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

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NCERT Solutions for Class 8 Maths Chapter 12 (EX 12.1)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1 (EX 12.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Register Online for Class 8 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 12 - Exponents and Powers

Exercise:

Exercise - 12.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

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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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 12 Exercise 12.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 12 Exercise 12.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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FAQs on NCERT Solutions for Class 8 Maths Chapter 12: Exponents and Powers - Exercise 12.1

1. What is meant by the term exponent?

An exponent of a number depics the counter of time then the number is multiplied to itself. If it is multiplied by 8 for n times then it is represented as 

8×8×8×….n times = 8^n

The above expression, 8^n, is said as 8^n. Therefore exponents are also known as power or sometimes indices.

2. What are powers and exponents?

Power is an expression that represents the repeated multiplication of a digit or any Integer. Usually a raise to power and is a power where a is the base and un is the exponent. For instance 6³ is the power which shows that 6 is multiplied by itself by 3 times.

3. Define negative exponents?

A negative exponent is used when one is divided by repeated multiplication of a factor. Like, 1 ∕ n is given by minus 1 we are -1 is the exponent. For example, 3²  is represented by 3².

4. Why should one of four NCERT solutions for class 8 Maths chapter 12 exponents and powers exercise 12.1 by vedantu?

When it comes to exam preparation, the NCERT solution for class 8 exponent and power exercise 12.1 from class 12 math from vedantu is thought to be the best choice for CBSE students. There are numerous exercises in this chapter. On this page in PDF format we have the exercise 10.1 class 8 math NCERT solution. This solution is available for download at your convenience where you can assist interactly from the Vedant website or app to study it.

5. Why should I practice class 8 math NCERT chapter 12 exponents and power exercise 12.1?

The NCERT book and solutions have been prepared by the best and most highly skilled educators and Scholars and have created the content in such a way that all the math concepts could be understood by each student. Practice Singh exercise 12.1 will have you in clearing your basics about the concept of exponents and powers will help you as you go through the chapter. The NCERT math book has explained the concept of exponents on which the first exercise is based with the help of solved examples which are written in the easiest way.