# NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers (EX 12.1) Exercise 12.1

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers (EX 12.1) Exercise 12.1

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## Access NCERT Solution for Class 8 Maths Chapter 12- Exponents and Powers

### Exercise 12.1

1. Evaluate

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

Ans: Using property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$, to evaluate ${3^{ - 2}}$.

Therefore,

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

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

Thus, the final value of ${3^{ - 2}}$ is $\dfrac{1}{9}$.

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

Ans:

Using property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$,to evaluate ${\left( { - 4} \right)^{ - 2}}$.

Therefore,

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

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

Using property $\left( { - a} \right) \times \left( { - a} \right) = {a^2}$

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

$\Rightarrow\dfrac{1}{{{4^2}}} = \dfrac{1}{{16}} \\$

Thus, the final value of ${\left( { - 4} \right)^{ - 2}}$ is $\dfrac{1}{{16}}$.

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

Ans.

Using property ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$, to evaluate ${\left( {\dfrac{1}{2}} \right)^{ - 5}}$.

Therefore,

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

$\Rightarrow\dfrac{{{1^{ - 5}}}}{{{2^{ - 5}}}} = \dfrac{1}{{{2^{ - 5}}}} \\$

$\Rightarrow\dfrac{1}{{{2^{ - 5}}}} = {2^5} \\$

$\Rightarrow{2^5} = 2 \times 2 \times 2 \times 2 \times 2 = 32 \\$

Thus the final value of ${\left( {\dfrac{1}{2}} \right)^{ - 5}}$ is 32.

2. Simplify and express the result in power notation with positive exponent.

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

Ans: To solve this problem, it has to use property ${a^m} \div {a^n} = {a^{m - n}}$

Therefore,

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

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

Again, using property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$

Therefore,

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

So, the simplified form of ${\left( { - 4} \right)^5} \div {\left( { - 4} \right)^8}$ is $\dfrac{1}{{{{\left( { - 4} \right)}^3}}}$.

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

Ans: To solve this problem, it has to use property ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$

Therefore,

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

$= \dfrac{1}{{{2^6}}} \\$

So, the simplified form of  ${\left( {\dfrac{1}{{{2^3}}}} \right)^2}$ is $\dfrac{1}{{{2^6}}}$.

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

Ans: To solve this problem, it has to use property ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$.

Therefore,

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

Again using the property ${\left( {ab} \right)^m} = {a^m} \times {b^m}$

Therefore, the above expression will be written as

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

Also,

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

Therefore,

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

$= {3^0} \times {5^4} \\$

Also,

${x^0} = 1$

Therefore,

${3^0} \times {5^4} = {5^4}$

So the simplified form of  ${\left( { - 3} \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4}$ is ${5^4}$.

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

Ans: To solve this problem, it has to use property

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

Therefore,

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

Again using property ${a^m} \times {b^m} = {\left( {ab} \right)^m}$.

Therefore,

Above expression will be written as,

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

$= \dfrac{1}{{{{\left( { - 14} \right)}^3}}} \\$

So, the simplified form of ${2^{ - 3}} \times {\left( { - 7} \right)^{ - 3}}$ is $\dfrac{1}{{{{\left( { - 14} \right)}^3}}}$.

3. Find the value of

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

Ans: To solve this problem, it has to use property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$ and ${x^0} = 1$.

Therefore,

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

$= \dfrac{5}{4} \times 4 \\$

$= 5 \\$

So, the value of $\left( {{3^0} + {4^{ - 1}}} \right) \times {2^2}$ is 5.

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

Ans:

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

To solve this expression, it has to use property ${\left( {{a^m}} \right)^n} = {a^{mn}}$.

Therefore,

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

Using property ${a^m} \times {a^n} = {a^{m + n}}$,

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

$= {2^{ - 3}} \div {2^{ - 2}} \\$

Using the property ${a^m} \div {a^n} = {a^{m - n}}$,

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

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

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

Using the property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$,

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

So the value of  $\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}$ is $\dfrac{1}{2}$.

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

Ans: To solve this problem above expression can be written as

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

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

$= 4 + 9 + 16 \\$

$= 29 \\$

Therefore the value of ${\left( {\dfrac{1}{2}} \right)^{ - 2}} + {\left( {\dfrac{1}{3}} \right)^{ - 2}} + {\left( {\dfrac{1}{4}} \right)^{ - 2}}$ is 29.

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

Ans: To solve this problem, it has to use property ${x^0} = 1$,

Therefore,

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

So the value of ${\left( {{3^{ - 1}} + {4^{ - 1}} + {5^{ - 1}}} \right)^0}$ is 1.

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

Ans: The given expression is

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

The above expression can be written as,

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

Now, using the property${\left( {{a^m}} \right)^n} = {a^{mn}}$,

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

Using the property  ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$,

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

$= \dfrac{{81}}{{16}} \\$

So the value of ${\left\{ {{{\left( {\dfrac{{ - 2}}{3}} \right)}^{ - 2}}} \right\}^2}$ is $\dfrac{{81}}{{16}}$.

4. Evaluate

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

Ans: To solve this question, using property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$,

So, the above expression becomes,

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

8 can be written as $2 \times 2 \times 2 = {2^3}$

Therefore,

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

Using the property ${a^m} \div {a^n} = {a^{m - n}}$, we get

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

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

$= 2 \times 125 \\$

$= 250 \\$

So the value of $\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}$ is 250.

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

Ans: To solve this problem, using property ${a^m} \times {b^m} = {\left( {ab} \right)^m}$,

Therefore,

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

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

Using the property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$, we get

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

$= \dfrac{1}{{10}} \times \dfrac{1}{6}$

$= \dfrac{1}{{60}}$

So, the value of $\left( {{5^{ - 1}} \times {2^{ - 1}}} \right) \times {6^{ - 1}}$ is $\dfrac{1}{{60}}$.

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

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

To solve this problem, use the property ${a^m} \div {a^n} = {a^{m - n}}$.

Therefore,

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

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

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

As the base of the power on both sides is the same, so their power must be equal.

Therefore,

$m + 3 = 5$

$\Rightarrow m = 5 - 3$

$\Rightarrow m = 2$

So the value  of $m$ is 2.

6. Evaluate

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

Ans: The given expression is

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

Above expression can be written as

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

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

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

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

Using property ${a^{ - n}} = \dfrac{1}{{{a^n}}}$, we get

${\left( { - 1} \right)^{ - 1}}$

$= \dfrac{1}{{ - 1}}$

$= - 1$

So the value of ${\left\{ {{{\left( {\dfrac{1}{3}} \right)}^{ - 1}} - {{\left( {\dfrac{1}{4}} \right)}^{ - 1}}} \right\}^{ - 1}}$ is $- 1$.

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

Ans: The given expression is

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

Above expression can be written as,

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

Using property ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$, we get,

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

Using property ${a^m} \div {a^n} = {a^{m - n}}$, we get

$\dfrac{{{8^7} \times {5^4}}}{{{5^7} \times {8^4}}} = \dfrac{{{8^{7 - 4}}}}{{{5^{7 - 4}}}}$

$= \dfrac{{{8^3}}}{{{5^3}}}$

$= \dfrac{{512}}{{125}}$

So the value of ${\left( {\dfrac{5}{8}} \right)^{ - 7}} \times {\left( {\dfrac{8}{5}} \right)^{ - 4}}$ is $\dfrac{{512}}{{125}}$.

7. Simplify

(i) $\dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}}\left( {t \ne 0} \right)$

Ans: 25 can be written as ${5^2}$.

10 can be written as $2 \times 5$.

Therefore, above expression can be written as,

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

Using the property ${a^m} \div {a^n} = {a^{m - n}}$ and ${a^m} \times {a^n} = {a^{m + n}}$, we get

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

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

$= \dfrac{{{5^2} \times {t^4}}}{{{5^{ - 2}} \times 2}}$

$= \dfrac{{{5^{\left( {2 - \left( { - 2} \right)} \right)}} \times {t^4}}}{2}$

$= \dfrac{{{5^4} \times {t^4}}}{2}$

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

So the  value of  $\dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 8}}}}$ is $\dfrac{{625{t^4}}}{2}$.

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

Ans: To solve this problem, 125 can be written as ${5^3}$.

Therefore, to simplify the expression, we will write the given numbers in terms of 2 and 5 using exponent property.

Hence,

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

Using property ${\left( {ab} \right)^m} = {a^m} \times {b^m}$, we get

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

Using property ${a^m} \div {a^n} = {a^{m - n}}$, we get

$\dfrac{{{3^{ - 5}} \times {2^{ - 5}} \times {5^{ - 5}} \times {5^3}}}{{{5^{ - 7}} \times {2^{ - 5}} \times {3^{ - 5}}}}$

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

$= {3^0} \times {2^0} \times {5^5}$

Using property ${x^0} = 1$, we get

${3^0} \times {2^0} \times {5^5} = {5^5}$

So the value of $\dfrac{{{3^{ - 5}} \times {{10}^{ - 5}} \times 125}}{{{5^{ - 7}} \times {6^{ - 5}}}}$ is ${5^5}$.

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.1

Opting for the NCERT solutions for Ex 12.1 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.1 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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