}{4}\times 2^{2}\\ \\ =5\times 2^{2-2}\\ \\ =5\times 2^{0}=5\times 1=5$

(ii)$(2^{-1}\times 4^{-1})\div 2^{-2}$

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

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

$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}=(2^{-1})^{2}+(3^{-1})^{2}+(4^{-1})^{2}\\ \\ =2^{-1\times (-2)}+3^{-1\times (-2)}+4^{-1\times (-2)}\\ \\ =2^{2}+3^{2}+4^{2}=4+9+16=29$

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

$(3^{-1}+4^{-1}+5^{-1})^{0}=(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{0}\\ \\ =(\frac{20+15+12}{60})^{0}=(\frac{47}{60})^{0}= 1$

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

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

Q.4. Evaluate:

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

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

Solution:

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

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

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

$(5^{-1}\times 2^{-1})\times 6^{-1}=(\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}\\ \\ =\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$

Q.5. Find the value of m for which$5^{m}\div 5^{-3}=5^{5}$.

Solution:

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

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

Q.6. Evaluate:

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

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

Solution:

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

$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}=[(\frac{3}{1})^{1}-(\frac{4}{1})^{1}]\, (\ because a^{-m}=\frac{1}{m})\\ \\ =[3-4]=-1$

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

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

Q.7. Simplify:

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

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

Solution:

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

$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{5^{2}\times t^{-4}}{5^{-3}\times 5\times2\times t^{-8}}\\ \\ =\frac{5^{2-(-3)-1}\times t^{-4(-8)}}{2}\\ \\ =\frac{5^{2+3-1}\times t^{-4+8}}{2}=\frac{5^{4}\times t^{4}}{2}\\ \\ =\frac{625}{2}t^{4}$

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

$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times (2\times 5)^{-5}\times 5^{3}}{5^{-7}\times (2\times 3)^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5}\times 5^{3}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5+3}}{5^{-7}\times 2^{-5}\times 3^{-5}}=\frac{3^{-5}\times 2^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =3^{-5-(-5)}\times 2^{-5-(-5)}\times 5^{-2-(-7)}\\ \\ =3^{-5+5}\times 2^{-5+5}\times 5^{-2+7}=3^{0}\times 2^{0}\times 5^{5}=1\times 1\times 3125\\ \\ =3125$

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

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

(ii)0.00000000000942

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

(iii)6020000000000000

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

(iv)0.00000000837

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

(v)31860000000

$31860000000 =31860000000 \times \frac{10^{10}}{10^{10}}=3.186\times 10^{10}$

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

(i)$3.02\times 10^{-6}$

(ii)$4.5\times 10^{4}$

(iii)$3\times 10^{-8}$

(iv)$1.0001\times 10^{9}$

(v) $5.8\times 10^{12}$

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

Solution:

(i)$3.02\times 10^{-6}$

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

(ii)$4.5\times 10^{4}$

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

(iii) $3\times 10^{-8}$

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

(iv) $1.0001 \times 10^{9}$

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

(v) $5.8\times 10^{12}$

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

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

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

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

(i) 1 micron is equal to $\frac{1}{1000000}$ 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 =$\frac{1}{1000000}=\frac{1}{10^{6}}=1\times 10^{-6}\: m$

(ii) Charge of an electron is 0.00000000000000000028 coulombs.

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

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

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

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

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 =$20\times 6=120\: mm$

Thickness of one paper = 0.016 mm

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

Total thickness of a stack =120+0.08

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

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