# NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - If someone asks you about the mass of your bag? Probably your answer is 7 kg but what if someone asks you about the mass of the earth. Mass of the earth is 5,970,000,000,000,000,000,000, 000 kg,(no need to remember) which is a very big data. You can write such large numbers more conveniently using exponents. In this article, you will get NCERT solutions for class 8 maths chapter 12 exponents and powers explained in a detailed manner to get a better understanding of the chapter. In this chapter, you will learn how to make the representation of bigger numbers easier using about exponents and powers. This is a new concept but if you go through solutions of NCERT for class 8 maths chapter 12 exponents and powers, you will understand it completely. Let’s look at a scenario from our daily life to understand this chapter.

Example-

 Questions Assume it’s your answer Distance between your home and school 15 KM Distance between Bangalore and Delhi 1500 KM Distance between India and South America 15000 KM Distance between Earth and the Sun 150000000 KM

In the above example, you can see that a very big number like 150000000 KM is difficult to understand and becomes more complex or tough to understand, read and operate. So it can be represented in exponent form as 1.5x108 km. The exponent of a number, describe that for how many times a number(base) will be multiplied. The base is the number that is going to be multiplied.

Example- $6^3$, where 6 is the base and 3 is the power. Multiplying three sixes together gives 6*6*6 = 216.

There are 11 questions in 2 exercises of the textbook. All these questions are explained in CBSE NCERT solutions for class 8 maths chapter 12 exponents and powers. You can get NCERT solutions from class 6 to 12 by clicking on the above link.

## Topics of NCERT Class 8 Maths Chapter 12 Exponents and Powers-

• 12.1 Introduction
• 12.2 Powers with Negative Exponents
• 12.3 Laws of Exponents
• 12.4  Use of Exponents to Express Small Numbers in Standard Form
• 12.5 Comparing very large and very small numbers

## Solutions of NCERT for class 8 maths chapter 12 topic 12.2 powers with negative exponents

$2^{-4}$

The detailed explanation for the question is written below,

The multiplicative inverse is $\frac{1}{a^{m}}$

So, the multiplicative inverse of  $2^{-4}$ is $2^{4}$

Question:(ii) Find the multiplicative inverse of the following.

$10^{-5}$

Here is the detailed solution for the above question,

As we know,

The multiplicative inverse of    $a^{m}$ is   $\frac{1}{a^{m}}$

So, the multiplicative inverse of  $10^{-5}$  is $10^{5}$

Question:(iii) Find the multiplicative inverse of the following.

$7^{-2}$

The multiplicative inverse of    $a^{m}$ is   $\frac{1}{a^{m}}$

So, the multiplicative inverse of $7^{-2}$ is $7^{2}$

Question:(iv) Find the multiplicative inverse of the following.

$5^{-3}$

we know,

The multiplicative inverse of    $a^{m}$ is   $\frac{1}{a^{m}}$

So, for $5^{-3}$ multiplicative inverse is $5^{3}$

$10^{-100}$

The multiplicative inverse of    $a^{m}$ is   $\frac{1}{a^{m}}$

So, the multiplicative inverse of $10^{-100}$ is $10^{100}$

CBSE NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.2 powers with negative exponents

Question:(i) Expand the following numbers using exponents.

1025.63

1025.63 = $1\times 10^{3}+o\times 10^{2}+2\times 10^{1}+5\times 10^{0}+6\times 10^{-1}+3\times 10^{-2}$

Question:(ii) Expand the following numbers using exponents.

1256.249

1256.249

$= 1\times 10^{3}+2\times 10^{2}+5\times 10^{1}+6\times 10^{0}+2\times 10^{-1}+4\times 10^{-2}+9\times 10^{-3}$

NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.3 laws of exponents

Question:1(i) Simplify and write in exponential form.

$(-2)^{-3}\times (-2)^{-4}$

this is simplified as follows

$(-2)^{-3}\times (-2)^{-4}$

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

$= \frac{1}{(-2)^{3+4}}=\frac{1}{(-2)^{7}}$

$= (-2)^{-7}$

Question:1(ii)

$p^3\times p^{-10}$

this is simplified as follows

$p^3\times p^{-10}$

$p^{3-10}$.............$[a^{m}\times a^{n}=a^{m+n}]$

$=p^{-7}$

Question:1(iii) Simplify and write in exponential form.

$3^2 \times 3^{-4}\times 3^6$

this can be simplified as follows

$3^2 \times 3^{-4}\times 3^6$

$= 3^{2+(-4)+6}$.............$[a^{m}\times a^{n}\times a^{o}=a^{m+n+o}]$

$= 3^{4}= 81$

Solutions of NCERT for class 8 maths chapter 12 exponents and powers-Exercise: 12.1

Question:1(i) Evaluate.

$3^{-2}$

The detailed explanation for the above-written question is as follows,

We know that,

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

So,    here m  =2 and a = 3

$3^{-2}=\frac{1}{3^{2}} = \frac{1}{3}\times \frac{1}{3} = \frac{1}{9}$

Question:1(ii) Evaluate.

$(-4)^{-2}$

The detailed explanation for the above-written question is as follows

We know that,

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

So, here (a = -4) and (m = 2)

Then according to the law of exponent

$(-4)^{-2}= \frac{1}{(-4)^{2}} = \frac{1}{(-4)}\times \frac{1}{(-4)} = \frac{1}{16}$  [  negative $\times$negative = positive]

Question:1(iii) Evaluate.

$\left(\frac{1}{2}\right )^{-5}$

The detailed solution for the above-written question is as follows

We know that,

$(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$         $\&$         $a^{-m} = \frac{1}{a^{m}}$

So, here

a = 1 and b = 2 and m =-5

According to the law of exponent

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

$=(-1)\times -2\times -2\times -2\times -2\times -2 = 32$

$(-4)^{5}\div (-4)^{8}$

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^{m}}{a_{n}} = a^{m-n}$        and   $a^{-m}= \frac{1}{a^{m}}$

So according to this

a = -4, m = 5 and n = 8

$\frac{4^{5}}{4_{8}} = -4^{5-8} = -4^{-3}$

$= (-\frac{1}{4})\times -(\frac{1}{4})\times(-\frac{1}{4}) = -\frac{1}{64}$

$\left (\frac{1}{2^3} \right )^2$

The detailed solution for the above-written question is as follows

We know the exponential formula

$\frac{a^{m}}{b^{m}} = (\frac{a}{b})^{m}$        and   $a^{-m}= \frac{1}{a^{m}}$  and  $(a^{m})^{n} = a^{mn}$

So, we have given

a = 1, b=2

By using above exponential law,

$\frac{a^{m}}{b^{m}} = \frac{1}{(2^{3})^{2}} = \frac{1}{2^{6}}$

$= \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} = \frac{1}{64}$

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

The detailed solution for the above-written question is as follows,

We know the exponential formula

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

So, $(-3)^{4}\times (\frac{5}{3})^{4}= (-3)^{4}\times \frac{5^{4}}{3^{4}} = \frac{625}{1}$

$(3^{-7}\div 3^{-10})\times 3^{-5}$

The detailed explanation for the above-written question is as follows

As we know the exponential form

$\frac{a^{m}}{b^{n}}= a^{m-n} \& (a^{m}\times a^{n})=a^{m+n}$

By using these two form we get,

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

$=3^{3} \times (3)^{-5}= 3^{-2}$

$=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$

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

The detailed solution for the above-written question is as follows,

we know the exponential forms

$a^{-m}=\frac{1}{a^{m}}$         &         $a^{m}\times b^{m}= (ab)^{m}$

So, according to our data,

here initially we use first forms and then the second one.

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

$= \frac{1}{(2\times -7)^{3}}= \frac{1}{(-14)^{3}}$

Question:3(i) Find the value of.

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

The detailed explanation for the above-written question is as follows,

As we know that $a^{0}=1$

So, $3^{0}=1$

now,

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

$=\frac{5}{4}\times 2^{2}\Rightarrow \frac{5}{2^{2}}\times 2^{2}=5/1$

Question:3(ii) Find the value of.

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

The detailed explanation for the above-written question is as follows

Rewrite the equation

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

$=(2^{-1+(-2)})\div 2^{-2}$    .................................$a^{m}\times a^{n}= a^{(m+n)}$

$=(2^{-3})\div 2^{-2} = 2^{-3-(-2)}$........................$a^{m}\div a^{n}= a^{(m-n)}$

$= 2^{-1}= \frac{1}{2}$

Question:3(iii) Find the value of.

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

The detailed explanation for the above-written question is as follows,

This is the exponential form

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

So,  $\left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}$

$=\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{4^{-2}}$.......................using this form $a^{m}= \frac{1}{a^{m}}$

$=2^{2}+3^{2}+4^{2}$

= 4+9+16

= 29

Question:3(iv) Find the value of.

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

since we know that

$a^{0}=1$

$(3^{-1} + 4^{-1} + 5^{-1})^0$ $=1$

Question:3(v) Find the value of.

$\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}$

The detailed explanation for the above-written question is as follows

$\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}$

$=(-2/3)^{-2\times 2}$.............. BY using these form of exponential $(a^{m})^{n}=a^{mn}$

$(-2/3)^{-4}=(-3/2)^{4}$......... use this  $a^{-m}=\frac{1}{a^{m}}$

$=\frac{81}{16}$

Question:4(i) Evaluate

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

The detailed explanation for the above written question is as follows

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

after rewriting the above equation we get,

$\frac{2^{-3}\times 5^{3}}{2^{-4}}$       $=2^{-3-(-4)}\times 5^{3}$  ...........as we know that  $\frac{a^{m}}{a^{n}}= a^{m-n}$

$\\=2^{1}\times 5^{3}\\ = 2\times 125 = 250$

An alternate method,

$= \frac{5^{3}}{2^{-4}\times 2^{3}}$

here you can use first $a^{-m}= \frac{1}{a^{m}}$  and after that use $a^{m}\times a^{n}= a^{m+n}$

Question:4(ii) Evaluate

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

The detailed explanation for the above-written question is as follows

We clearly see that this is in the form of  $a^{-m}=\frac{1}{a^{m}}$

So, $(5^{-1}\times 2^{-1})\times 6^{-1}= (\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}$

$= \frac{1}{60}$

$5^m \div 5^{-3} = 5^5$

We have,

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

Here a = 5 and n =-3 and m-n = 5

therefore,

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

By comparing from both sides we get

m+3 = 5

m= 2

Question:6(i) Evaluate

$\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}$

The detailed solution for the above-written question is as follows

$\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}$

$= [(1\times 3)-(1\times 4)]^{-1}$.............by using     $a^{-m}= \frac{1}{a^{m}}$

$= [3-4]^{-1}$

$\\= [-1]^{-1}\\=-1$

Question:6(ii) Evaluate

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

The detailed solution for the above-written question is as follows

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

$=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}$................ using the form    $\frac{a^{m}}{b^{m}}= (a/b)^{m}$

$=5^{-7+4}\times 8^{-4+7}$............using   $a^{m} \div a^{n} = a^{m-n}$

$= 5^{-3}\times 8^{+3}$

$\\= \frac{8^{3}}{5^{3}}\\\\=\frac{512}{125}$

Question:7(i) Simplify

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

The detailed solution for the above-written question is as follows

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

we can write $25= 5^{2}$

So, after rewriting the equation,

$\frac{5^{2}\times t^{-4}}{5^{-3}\times 10\times t^{-8}}$

$= \frac{5^{2+3}\times t^{-4+8}}{10}$.................using the form   $[a^{m}\div a^{n}= a^{m-n}]$

$= \frac{5^{5}\times t^{4}}{10}$ .............(By expanding we have now)

$= \frac{625t^{4}}{2}$

Question:7(ii) Simplify.

$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}$

The detailed solution for the above-written question is

$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}$

we can write 125 = $5^{3}$ and $6^{-5}$ can be written as $(2\times 3)^{-5}$

Now, rewriting the equation, we get

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

$=\frac{3^{-5}\times 10^{-5}\times 5^{3+7}}{(2\times 3) ^{-5}}$.............by using $[a^{m}\div a^{n}=a^{m-n}]$

$=\frac{ 10^{-5}\times 5^{10}}{(2) ^{-5}}$  .....................Use $[a^{m}\div a^{n}=a^{m-n}]$

$5^{10-5}= 5^{5}=3125$.........................As $[10^{-5} = (2\times 5)^{-5}=2^{-5}\times 5^{-5}]$ . $2^{-5}$ can be cancelled out with the denominator $2^{-5}$

CBSE NCERT solutions for class 8 maths chapter 12 exponents and powers topic 12.4 use of exponents to express small numbers in standard form

Question:1(i) Write the following numbers in standard form.

0.000000564

the standard form of 0.000000564 is

$\frac{564}{1000000000}=5.64\times10^{-7}$

Question:1(ii) Write the following numbers in standard form.

0.0000021

The  standard form 0.0000021 is

$=\frac{21}{10000000}=2.1\times 10^{-6}$

Question:1(iii) Write the following numbers in standard form.

21600000

The standard form 21600000 is

$=2.16\times 10^{7}$

Question:1(iv) Write the following numbers in standard form.

15240000

The standard form 15240000

$=1.524\times 10^{7}$

1. Distance between sun and earth $1.496\times10^{11}m$

2. speed  of light is $3\times10^{8}m/s$

3. The avg. diameter of red blood cells is $(7\times10^{-6}mm)$

4. the distance of the moon from the earth is $(3.84467\times10^{8}m)$

5. size of the plant cell is $(1.275\times10^{-5}m)$

6. The diameter of the wire on a computer chip is $(3\times10^{-6}m)$

7. the height of the Mount Everest is $(8.848\times10^{3}m)$

NCERT solutions for class 8 maths chapter 12 exponents and powers-Exercise: 12.2

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

0.0000000000085

The standard form is $8.5\times 10^{-12}$

Question:1(ii) Express the following numbers in standard form.

0.00000000000942

The standard form is $9.42\times 10^{-12}$

Question:1(iii) Express the following numbers in standard form.

6020000000000000

The standard form is $6.02\times 10^{15}$

Question:1(iv) Express the following numbers in standard form.

0.00000000837

The standard form of the given number is $8.37\times10^{-9}$

Question:1(v) Express the following numbers in standard form.

31860000000

The standard form is $3.186\times10^{10}$

Question:2(i) Express the following numbers in usual form.

$3.02\times 10^{-6}$

$3.02\times 10^{-6}$

$=\frac{3.02}{1000000}$

$=0.00000302$

this is the usual form

Question:2(ii) Express the following numbers in usual form.

$4.5 \times 10^4$

$4.5 \times 10^4=4.5\times 10000$

$=45000$

this is the usual form

Question:2(iii) Express the following numbers in the usual form.

$3\times 10^{-8}$

$3\times 10^{-8}$

$\frac{3}{100000000} = 0.000000030$

this is the usual form

Question:2(iv) Express the following numbers in usual form.

$1.0001\times 10^9$

$1.0001\times 10^9$

$=1.0001\times 100000000$

$=1000100000$

this is the usual form

Question:2(v) Express the following numbers in usual form.

$5.8\times 10^{12}$

$5.8\times 10^{12}$

$=5.8\times 100000000000$

$=5800000000000$

this is the usual form

Question:2(vi) Express the following numbers in usual form.

$3.61492 \times 10^6$

$3.61492 \times 10^6$

$= 3.61492\times 1000000$

$= 3614920$

17155

1 micron is equal to $\frac{1}{1000000}m$.

1 micron is equal to

$\frac{1}{1000000}m$

$= 1\times 10^{-6}$

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

$=1.6\times 10^{-19}$ coulomb.

Size of a bacteria is 0.0000005 m.

Size of a bacteria is 0.0000005 m

$\frac{5}{10000000}=5\times 10^{-7}m$

Size of a plant cell is 0.00001275 m.

Size of a plant cell is0.00001275m

$=\frac{1275}{10000}=1.275\times10^{-5}m$

Thickness of a thick paper is 0.07 mm

The thickness of a thick paper is 0.07

$=\frac{7}{100}=7\times 10^{-2}mm$

the thickness of each book = 20mm

So, the thickness of 5 books = $(5\times 20)=100 mm$

the thickness of one paper sheet =0.016mm

So, the thickness of 5 paper sheet =  $(5\times 0.016)=0.08 mm$

the total thickness of the stack = (100+0.08)mm

=100.08 mm or

$(1.008\times10^{-2}mm)$

## NCERT solutions for class 8 maths: Chapter-wise

 Chapter -1 NCERT solutions for class 8 maths chapter 1 Rational Numbers Chapter -2 Solutions of NCERT for class 8 maths chapter 2 Linear Equations in One Variable Chapter-3 CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Chapter-4 NCERT solutions for class 8 maths chapter 4 Practical Geometry Chapter-5 Solutions of NCERT for class 8 maths chapter 5 Data Handling Chapter-6 CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots Chapter-7 NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots Chapter-8 Solutions of NCERT for class 8 maths chapter 8 Comparing Quantities Chapter-9 NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities Chapter-10 CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes Chapter-11 NCERT solutions for class 8 maths chapter 11 Mensuration Chapter-12 NCERT Solutions for class 8 maths chapter 12 Exponents and Powers Chapter-13 CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions Chapter-14 NCERT solutions for class 8 maths chapter 14 Factorization Chapter-15 Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs Chapter-16 CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers

## NCERT solutions for class 8: Subject-wise

Some important result from NCERT solutions for class 8 maths chapter 12 exponents and powers-

• $a^{m} \times a^{n}=a^{m+n}$
• $a^{m} \div a^{n}=a^{m-n}$
• $\left(a^{m}\right)^{n}=a^{m n}$
• $a^{m} \times b^{m}=(a b)^{m}$
• $a^{0}=1$
• $\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}$

The above results are valid for both positive and negative powers. You don't have to remember the results. If you understood the concepts you can derive all the above results. You can go through NCERT solutions for class 8 maths chapter 12 exponents and powers to get conceptual clarity.