NCERT Solutions for Class 8 Maths Chapter 12 - Exponents And Powers

Exercise 12.1 

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 12. 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 $1.0008\times {{10}^{2}}\text{mm}$.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - PDF Download

NCERT Solutions for Class 8 Maths- Free PDF Download 

If you’re searching for essential exponents class 8 guide, Vedantu is just what you’re looking for. Now get detailed guides on Class 8 chapter 12 maths to expand your knowledge on distinguishing concepts of exponents and powers. Get premium access to thoroughly explained PDFs free of cost. 

NCERT Solutions for Class 8 Maths - Chapterwise Solutions

• Chapter 1 - Rational Numbers

• Chapter 2 - Linear Equations in One Variable

• Chapter 3 - Understanding Quadrilaterals

• Chapter 4 - Practical Geometry

• Chapter 5 - Data Handling

• Chapter 6 - Squares and Square Roots

• Chapter 7 - Cubes and Cube Roots

• Chapter 8 - Comparing Quantities

• Chapter 9 - Algebraic Expressions and Identities

• Chapter 10 - Visualising Solid Shapes

• Chapter 11 - Mensuration

• Chapter 13 - Direct and Inverse Proportions

• Chapter 14 - Factorisation

• Chapter 15 - Introduction to Graphs

• Chapter 16 - Playing with Numbers

NCERT Solutions for Class 8 Maths Chapter 12 Exercises

Topics Included in the Class 8 Maths Chapter 12 Exponents and Powers

Below-given is the list of topics discussed in Chapter 12 Exponents and Powers.

Now let us discuss each topic in detail.

Chapter 12- Exponents and Powers 

12.1 Introduction 

Chapter 12 aims at enhancing students’ knowledge of how they can write large numbers with the help of exponents. In 7th grade, students came across the concept of the non-zero integer a. In this chapter, students will learn how they can express small and large numbers with the help of exponents. In brief, the exponents and powers class 8 PDF consist of concise explanations of every section and solutions to the exercise questions. 

The first section of the chapter is the introduction. This section is a small yet crucial section that introduces students to the concept of exponents. The introduction starts with a fact about the mass of the earth which is 5,970,000,000,000,000,000,000,000 kg. Now, students already know how such a huge number can be written in exponents- 5.87 x 1024 kg. 

In this chapter, you’ll learn more about topics like negative integers and their exponents, and so on. This chapter helps students analyze how exactly they can use exponential powers in mathematics. 

12.2 Powers with Negative Exponents

Power and exponents class 8 is a comparatively easy chapter that involves minimal practice. If you understand the topics thoroughly, you can effortlessly ace your exams. The second section of the chapter consists of defining powers that comprise negative exponents. 

Throughout this section, you will go through several examples and questions that will help you grasp the concept better. This section of the chapter is remotely easier compared to the rest. It only involves finding the value of certain integers with negative powers. Make sure to solve a good amount of questions to understand powers with negative numbers better.

12.3 Laws of Exponents 

You will also have access to different laws of exponents in the maths exponents and powers class 8 PDF. In this guide, you will get a much more extensive understanding of different laws of exponents. Every student should ensure going through these laws regularly and even jotting them down while revising.

The section also includes questions based on these laws. Make sure that you go through these questions carefully to ensure that you are familiar with the laws of exponents. Solutions offered by Vedantu will thus give you access to varied questions that can improve your grasping skills, thereby helping you remember better. 

12.4 Use of Exponents to Express Small Numbers in Standard Form 

The fourth section covered the use of exponents for expressing small numbers in standard forms. This section offers a wide range of facts that you should go through to precisely learn about this topic. Some facts that are mentioned in this section include- the size of a plant cell (0.00001275 m), speed of light (300,000,000 m/sec), the average radius of the sun (695000 km), and so on. 

You are asked to observe how a few numbers are smaller compared to others. These facts thus help students understand how they can convert small numbers in standard form. Students are advised to go through the ‘try these ’ sections and solve different problems to learn more about this topic. 

12.4.1 Comparing Very Large And Very Small Numbers 

The NCERT Class 8 Maths Chapter 12 includes helping students learn extensively about every topic and subtopic, including comparing large and small numbers. This subheading may be comparatively difficult, and thus, students are asked to practice regularly. 

Exercises 

The chapter consists of 2 exercises in total. Exercise 12.1 comprises seven questions that test the brief understanding of the student on the laws of exponents. The second Exercise 12.2 consists of four questions on the concepts mentioned above. Students can refer to the exponents and powers class 8 CBSE guide to learn how to solve the exercise questions with ease. 

Important Points Covered in NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

Exponents and Powers is a very crucial concept in Mathematics. NCERT Solutions Maths Class 8 Chapter 12 mainly covers concepts like the basic understanding of exponents, their properties, types of exponents, laws of exponents, the use of exponents to express small numbers in standard form and comparing very large and small numbers. Below are some important points covered in this chapter.

• An exponent of any given number helps to express the number of times a base number is multiplied by itself to get the given number. The base number is raised to a certain power whereas the exponent is the power to which the base number is raised. An exponent is written as a superscript to the base number.

• Four types of exponents are there; these are zero exponents, positive exponents, negative exponents, and rational exponents.

• Laws of Exponents: The laws of exponents are as follows.

1. Product Law: To multiply two numbers expressed as the same bases raised to different powers, we have to add the powers and raise the base number to the resultant power. For example, z^a × z^b = z^a+b.

2. Quotient Law: To divide two numbers expressed as the same bases raised to different powers, we have to subtract the powers and raise the base number to the resultant power. For example, z^a ÷ z^b = z^a-b.

3. Power Law: When a base number having an exponent is raised to another exponent, we multiply the exponents and raise the base number to the resultant exponent. For example, (z^a)^b = z^ab.

4. Power of Product Law: The power of product law is expressed as, \[(yz)^a = y^a \times Z^a\]. 

5. Power of Quotient Law: The power of quotient law is expressed as, (x ÷ y)^a = x^a ÷ y^a.

6. Zero Power Law: When any base is raised to the power zero, its value will always be equal to one. For example, z⁰ = 1.

7. Negative Exponent Law: When a number has a negative exponent, flip it into a reciprocal to make express it as a positive exponent. For example, z^-a= 1/z^a.

List of Formulas of Class 8 Maths Chapter 12 Exponents and Powers

Find here the list of all important formulas discussed in the chapter. 

Importance of Exponents and Powers

Exponents and Powers are very significant concepts for Class 8 and have applications even in higher classes. Therefore, having a deep understanding of the concepts covered in this chapter is very crucial for students and it will also help to build a strong foundation for advanced studies in Maths. So, learning these basic concepts will boost students’ practical knowledge to apply exponents and powers in real-world situations.

Key Features of NCERT Solutions for Class 8 Chapter 12 Exponents and Powers

Exponents and Powers is a small yet crucial chapter of Class 8 Maths. NCERT Solutions for Class 8 Chapter 12 Exponents and Powers can help students prepare well for their upcoming exams. Here are some prime benefits of Vedantu's NCERT Solutions for Class 8 Maths Chapter 12:

• Gain a comprehensive knowledge of vital topics of exponents and powers. Vedantu's solutions cover all the concepts in the chapter in a comprehensive and easy-to-understand manner.

• Get access to important questions and answers and revision notes to improve your problem-solving skills. Vedantu's solutions include step-by-step solutions to all the problems in the chapter, as well as revision notes for last minute practice.

• Become familiar with the laws of exponents and how you can use them to solve questions. Vedantu's solutions explain the laws of exponents in detail and provide examples of how to use them to solve problems.

• Get access to Vedantu's personalized online tuition classes, live interactive learning, and doubt-solving sessions to get extra help with the concepts and problems. Vedantu's experienced and qualified teachers can help students to understand the concepts and solve problems efficiently and accurately.

Conclusion

Exponents and Powers is a small yet crucial chapter of Class 8 Maths. NCERT Solutions for Class 8 Maths Chapter 12 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 Class 8 Maths Chapter 12 Exponents and Powers 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.