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NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers

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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.

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Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 8

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


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.



Important Related Links for CBSE Class 8 Maths

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

1. What are NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Powers?

NCERT Solutions for Class 8 Maths Chapter 10 provide detailed, step-by-step answers to all textbook questions on exponents and powers. These solutions explain essential concepts, properties of exponents, and standard forms in a student-friendly manner, following CBSE 2025–26 guidelines.

2. How do NCERT Solutions help in understanding the laws of exponents in Class 8 Maths?

NCERT Solutions clarify the laws of exponents by giving solved examples and explanations for each law, such as the product and quotient of powers, power of a power, power of a product, zero exponent rule, and negative exponent rule. This helps students practise and understand their applications as per CBSE Class 8 Maths syllabus.

3. What is the significance of negative exponents and how are they solved in Class 8 NCERT Solutions?

Negative exponents indicate the reciprocal of the base raised to the corresponding positive exponent. NCERT Solutions for Class 8 Maths Chapter 10 provide stepwise solutions demonstrating how to simplify expressions and convert negative exponents into positive forms, which is crucial for solving CBSE pattern questions.

4. Why is it important to learn expressing numbers in standard form as per Class 8 Chapter 10 NCERT Solutions?

Expressing numbers in standard form simplifies very large or very small numbers, making calculations easier and more readable. The NCERT Solutions teach this skill with examples, which is essential for both academic and real-life scientific calculations, aligned with the CBSE 2025–26 syllabus.

5. What types of problems are covered in NCERT Solutions for Class 8 Maths Exercise 10.1 and 10.2?

Exercise 10.1 covers definitions, properties, and basic applications of exponents, including real-life examples and pattern recognition. Exercise 10.2 focuses on applying laws of exponents to simplify, evaluate, and compare numbers, as per CBSE exam requirements for Class 8 students.

6. Are there any common misconceptions in Class 8 Exponents and Powers addressed by the NCERT Solutions?

Yes, NCERT Solutions address misconceptions such as confusing the product and power rules, incorrectly handling negative exponents, and misapplying zero exponent rule. The solutions provide clarifying examples to ensure students avoid common calculation errors during CBSE exams.

7. How does understanding exponents help in solving higher-level Maths problems in Class 8 and beyond?

Mastering exponents builds foundational skills for algebra, scientific notation, and advanced mathematical concepts. NCERT Solutions enable students to manipulate exponential expressions effectively, supporting future topics in higher classes aligned with CBSE and competitive exam patterns.

8. In what ways can NCERT Solutions for Class 8 Chapter 10 help improve exam performance?

NCERT Solutions offer systematic, CBSE-patterned explanations, helping students develop accuracy and speed in solving problems. Practising these solutions boosts conceptual clarity and confidence, ensuring better scores in school and board exams.

9. What is the correct approach to solving problems with fractional and decimal exponents in Class 8 NCERT Solutions?

The NCERT Solutions explain that for fractional exponents, the numerator represents the power and the denominator indicates the root, while for decimals, numbers are expressed in standard form using powers of 10. Worked-out solutions illustrate each conversion and simplification method per CBSE guidelines.

10. How are the deleted topics from the CBSE syllabus handled in the latest NCERT Solutions for Class 8 Maths Chapter 10?

The latest NCERT Solutions are updated to exclude questions and explanations from removed syllabus sections, such as certain preliminary topics on fractions and decimals. They only focus on the current official CBSE 2025–26 portion of Exponents and Powers for clarity and exam relevance.

11. Can NCERT Solutions for Class 8 Maths Chapter 10 be used for quick revision before exams?

Yes, the solutions summarize key concepts, rules, and typical problem types, making them ideal for revision. Reviewing these stepwise answers helps reinforce understanding and recall, which is critical for last-minute exam preparation.

12. How do the NCERT Solutions demonstrate the use of exponents in real-world applications?

Solutions include problems where exponents are applied to real-life scenarios, such as measurements in science and comparisons of magnitudes, showing how mathematical concepts are useful outside the classroom, in alignment with CBSE’s practical approach for Class 8.

13. What advanced reasoning or HOTS (Higher Order Thinking Skills) questions are addressed by NCERT Solutions in this chapter?

NCERT Solutions incorporate HOTS questions such as comparing large numbers using exponents, simplifying nested exponents, reasoning with negative powers, and conversions between forms, challenging students to think critically, as encouraged by CBSE for exam success.