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# NCERT Solutions for Class 8 Maths Chapter 10 - Exponents and Powers

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers - Free PDF Download

The NCERT Solutions for class 8 maths Chapter 10 focus on helping students get better clarity on essential topics of exponents and powers. Exponents and powers class 8 is one of the most critical chapters of Class 8. Students can now avail the exponents and powers class 8 solutions PDF to gain relevant insights on how to ace their exams. The study materials offered by Vedantu will thus help students learn how to prepare for their upcoming exams. The NCERT Solution PDFs provided here contain specific techniques through which students can learn about Chapter 8.

Table of Content
1. NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers - Free PDF Download
2. Glance of NCERT Solutions for Class 8 Maths Chapter Exponents and Powers | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers
5. Access NCERT Solutions for Class 8 Maths Chapter 10 – Exponents and Powers
5.1Exercise 10.1
5.2Exercise 10. 2
6. Overview of Deleted Syllabus for CBSE Class 8 Maths Chapter 10 Exponents and Powers
7. NCERT Solutions for Class 8 Maths Chapter 10 Exercises
8. CBSE Class 8 Maths Chapter 10 Other Study Materials
9. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

## Glance of NCERT Solutions for Class 8 Maths Chapter Exponents and Powers | Vedantu

• This chapter focuses on how to write large numbers more conveniently using exponents and powers, express numbers in standard form, deal with negative exponents, apply various laws of exponents, and much more.

• Powers with Negative Exponents will help us understand how to handle and simplify expressions with negative exponents by using formulas.

• Laws of exponents, like learning the rules for multiplying, dividing, and raising powers to powers, which are essential for simplifying complex expressions.

• This chapter reinforces our understanding of exponent rules and their applications.

• This chapter helps to enable students to simplify mathematical expressions and solve real-world problems using scientific notation.

• It also helps to build a foundation for higher-level algebra and mathematics.

• There are links to video tutorials explaining class 8 chapter Exponents and Powers for better understanding.

• There are two exercises (11 fully solved questions) in Class 12th Maths, Chapter 8, Exponents and Powers.

## Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 8

 S.No. Current Syllabus Exercises of Class 8 Maths Chapter 10 1 Class 8 Maths Exponents and Powers Exercise 10.1 2 Class 8 Maths Exponents and Powers Exercise 10.2

## Exercises Under NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers

### Exercise 10.1: Introduction to Exponents

This exercise introduces the concept of exponents, where a^n means multiplying the base a by itself n times. Students learn the basic properties of exponents and how to identify patterns. The exercise includes real-life applications of exponents, like calculating areas and volumes, to show the practical use of these mathematical concepts.

### Exercise 10.2: Laws of Exponents

The product of powers rule is explained, where multiplying powers with the same base results in adding the exponents. The quotient of powers rule shows that dividing powers with the same base results in subtracting the exponents. The power of a power rule demonstrates that raising a power to another power multiplies the exponents. The power of a product rule illustrates that raising a product to a power distributes the exponent to each factor. The power of a quotient rule explains that raising a quotient to a power applies the exponent to both the numerator and the denominator.

## Access NCERT Solutions for Class 8 Maths Chapter 10 – Exponents and Powers

### Exercise 10.1

1. Evaluate the Following:

(i) ${{3}^{-2}}$

Ans: We have to evaluate ${{3}^{-2}}$.

We will apply the identity of indices ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ , we get

${{3}^{-2}}=\dfrac{1}{{{3}^{2}}}$

$\Rightarrow {{3}^{-2}}=\dfrac{1}{3\times 3}$

$\therefore {{3}^{-2}}=\dfrac{1}{9}$

(ii) ${{\left( -4 \right)}^{-2}}$

Ans: We have to evaluate ${{\left( -4 \right)}^{-2}}$.

We will apply the identity of indices ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ , we get

${{\left( -4 \right)}^{-2}}=\dfrac{1}{{{\left( -4 \right)}^{2}}}$

$\Rightarrow {{\left( -4 \right)}^{-2}}=\dfrac{1}{-4\times -4}$

$\therefore {{\left( -4 \right)}^{-2}}=\dfrac{1}{16}$

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

Ans: We have to evaluate ${{\left( \dfrac{1}{2} \right)}^{-5}}$.

We will apply the identity of indices ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$ , we get

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

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-5}}=\dfrac{1}{{{2}^{-5}}}$

We can apply the identity $\dfrac{1}{{{a}^{-n}}}={{a}^{n}}$, we get

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-5}}={{2}^{5}}$

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-5}}=2\times 2\times 2\times 2\times 2$

$\therefore {{\left( \dfrac{1}{2} \right)}^{-5}}=32$

2. Simplify and Express the Result in Power Notation With Positive Exponent.

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

Ans: We have to simplify the expression ${{\left( -4 \right)}^{5}}\div {{\left( -4 \right)}^{8}}$.

According to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

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

Now, applying the above identity to the given expression, we get

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

$\Rightarrow {{\left( -4 \right)}^{5}}\div {{\left( -4 \right)}^{8}}={{4}^{-3}}$

We have to express the result with positive exponent, we can write

${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$

$\therefore {{\left( -4 \right)}^{5}}\div {{\left( -4 \right)}^{8}}=\dfrac{1}{{{4}^{3}}}$

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

Ans: We have to simplify the expression ${{\left( \dfrac{1}{{{2}^{3}}} \right)}^{2}}$.

We know that we can apply the identity of indices ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$ , we get

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

Now, by applying the identity power of power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, we get

$\Rightarrow {{\left( \dfrac{1}{{{2}^{3}}} \right)}^{2}}=\dfrac{{{1}^{2}}}{{{2}^{3\times 2}}}$

$\therefore {{\left( \dfrac{1}{{{2}^{3}}} \right)}^{2}}=\dfrac{{{1}^{2}}}{{{2}^{6}}}$

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

Ans: We have to simplify the expression ${{\left( -3 \right)}^{4}}\times {{\left( \dfrac{5}{3} \right)}^{4}}$.

We can apply the identity of indices ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$ , we get

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

$\Rightarrow {{\left( -3 \right)}^{4}}\times {{\left( \dfrac{5}{3} \right)}^{4}}={{\left( -1 \right)}^{4}}\times {{3}^{4}}\times \left( \dfrac{{{5}^{4}}}{{{3}^{4}}} \right)$

$\Rightarrow {{\left( -3 \right)}^{4}}\times {{\left( \dfrac{5}{3} \right)}^{4}}={{\left( -1 \right)}^{4}}\times {{5}^{4}}$

$\Rightarrow {{\left( -3 \right)}^{4}}\times {{\left( \dfrac{5}{3} \right)}^{4}}=1\times {{5}^{4}}$

$\therefore {{\left( -3 \right)}^{4}}\times {{\left( \dfrac{5}{3} \right)}^{4}}={{5}^{4}}$

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

Ans: We have to simplify the expression $\left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}$.

According to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

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

Now, applying the above identity to the given expression, we get

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

$\Rightarrow \left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}={{3}^{-7+10}}\times {{3}^{-5}}$

$\Rightarrow \left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}={{3}^{3}}\times {{3}^{-5}}$

Now, according to the product of power rule of exponents

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

Now, applying the above identity , we get

$\Rightarrow \left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}={{3}^{3+\left( -5 \right)}}$

$\Rightarrow \left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}={{3}^{-2}}$

We have to express the result with positive exponent, we can write

${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$

$\therefore \left( {{3}^{-7}}\div {{3}^{-10}} \right)\times {{3}^{-5}}=\dfrac{1}{{{3}^{2}}}$

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

Ans: We have to simplify the expression ${{2}^{-3}}\times {{\left( -7 \right)}^{-3}}$.

We know that ${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$

Now, applying the above identity to the given expression, we get

$\Rightarrow {{2}^{-3}}\times {{\left( -7 \right)}^{-3}}={{\left( 2\times \left( -7 \right) \right)}^{-3}}$

$\Rightarrow {{2}^{-3}}\times {{\left( -7 \right)}^{-3}}={{\left( -14 \right)}^{-3}}$

We have to express the result with positive exponent, we can write

${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$

$\therefore {{2}^{-3}}\times {{\left( -7 \right)}^{-3}}=\dfrac{1}{{{\left( -14 \right)}^{3}}}$

3. Find the Value of Following:

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

Ans: We have to find the value of $\left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}$.

We will apply the identity ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$, we get

$\left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}=\left( {{3}^{0}}+\dfrac{1}{{{4}^{1}}} \right)\times {{2}^{2}}$

Now, we know that ${{a}^{0}}=1$, we get

$\Rightarrow \left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}=\left( 1+\dfrac{1}{4} \right)\times {{2}^{2}}$

$\Rightarrow \left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}=\left( \dfrac{4+1}{4} \right)\times 2\times 2$

$\Rightarrow \left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}=\left( \dfrac{5}{4} \right)\times 4$

$\therefore \left( {{3}^{0}}+{{4}^{-1}} \right)\times {{2}^{2}}=5$

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

Ans: We have to find the value of $\left( {{2}^{-1}}\times {{4}^{-1}} \right)\div {{2}^{-2}}$.

The given expression can be written as

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

We will apply the identity ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, we get

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

Now, we know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, we get

$\Rightarrow \left( {{2}^{-1}}\times {{4}^{-1}} \right)\div {{2}^{-2}}=\left( {{2}^{-1+\left( -2 \right)}} \right)\div {{2}^{-2}}$

$\Rightarrow \left( {{2}^{-1}}\times {{4}^{-1}} \right)\div {{2}^{-2}}={{2}^{-3}}\div {{2}^{-2}}$

We will apply the identity ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$, we get

$\Rightarrow \left( {{2}^{-1}}\times {{4}^{-1}} \right)\div {{2}^{-2}}={{2}^{-3-\left( -2 \right)}}$

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

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

$\therefore \left( {{2}^{-1}}\times {{4}^{-1}} \right)\div {{2}^{-2}}=\dfrac{1}{2}$

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

Ans: We have to find the value of ${{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}$.

We will apply the identity ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$, we get

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

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}=\left( \dfrac{1}{{{2}^{-2}}} \right)+\left( \dfrac{1}{{{3}^{-2}}} \right)+\left( \dfrac{1}{{{4}^{-2}}} \right)$

We can apply the identity $\dfrac{1}{{{a}^{-n}}}={{a}^{n}}$, we get

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}={{2}^{2}}+{{3}^{2}}+{{4}^{2}}$

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}=2\times 2+3\times 3+4\times 4$

$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}=4+9+16$

$\therefore {{\left( \dfrac{1}{2} \right)}^{-2}}+{{\left( \dfrac{1}{3} \right)}^{-2}}+{{\left( \dfrac{1}{4} \right)}^{-2}}=29$

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

Ans: We have to find the value of ${{\left( {{3}^{-1}}+{{4}^{-1}}+{{5}^{-1}} \right)}^{0}}$.

We know that ${{a}^{0}}=1$, then we get the value of the given expression

$\therefore {{\left( {{3}^{-1}}+{{4}^{-1}}+{{5}^{-1}} \right)}^{0}}=1$

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

Ans: We have to find the value of ${{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}$.

We can apply the identity ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, we get

${{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}={{\left( \dfrac{-2}{3} \right)}^{-2\times }}^{2}$

$\Rightarrow {{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}={{\left( \dfrac{-2}{3} \right)}^{-4}}$

Now, applying the identity ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$, we get

$\Rightarrow {{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}=\left( \dfrac{-{{2}^{-4}}}{{{3}^{-4}}} \right)$

Now, we know that $\dfrac{1}{{{a}^{-n}}}={{a}^{n}}$, we get

$\Rightarrow {{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}=\left( \dfrac{{{3}^{4}}}{-{{2}^{4}}} \right)$

$\Rightarrow {{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}=\left( \dfrac{3\times 3\times 3\times 3}{-2\times -2\times -2\times -2} \right)$

$\therefore {{\left\{ {{\left( \dfrac{-2}{3} \right)}^{-2}} \right\}}^{2}}=\dfrac{81}{16}$

4. Evaluate the Following:

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

Ans: We have to evaluate the given expression $\dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}$.

We can write the given expression as $\dfrac{{{\left( {{2}^{3}} \right)}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}$.

$\Rightarrow \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}=\dfrac{{{2}^{-3}}\times {{5}^{3}}}{{{2}^{-4}}}$

Now, we know that ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$, we get

$\Rightarrow \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}={{2}^{-3-\left( -4 \right)}}\times {{5}^{3}}$

$\Rightarrow \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}={{2}^{-3+4}}\times {{5}^{3}}$

$\Rightarrow \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}={{2}^{1}}\times {{5}^{3}}$

$\Rightarrow \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}=2\times 5\times 5\times 5$

$\therefore \dfrac{{{8}^{-1}}\times {{5}^{3}}}{{{2}^{-4}}}=250$

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

Ans: We have to evaluate the given expression $\left( {{5}^{-1}}\times {{2}^{-1}} \right)\times {{6}^{-1}}$.

We know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$, we get

$\left( {{5}^{-1}}\times {{2}^{-1}} \right)\times {{6}^{-1}}=\left( \dfrac{1}{5}\times \dfrac{1}{2} \right)\times \dfrac{1}{6}$

$\Rightarrow \left( {{5}^{-1}}\times {{2}^{-1}} \right)\times {{6}^{-1}}=\dfrac{1}{10}\times \dfrac{1}{6}$

$\therefore \left( {{5}^{-1}}\times {{2}^{-1}} \right)\times {{6}^{-1}}=\dfrac{1}{60}$

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

Ans: The given expression is ${{5}^{m}}\div {{5}^{-3}}={{5}^{5}}$.

Now, according to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

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

Applying the above identity to the given expression, we get

${{5}^{m-\left( -3 \right)}}={{5}^{5}}$

$\Rightarrow {{5}^{m+3}}={{5}^{5}}$

Since the bases are same on both sides, therefore the exponents must be equal to each other, we get

$\Rightarrow m+3=5$

$\Rightarrow m=5-3$

$\therefore m=2$

Therefore, we get the value of $m=2$.

6. Evaluate the Following:

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

Ans: Given expression is ${{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}$.

We know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$, applying to the given expression we get

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

$\Rightarrow {{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}={{\left\{ 3-4 \right\}}^{-1}}$

$\Rightarrow {{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}={{\left( -1 \right)}^{-1}}$

Again applying the identity ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$, we get

$\Rightarrow {{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}=\dfrac{1}{-1}$

$\therefore {{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}=-1$

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

Ans: Given expression is ${{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}$.

Now, applying the identity ${{\left( \dfrac{1}{a} \right)}^{m}}=\dfrac{{{1}^{m}}}{{{a}^{m}}}$, we get

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

Now, we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$, we get

$\Rightarrow {{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}=\dfrac{{{8}^{7}}}{{{5}^{7}}}\times \dfrac{{{5}^{4}}}{{{8}^{4}}}$

Now, applying the identity ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$, we get

$\Rightarrow {{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}=\dfrac{{{8}^{7-4}}}{{{5}^{7-4}}}$

$\Rightarrow {{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}=\dfrac{{{8}^{3}}}{{{5}^{3}}}$

$\Rightarrow {{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}=\dfrac{8\times 8\times 8}{5\times 5\times 5}$

$\therefore {{\left( \dfrac{5}{8} \right)}^{-7}}\times {{\left( \dfrac{8}{5} \right)}^{-4}}=\dfrac{512}{125}$

7. Simplify the Given Expressions:

(i) $\dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}$

Ans: Given expression $\dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}$ can be written as

$\dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{{{5}^{2}}\times {{t}^{-4}}}{{{5}^{-3}}\times 5\times 2\times {{t}^{-8}}}$

$\Rightarrow \dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{{{5}^{2}}\times {{t}^{-4}}}{{{5}^{-3+1}}\times 2\times {{t}^{-8}}}$

Now, applying the identity ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$, we get

$\Rightarrow \dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{{{5}^{2-\left( -2 \right)}}\times {{t}^{-4-\left( -8 \right)}}}{2}$

$\Rightarrow \dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{{{5}^{4}}\times {{t}^{4}}}{2}$

$\Rightarrow \dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{5\times 5\times 5\times 5\times {{t}^{4}}}{2}$

$\therefore \dfrac{25\times {{t}^{-4}}}{{{5}^{-3}}\times 10\times {{t}^{-8}}}=\dfrac{625{{t}^{4}}}{2}$

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

Ans: Given expression $\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}$ can be written as

$\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}=\dfrac{{{3}^{-5}}\times {{\left( 2\times 5 \right)}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{\left( 2\times 3 \right)}^{-5}}}$

$\Rightarrow \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}=\dfrac{{{3}^{-5}}\times {{2}^{-5}}\times {{5}^{-5}}\times {{5}^{3}}}{{{5}^{-7}}\times {{2}^{-5}}\times {{3}^{-5}}}$

Now, applying the identity ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$, we get

$\Rightarrow \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}={{3}^{-5-\left( -5 \right)}}\times {{2}^{-5-\left( -5 \right)}}\times {{5}^{-5+3-\left( -7 \right)}}$

$\Rightarrow \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}={{3}^{0}}\times {{2}^{0}}\times {{5}^{5}}$

We know that ${{a}^{0}}=1$, we get

$\Rightarrow \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}=1\times 1\times {{5}^{2}}$

$\therefore \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}={{5}^{5}}$

### Exercise 10. 2

1. Express the Following Numbers in Standard Form.

(i) $0.0000000000085$

Ans: Given number is $0.0000000000085$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

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

(ii) $0.00000000000942$

Ans: Given number is $0.00000000000942$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

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

(iii) $6020000000000000$

Ans: Given number is $6020000000000000$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

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

(iv) $0.00000000837$

Ans: Given number is $0.00000000837$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

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

(v) $31860000000$

Ans: Given number is $31860000000$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

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

2. Express the Following Numbers in Usual Form.

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

Ans: Given number is $3.02\times {{10}^{-6}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros before the number to remove the negative exponent.

We get

$3.02\times {{10}^{-6}}=.00000302$

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

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

Ans: Given number is $4.5\times {{10}^{4}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros after the number to remove the positive exponent.

We get

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

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

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

Ans: Given number is $3\times {{10}^{-8}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros before the number to remove the negative exponent.

We get

$3\times {{10}^{-8}}=.00000003$

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

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

Ans: Given number is $1.0001\times {{10}^{9}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros after the number to remove the positive exponent.

We get

$1.0001\times {{10}^{9}}=1.0001\times 1000000000$

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

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

Ans: Given number is $5.8\times {{10}^{12}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros after the number to remove the positive exponent.

We get

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

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

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

Ans: Given number is $3.61492\times {{10}^{6}}$.

To write the given number in usual form we will add number of zeros equal to the exponent number.

We will add zeros after the number to remove the positive exponent.

We get

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

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

3. Express the Number Appearing in the Following Statements in Standard Form.

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

Ans: Here, the given number is $\dfrac{1}{1000000}m$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

The number of zeros in denominator is $6$ in the number $\dfrac{1}{1000000}m$.

Now, expressing the given number in standard form, we get

$\therefore \dfrac{1}{1000000}m=1\times {{10}^{-6}}$.

(ii) Charge of an Electron is $0.000,000,000,000,000,000,16$ Coulomb.

Ans: Here, the given number is $0.000,000,000,000,000,000,16$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

$\therefore 0.000,000,000,000,000,000,16=1.6\times {{10}^{-19}}$.

(iii) Size of a bacteria is $0.0000005\text{ m}$.

Ans: Here the given number is $0.0000005\text{ m}$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

$\therefore 0.0000005\text{ m}=5\times {{10}^{-7}}\text{ m}$.

(iv) Size of a plant cell is $0.00001275\text{ m}$.

Ans: Here the number is $0.00001275\text{ m}$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

$\therefore 0.00001275\text{ m}=1.275\times {{10}^{-5}}\text{ m}$.

(v) Thickness of a thick paper is $0.07\text{ m}m$.

Ans: Here, the number is $0.07\text{ m}m$.

A number can be expressed in standard decimal form of $a\times {{10}^{b}}$.

Where, $a$ is any number between $1.0$ and $10.0$. Value of $b$ can be positive or negative.

When we shift the decimal to the right the exponent is negative and when we shift the decimal to the left the exponent is positive.

Now, expressing the given number in standard form, we get

$\therefore 0.07\text{ m}m=7\times {{10}^{-2}}\text{ m}m$.

4. In a Stack There are $5$ books each of thickness $20\text{ mm}$ and $5$ paper sheets each of thickness $0.016\text{ mm}$. What is the total thickness of the stack?

Ans: Given that there are $5$ books each of thickness $20\text{ mm}$ and $5$ paper sheets each of thickness $0.016\text{ mm}$ in a stack.

We have to find the total thickness of the stack.

Total thickness of the stack will be the sum of thickness of $5$ books and thickness of $5$ paper sheets.

Thickness of $5$ books is $=5\times 20=100\text{ mm}$.

Thickness of $5$ paper sheets is $=5\times 0.016=0.080\text{ mm}$

Now,

Total thickness of stack is

$\Rightarrow 100+0.080$

$\Rightarrow 100.08\text{ mm}$

$\therefore total\text{ thickness}=1.0008\times {{10}^{2}}\text{mm}$

Therefore, the total thickness of the stack is $1.0008\times {{10}^{2}}\text{mm}$.

## Overview of Deleted Syllabus for CBSE Class 8 Maths Chapter 10 Exponents and Powers

 Chapter Dropped Topics Exponents and Powers 10.1 Introduction 10.2 How well have you learned about fractions 10.5 How well have you learned about decimals.

## NCERT Solutions for Class 8 Maths Chapter 10 Exercises

 Exercise Number of Questions Exercise 10.1 7 Questions & Solutions Exercise 10.2 4 Questions & Solutions

## Conclusion

Exponents and Powers is a small yet crucial chapter of Class 8 Maths. NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers can help students prepare well for their upcoming exams. We encourage all Class 8 students to download and use Vedantu's NCERT Solutions for Maths Chapter 10 Exponents and Powers Class 8 PDF to learn and excel in this chapter. With Vedantu's help, students can develop a strong understanding of the concepts of exponents and powers and prepare well for their exams.

## CBSE Class 8 Maths Chapter 10 Other Study Materials

 S.No Important Links for Chapter 10 Exponents and Powers 1 Class 8 Exponents and Powers Important Questions 2 Class 8 Exponents and Powers Revision Notes 3 Class 8 Exponents and Powers Formula 4 Class 8 Exponents and Powers NCERT Exemplar Notes

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 10 - Exponents and Powers

1. What is included in the NCERT study guides?

The exponents and powers class 8 NCERT  solutions focus on offering in-depth explanations and questionnaires based on exponents and powers to students. These explanations are easy to understand so that students can relevantly prepare for their exams. These PDFs offer meticulous explanations on different topics and also test your understanding of the topics. Students can make use of these PDFs to learn which factors are involved in helping students understand the chapter in-depth.

2. What does this chapter summarise?

Exponents and powers is a vast chapter that deals with helping students learn how to write integers in the form of exponents. Further, students will also learn the exponential powers of negative integers and how they can convert different negative integers into standard form. Additionally, students will also learn about large numbers and their comparison to small numbers.

3. What is the Importance of NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers?

NCERT Solutions provide a clear and concise view of the concepts of Class 8 Maths Chapter 12 Exponents and Powers. This chapter deals with representing very large numbers or very small numbers in a standard form. Vedantu offers solutions for all the exercises of this chapter. You can find it in the link NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers. You can refer to these solutions online and also download the Solutions PDF to refer offline.

4. What are the Important Topics Covered in NCERT Solutions Class 8 Maths Chapter 12?

NCERT Solutions Class 8 Maths Chapter 12 covers many topics. It introduces students to the concept of exponents. It also briefs you about defining powers that comprise negative exponents. You will also get a much more extensive understanding of the different laws of exponents. It is a crucial chapter, and you can become familiar with the laws of exponents from it.

5. How do I get the Solutions of Class 8 Maths Exercise?

The Vedantu Class 8 NCERT Maths Solutions are available in PDF format. Here is how you can save them:

1. Visit the link NCERT Solutions Class 8 Maths Chapter 12 on the Vedantu website.

2. Scroll down and find the question for which you want the solution.

3. You will find your desired solution. The solutions are free of cost and available on the Vedantu Mobile app.

6. Give a brief summary about Chapter 12 Class 8 Maths?

Chapter 12 seeks to improve students' understanding of how to write big numbers using exponents. Students learned about the non-zero integer in seventh grade. In this chapter, students will learn how to use exponents to express tiny and big numbers. In summary, the Exponents and Powers Class 8 PDF contains clear explanations of each unit as well as answers to the practise questions.

7. Is Class 8 Chapter 12 Exponents and Powers difficult?

Class 8 Power and Exponents is a relatively simple chapter with little exercise. You can ace your examinations with ease if you completely comprehend the content. The chapter's second portion is dedicated to defining powers that have negative exponents. You'll go through numerous examples and questions in this part to help you understand the topic better.

8. How to Understand Powers with Negative Exponents in class 8 exponents and powers ?

In Chapter 10 of Class 8 Maths, understanding powers with negative exponents is essential. A negative exponent indicates that the base should be taken as the reciprocal and then raised to the corresponding positive exponent. This means that any number with a negative exponent can be rewritten as one divided by the number raised to the positive version of that exponent. This fundamental rule simplifies working with negative exponents, making it easier to manage and solve various mathematical problems involving exponents.

9. What are exponents and powers in class 8 maths chapter exponents and powers?

In exponents and powers are mathematical concepts used to express repeated multiplication of a number by itself. An exponent, also known as a power, indicates how many times the base number is multiplied by itself. Exponents make it easier to write and work with large numbers or repeated multiplications. They are used in various areas of mathematics and science to simplify expressions and calculations.

10. How to calculate exponents in class 8 exponents and powers?

Calculating exponents involves multiplying the base number by itself as many times as the exponent specifies. Here’s how to do it step-by-step:

• Identify the base and the exponent.

• Multiply the base by itself the number of times indicated by the exponent.

• For higher exponents, repeat the multiplication process.

• This method applies to any base and exponent, making it easier to handle repeated multiplications efficiently.

11. What is the rule of exponents and powers in class 8 chapter exponents and powers?

The rules of exponents and powers help simplify mathematical expressions and perform calculations more easily. Key rules include:

• Product of Powers Rule: When multiplying two expressions with the same base, add their exponents.

• Quotient of Powers Rule: When dividing two expressions with the same base, subtract the exponent of the denominator from the exponent of the numerator.

• Power of a Power Rule: When raising an exponent to another exponent, multiply the exponents.

• Power of a Product Rule: When raising a product to an exponent, apply the exponent to each factor.

• Zero Exponent Rule: Any non-zero base raised to the power of zero equals one.

• Negative Exponent Rule: A negative exponent means taking the reciprocal of the base and then applying the positive exponent.

12. How do you multiply exponents in class 8 exponents and powers?

To multiply exponents with the same base, use the Product of Powers Rule, which states that you should add the exponents together. Here’s how it works:

• Identify the base and the exponents.

• Add the exponents together while keeping the base the same.

• Simplify if necessary.

• This rule simplifies the process of multiplying exponents, especially when dealing with larger numbers or more complex expressions.