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

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

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