# NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers (EX 13.2) Exercise 13.2

Every student has a different way of learning. In a classroom context, some students might not receive the appropriate guidance. It is suggested that students use the online resources that best fit their learning preferences to get ready for their examinations. Students need to obtain study materials from reputable sources to make sure their exam preparation is on track. Students can enhance their self-study by using online resources. It’s true that online learning resources can’t entirely take the place of traditional learning methods. Students from all study backgrounds can now easily access study tools thanks to technology and the internet. Students may have additional learning opportunities when traditional and online teaching approaches are combined. Online learning platforms allow students from faraway places to study efficiently. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are available online for students to download. The Extramarks learning platform has very authentic NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2. Class 7 students will be able to practice all the questions of Exercise 13.2 by referring to the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2.

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The NCERT Solutions For Class 7 Maths Chapter 13 Exercise 13.2 are designed by expert Mathematics teachers and students can utilise them to prepare well for the Mathematics examination. Class 7 students preparing for the Mathematics examination are advised to access the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 from Extramarks. The Class 7 Maths Chapter 13 Exercise 13.2 Solutions are useful in finding solutions to Class 7 Maths Ch 13 Ex 13.2. To prepare for the Mathematics examination it is necessary to keep referring to the NCERT textbook of Mathematics.From the standpoint of the mathematics examination, all of the chapters in the textbook are important.Students must practise exercises after learning chapters. Practice exercises help in understanding the theoretical parts of chapters in detail. Class 7 Students will be able to practise questions given in the NCERT textbook by utilising NCERT solutions.  The Extramarks learning platform provides NCERT solutions for all classes. They are very essential for improving the answering skills of students. Students will learn to give answers in detail with the assistance of NCERT solutions provided by Extramarks. It is necessary to keep revising the topics in order to get a deeper understanding of the concepts. Revision notes are also available on Extramarks. The revision notes are essential for quickly revising the topics before examinations. The National Council of Educational Research and Training was established with the purpose of standardising the level of education in India. Innovative ideas and methods of NCERT need to be followed by each school in India. The NCERT is the only body with the authority to publish textbooks for the CBSE Board, which is followed by all CBSE Board schools in India.It is also followed by many other Indian state boards for teaching subjects.

Mathematics is one of the most important subjects in Class 7. Students think of Mathematics as a difficult subject, but it can be easy to score higher marks in Mathematics with a guided strategy and regular practise of questions.  Students are required to practise questions related to each topic to efficiently prepare for the Mathematics examination. Class 7 Mathematics consists of 15 chapters that students must study in order to pass the final mathematics examination.It is crucial to practise all the exercises in each chapter in order to do well in the Mathematics examination. The central theme of Chapter 13 is Exponents and Powers. The new topics of the chapter can be difficult for Class 7 students to master. They are advised to make use of the live classes offered by Extramarksto get a thorough understanding of the theories related to Chapter 13 topics. Students will get guidance from the best teachers of Mathematics during the live lectures. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are available on the Extramarks website and mobile application in PDF format. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are helpful in practising questions given in NCERT Maths Class 7 Chapter 13 Exercise 13.2. The questions asked in Maths Class 7 Chapter 13 Exercise 13.2 are based on the topic titled Laws of Exponents. Students must go through all the derivations given under the topic to solve questions based on them. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are essential for solving questions related to laws of exponents.

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Finding weaker areas in a subject is essential. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 will assist students to overcome weaker areas of Chapter 13. To assist students in fully preparing for the Mathematics examination, all of the solutions to each exercise in Chapter 13 are provided by Extramarks. One of the key chapters in Class 7 Mathematics is Exponents and Powers. It is crucial to practise all the questions in the chapter. Students can benefit from the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 while solving exercises of the chapter. They must access the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 from the Extramarks learning platform. Students having difficulties in solving any question of Exercise 13.2 can refer to the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2.

## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers (EX 13.2) Exercise 13.2

Theoretical concepts in a subject like Mathematics must be learned at regular intervals by students. Since questions from every chapter are covered in the Mathematics examination, it is crucial for students to practise questions based on each chapter. The formulae, definitions, theorems, etc., related to the chapters should be revised from time to time. The Extramarks learning platform has all the study materials that are required for scoring well in the examinations. Each topic in the Mathematics textbook is explained with the aid of solved examples. It is advised to go through each solved example after learning  the theories of the topics. To better comprehend the topics, students should read through the examples provided in the textbook. There are 5 questions in Exercise 13.2. Students should practise each question to prepare effectively for the examination. All of them can be solved with the help of the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2. Students need to refer to the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 whenever they get stuck while solving questions from Exercise 13.2.

Question 1 of Exercise 13.2 is about simplifying the given expressions using the laws of exponents and writing the answers in exponential form. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are useful for solving question 1 of Exercise 13.2. Students facing challenges in solving question 2 of Exercise 13.2 should refer to the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2. Question 3 requires students to simplify and express each of the expressions in exponential form. Some students may not be able to practise this question, so they must utilise the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2.

Students can download the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 in PDF format. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 must be downloaded in PDF format by students from the Extramarks website and mobile application. Before working through the Class 7 Mathematics problems, it is important to have a firm grasp of the concepts. To achieve higher scores in the final examination, every exercise problem must be solved in detail. From the standpoint of the final examination, each of the topics of Chapter 13 is significant. To effectively practise questions, students must become well-versed in the theories of Chapter 13. While reading the NCERT textbooks, attention must be paid to the topics. It is recommended that all Class 7 students continue to practise the exercise questions from the Mathematics chapters to increase their confidence. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 will assist students in gaining confidence when they practice questions. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 will aid in retaining all the key formulae and concepts specific to the exercise questions.Students will benefit from learning how to solve problems when answering mathematics questions.It is also recommended for students to practise questions given in past years’ papers and sample papers of Mathematics. Solving past years’ papers is necessary for getting familiar with the distribution of Mathematics. Students must focus more on practising questions that are based on higher-weightage topics. Past years’ papers are available on the Extramarks learning platform. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 may also help students solve papers from past years. Students need to practise sample papers in mathematics as well. Sample papers are helpful in enhancing the exam preparation of students.

### Access NCERT Solution For Class 7 Maths Chapter 13- Exponents and Powers

The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 can be downloaded from the Extramarks website and mobile application in PDF format. Class 7 students can download NCERT solutions for other chapters of Mathematics from Extramarks as well. Practising questions is essential for boosting the performance of students in the Mathematics examination.

### NCERT Solutions For Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2

There are many advantages of referring to the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2. Students who are having difficulty solving questions can refer to the NCERT solutions provided by the Extramarks learning portal. Students are also encouraged to access other resources from Extramarks, like revision notes, past years’ papers, sample papers, etc. They are updated regularly and are error-free. Practising questions with the assistance of the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 is helpful in boosting the confidence of Class 7 students. The NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2 are beneficial for learning the best methods of practising questions. Students will learn to give appropriate solutions to questions in Mathematics with the help of study materials like the NCERT Solutions Class 7 Maths Chapter 13 Exercise 13.2.

Q.1

$\begin{array}{l}\text{Using laws of exponents, simplify and write the answer}\\ \text{in exponential form:}\\ \left(\text{i}\right){\text{ 3}}^{\text{2}}{\text{× 3}}^{\text{4}}{\text{× 3}}^{\text{8}}\text{ }\left(\text{ii}\right){\text{ 6}}^{\text{15}}{\text{÷6}}^{\text{10}}\text{ }\left(\text{iii}\right){\text{ a}}^{\text{3}}{\text{×a}}^{\text{2}}\\ \left(\text{iv}\right){\text{ 7}}^{\text{x}}{\text{×7}}^{\text{2}}\text{ }\left(\text{v}\right){\text{ 5}}^{{\text{2}}^{\text{3}}}{\text{÷5}}^{\text{3}}\text{ }\left(\text{vi}\right){\text{ 2}}^{\text{5}}{\text{× 5}}^{\text{5}}\\ \left(\text{vii}\right){\text{ a}}^{\text{4}}{\text{×b}}^{\text{4}}\text{ }\left(\text{viii}\right)\text{ }{\left({\text{3}}^{\text{4}}\right)}^{\text{3}}\text{ }\left(\text{ix}\right)\text{ }\left({\text{2}}^{\text{20}}{\text{÷2}}^{\text{15}}\right){\text{×2}}^{\text{3}}\\ \left(\text{x}\right){\text{ 8}}^{\text{t}}{\text{÷8}}^{\text{2}}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{ }{\text{3}}^{\text{2}}×{\text{3}}^{\text{4}}×{\text{3}}^{\text{8}}\\ ={\left(3\right)}^{2+4+8}\left({\text{using a}}^{\text{m}}×{a}^{\text{n}}={a}^{\text{m+n}}\right)\\ ={3}^{14}\\ \\ \left(\text{ii}\right){\text{ 6}}^{\text{15}}÷{\text{6}}^{\text{1}0}\\ ={\left(6\right)}^{15-10}\left({\text{using a}}^{\text{m}}÷{a}^{\text{n}}={a}^{\text{m}-\text{n}}\right)\\ ={6}^{5}\\ \\ \left(\text{iii}\right)\text{ }{\text{a}}^{\text{3}}×{\text{a}}^{\text{2}}\\ ={\left(a\right)}^{3+2}\left({\text{using a}}^{\text{m}}×{a}^{\text{n}}={a}^{\text{m+n}}\right)\\ ={a}^{5}\end{array}$

$\begin{array}{l}\left(\text{iv}\right){\text{7}}^{\text{x}}×{\text{7}}^{\text{2}}\\ ={\left(7\right)}^{\mathrm{x}+2}\left({\text{using a}}^{\text{m}}×{\mathrm{a}}^{\text{n}}={\mathrm{a}}^{\text{m+n}}\right)\\ ={7}^{\mathrm{x}+2}\\ \\ \left(\text{v}\right)\text{ }{\left({\text{5}}^{2}\right)}^{3}÷{5}^{3}\\ ={5}^{2×3}÷{5}^{3}\left(\text{using}{\left({\text{a}}^{\text{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\text{m}×\text{n}}\right)\\ ={5}^{6}÷{5}^{3}\\ ={5}^{6-3}\left({\text{using a}}^{\text{m}}÷{\mathrm{a}}^{\text{n}}={\mathrm{a}}^{\text{m-n}}\right)\\ ={5}^{3}\\ \\ \left(\text{vi}\right){\text{ 2}}^{\text{5}}×{\text{5}}^{\text{5}}\\ ={\left(2×5\right)}^{5}\left({\text{using a}}^{\text{m}}×{\mathrm{b}}^{\text{m}}={\left(\mathrm{a}×\mathrm{b}\right)}^{\text{m}}\right)\\ ={10}^{5}\\ \\ \left(\text{vii}\right){\text{ a}}^{\text{4}}×{\text{b}}^{\text{4}}\\ ={\left(\mathrm{a}×\mathrm{b}\right)}^{4}\left({\text{using a}}^{\text{m}}×{\mathrm{b}}^{\text{m}}={\left(\mathrm{a}×\mathrm{b}\right)}^{\text{m}}\right)\\ \\ \left(\text{viii}\right)\text{ }{\left({3}^{4}\right)}^{3}\\ ={\left(3\right)}^{4×3}\left(\text{using}{\left({\text{a}}^{\text{m}}\right)}^{\mathrm{n}}={\left(\mathrm{a}\right)}^{\text{m}×\text{n}}\right)\\ ={3}^{12}\end{array}$

$\begin{array}{l}\left(\text{ix}\right)\text{\hspace{0.17em} }\left({2}^{20}÷{2}^{15}\right)×{2}^{3}\\ =\left({2}^{20-15}\right)×{2}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\text{using a}}^{\text{m}}÷{\mathrm{a}}^{\text{n}}={\left(\mathrm{a}\right)}^{\text{m}-\text{n}}\right)\\ ={2}^{5}×{2}^{3}\\ ={2}^{5+3}\left({\text{using a}}^{\text{m}}×{\mathrm{a}}^{\text{n}}={\mathrm{a}}^{\text{m+n}}\right)\\ ={2}^{8}\\ \left(\text{x}\right)\text{\hspace{0.17em} }{8}^{\mathrm{t}}÷{8}^{2}\\ ={8}^{\mathrm{t}-2}\left({\text{using a}}^{\text{m}}÷{\mathrm{a}}^{\text{n}}={\mathrm{a}}^{\text{m}-\text{n}}\right)\end{array}$

Q.2

$\begin{array}{l}\text{Simplify and express each of the following in exponential form:}\\ \left(\text{i}\right)\text{ }\frac{{\text{2}}^{\text{3}}{\text{×3}}^{\text{4}}\text{×4}}{\text{3×32}}\text{ }\left(\text{ii}\right)\text{ }\left[{\text{5}}^{{\text{2}}^{\text{3}}}{\text{×5}}^{\text{4}}\right]{\text{÷5}}^{\text{7}}\text{ }\left(\text{iii}\right){\text{ 25}}^{\text{4}}{\text{÷5}}^{\text{3}}\\ \left(\text{iv}\right)\text{ }\frac{{\text{3×7}}^{\text{2}}{\text{×11}}^{\text{8}}}{{\text{21×11}}^{\text{3}}}\text{ }\left(\text{v}\right)\text{ }\frac{{\text{3}}^{\text{7}}}{{\text{3}}^{\text{4}}{\text{×3}}^{\text{3}}}\text{ }\left(\text{vi}\right){\text{ 2}}^{\text{0}}{\text{+3}}^{\text{0}}{\text{+4}}^{\text{0}}\\ \left(\text{vii}\right){\text{ 2}}^{\text{0}}{\text{×3}}^{\text{0}}{\text{×4}}^{\text{0}}\text{ }\left(\text{viii}\right)\text{ }\left({\text{3}}^{\text{0}}{\text{÷2}}^{\text{0}}\right)\text{ }\left(\text{ix}\right)\text{ }\frac{{\text{2}}^{\text{8}}{\text{×a}}^{\text{5}}}{{\text{4}}^{\text{3}}{\text{×a}}^{\text{3}}}\\ \left(\text{x}\right)\text{ }\left(\frac{{\text{a}}^{\text{5}}}{{\text{a}}^{\text{3}}}\right){\text{×a}}^{\text{8}}\text{ }\left(\text{xi}\right)\text{ }\frac{{\text{4}}^{\text{5}}{\text{×a}}^{\text{8}}{\text{b}}^{\text{3}}}{{\text{4}}^{\text{5}}{\text{×a}}^{\text{5}}{\text{b}}^{\text{2}}}\text{ }\left(\text{xii}\right)\text{ }{\left({\text{2}}^{\text{3}}\text{×2}\right)}^{\text{2}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\frac{{2}^{3}×{3}^{4}×4}{3×32}\\ =\frac{{2}^{3}×{3}^{4}×{2}^{2}}{3×{2}^{5}}\\ =\frac{{\left(2\right)}^{3+2}×{3}^{4}}{3×{2}^{5}}\left({\text{Using a}}^{\text{m}}×{\text{a}}^{\text{n}}={\text{a}}^{\text{m+n}}\right)\\ =\frac{{2}^{5}×{3}^{4}}{{2}^{5}×3}\\ ={2}^{5-5}×{3}^{4-1}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={2}^{0}×{3}^{3}\\ ={3}^{3}\\ \\ \left(\mathrm{ii}\right)\text{ }\left({\left({5}^{2}\right)}^{3}×{5}^{4}\right)÷{5}^{7}\\ =\left({5}^{6}×{5}^{4}\right)÷{5}^{7}\left(\text{Using}{\left({\text{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)\\ =\left({5}^{6+4}\right)÷{5}^{7}\left({\text{Using a}}^{\text{m}}×{\text{a}}^{\text{n}}={\text{a}}^{\text{m+n}}\right)\\ ={5}^{10}÷{5}^{7}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={5}^{3}\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{ }{25}^{4}÷{5}^{3}\\ ={\left({5}^{2}\right)}^{4}÷{5}^{3}\left(\text{Using}{\left({\text{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)\\ ={5}^{8}÷{5}^{3}\\ ={5}^{8-3}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={5}^{5}\\ \\ \left(\mathrm{iv}\right)\text{ }\frac{3×{7}^{2}×{11}^{8}}{21×{11}^{3}}\\ =\frac{3×{7}^{2}×{11}^{8}}{3×7×{11}^{3}}\\ ={3}^{1-1}×{7}^{2-1}×{11}^{8-3}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ =7×{11}^{5}\\ \\ \left(\mathrm{v}\right)\text{ }\frac{{3}^{7}}{{3}^{4}×{3}^{3}}\\ =\frac{{3}^{7}}{{3}^{7}}\left({\text{Using a}}^{\text{m}}×{\text{a}}^{\text{n}}={\text{a}}^{\text{m+n}}\right)\\ ={3}^{7-7}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={3}^{0}\text{or 1}\\ \\ \left(\mathrm{vi}\right)\text{ }{2}^{0}+{3}^{0}+{4}^{0}\\ =1+1+1\\ =3\\ \\ \left(\mathrm{vii}\right)\text{ }{2}^{0}×{3}^{0}×{4}^{0}\\ =1×1×1\\ =1\end{array}$

$\begin{array}{l}\left(\mathrm{viii}\right)\text{ }\left({3}^{0}+{2}^{0}\right)×{5}^{0}\\ =\left(1+1\right)×1\\ =2\\ \\ \left(\mathrm{ix}\right)\text{ }\frac{{2}^{8}×{\mathrm{a}}^{5}}{{4}^{3}×{\mathrm{a}}^{3}}\\ =\frac{{2}^{8}×{\mathrm{a}}^{5}}{{\left({2}^{2}\right)}^{3}×{\mathrm{a}}^{3}}\\ =\frac{{2}^{8}×{\mathrm{a}}^{5}}{{2}^{6}×{\mathrm{a}}^{3}}\left(\text{Using}{\left({\text{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)\\ ={2}^{8-6}×{\mathrm{a}}^{5-3}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={2}^{2}×{\mathrm{a}}^{2}\\ \\ \left(\mathrm{x}\right)\text{ }\left(\frac{{\mathrm{a}}^{5}}{{\mathrm{a}}^{3}}\right)×{\mathrm{a}}^{8}\\ =\left({\mathrm{a}}^{5-3}\right)×{\mathrm{a}}^{8}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={\mathrm{a}}^{2}×{\mathrm{a}}^{8}\\ ={\mathrm{a}}^{2+8}\left({\text{Using a}}^{\text{m}}×{\text{a}}^{\text{n}}={\text{a}}^{\text{m}+\text{n}}\right)\\ ={\mathrm{a}}^{10}\\ \\ \left(\mathrm{xi}\right)\text{ }\frac{{4}^{5}×{\mathrm{a}}^{8}{\mathrm{b}}^{3}}{{4}^{5}×{\mathrm{a}}^{5}{\mathrm{b}}^{2}}\\ =\frac{{4}^{5}×{\mathrm{a}}^{8}×{\mathrm{b}}^{3}}{{4}^{5}×{\mathrm{a}}^{5}×{\mathrm{b}}^{2}}\\ ={4}^{5-5}×{\mathrm{a}}^{8-5}×{\mathrm{b}}^{3-2}\left({\text{Using a}}^{\text{m}}÷{\text{a}}^{\text{n}}={\text{a}}^{\text{m}-\text{n}}\right)\\ ={4}^{0}×{\mathrm{a}}^{3}×\mathrm{b}\\ ={\mathrm{ba}}^{3}\end{array}$

$\begin{array}{l}\left(\mathrm{xii}\right)\text{ }{\left({2}^{3}×2\right)}^{2}\\ ={\left({2}^{3+1}\right)}^{2}\left({\text{Using a}}^{\text{m}}×{\text{a}}^{\text{n}}={\text{a}}^{\text{m}+\text{n}}\right)\\ ={\left({2}^{4}\right)}^{2}\\ ={2}^{8}\left(\text{Using}{\left({\text{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right)\end{array}$

Q.3

$\begin{array}{l}\text{Say true or false and justify your answer.}\\ \left(\text{i}\right){\text{ 10×10}}^{\text{11}}{\text{ = 100}}^{\text{11}}\\ \left(\text{ii}\right){\text{ 2}}^{\text{3}}{\text{> 5}}^{\text{2}}\\ \left(\text{iii}\right){\text{ 2}}^{\text{3}}{\text{× 3}}^{\text{2}}{\text{ = 6}}^{\text{5}}\\ \left(\text{iv}\right){\text{ 3}}^{\text{0}}\text{ = }{\left(\text{1000}\right)}^{\text{0}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)10×{10}^{11}={100}^{11}\\ 10×{10}^{11}={10}^{1+11}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={10}^{12}\ne {100}^{11}\\ \text{Therefore, the given statement is false.}\\ \\ \left(\mathrm{ii}\right){2}^{3}>{5}^{2}\\ {2}^{3}=2×2×2=8\\ {5}^{2}=5×5=25\\ \text{Therefore, the given statement is false.}\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right){2}^{3}×{3}^{2}={6}^{5}\\ {2}^{3}×{3}^{2}=8×9\\ =72\\ \text{and}\\ {\text{6}}^{5}=6×6×6×6×6\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7776\\ \text{Therefore, the given statement is false.}\\ \\ \left(\mathrm{iv}\right){3}^{0}={\left(1000\right)}^{0}\\ {3}^{0}=1\\ {1000}^{0}=1\\ \text{Therefore, the given statement is true.}\end{array}$

Q.4

$\begin{array}{l}\text{Express each of the following as a product of prime factors}\\ \text{only in exponential form:}\\ \left(\text{i}\right)\text{108 × 192}\\ \left(\text{ii}\right)\text{270}\\ \left(\text{iii}\right)\text{729 × 64}\\ \left(\text{iv}\right)\text{768}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{ 108 × 192}\\ \text{= }\left(\text{2×2×3×3×3}\right)\text{×}\left(\text{2×2×2×2×2×2×3}\right)\\ \text{= }\left({\text{2}}^{\text{2}}{\text{×3}}^{\text{3}}\right)\text{×}\left({\text{2}}^{\text{6}}\text{×3}\right)\\ {\text{= 2}}^{\text{6+2}}{\text{×3}}^{\text{3+1}}\left({\text{Using a}}^{\text{m}}{\text{ × a}}^{\text{n}}{\text{ = a}}^{\text{mn}}\right)\\ {\text{= 2}}^{\text{8}}{\text{×3}}^{\text{4}}\\ \\ \left(\text{ii}\right)\text{270}\\ \text{= 2×3×3×3×5}\\ {\text{= 2×3}}^{\text{4}}\text{×5}\\ \\ \left(\text{iii}\right)\text{ 729 × 64}\\ \text{= }\left(\text{3×3×3×3×3×3}\right)\text{×}\left(\text{2×2×2×2×2×2}\right)\\ {\text{= 3}}^{\text{6}}{\text{×2}}^{\text{6}}\\ \\ \left(\text{iv}\right)\text{768}\\ \text{= }\left(\text{2×2×2×2×2×2×2×2×3}\right)\\ {\text{= 2}}^{\text{8}}\text{×3}\end{array}$

Q.5

$\begin{array}{l}\text{Simplify:}\\ \left(\text{i}\right)\text{ }\frac{{\left({\text{2}}^{\text{5}}\right)}^{\text{2}}{\text{×7}}^{\text{3}}}{{\text{8}}^{\text{2}}\text{×7}}\\ \left(\text{ii}\right)\text{ }\frac{{\text{25×5}}^{\text{2}}{\text{×t}}^{\text{8}}}{{\text{10}}^{\text{3}}{\text{×t}}^{\text{4}}}\\ \left(\text{iii}\right)\text{ }\frac{{\text{3}}^{\text{5}}{\text{×10}}^{\text{5}}\text{×25}}{{\text{5}}^{\text{7}}{\text{×6}}^{\text{5}}}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{ }\frac{{\left({\text{2}}^{\text{5}}\right)}^{\text{2}}{\text{ × 7}}^{\text{3}}}{{\text{8}}^{\text{2}}\text{ × 7}}\\ \text{= }\frac{{\left({\text{2}}^{\text{5}}\right)}^{\text{2}}{\text{ × 7}}^{\text{3}}}{{\left({\text{2}}^{\text{3}}\right)}^{\text{2}}\text{ × 7}}\\ \text{= }\frac{{\text{2}}^{\text{10}}{\text{ × 7}}^{\text{3}}}{{\text{2}}^{\text{6}}\text{ × 7}}\text{ }\left(\text{Using}{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}{\text{ = a}}^{\text{mn}}\right)\\ {\text{= 2}}^{\text{10}-\text{6}}{\text{ × 7}}^{\text{3}-\text{1}}\text{ }\left({\text{Using a}}^{\text{m}}{\text{×a}}^{\text{n}}{\text{ = a}}^{\text{m+n}}\right)\\ {\text{= 2}}^{\text{4}}{\text{×7}}^{\text{2}}\\ \\ \left(\text{ii}\right)\text{ }\frac{{\text{25 × 5}}^{\text{2}}{\text{ × t}}^{\text{8}}}{{\text{10}}^{\text{3}}{\text{ × t}}^{\text{4}}}\\ \text{= }\frac{{\text{5}}^{\text{2}}{\text{ × 5}}^{\text{2}}{\text{ × t}}^{\text{8}}}{{\left(\text{2×5}\right)}^{\text{3}}{\text{ × t}}^{\text{4}}}\\ \text{= }\frac{{{\text{5}}^{\text{2+}}}^{\text{2}}{\text{ × t}}^{\text{8}}}{{\text{2}}^{\text{3}}{\text{ × 5}}^{\text{3}}{\text{ × t}}^{\text{4}}}\left({\text{Using a}}^{\text{m}}{\text{ × a}}^{\text{n}}{\text{ = a}}^{\text{m+n}}\right)\\ \text{= }\frac{{\text{5}}^{\text{4-3}}{\text{ × t}}^{\text{8-4}}}{{\text{2}}^{\text{3}}}\left({\text{Using a}}^{\text{m}}{\text{ ÷ a}}^{\text{n}}{\text{ = a}}^{\text{m}-\text{n}}\right)\\ \text{= }\frac{{\text{5t}}^{\text{4}}}{\text{8}}\\ \\ \left(\text{iii}\right)\text{ }\frac{{\text{3}}^{\text{5}}{\text{ × 10}}^{\text{5}}\text{ × 25}}{{\text{5}}^{\text{7}}{\text{×6}}^{\text{5}}}\\ \text{= }\frac{{\text{3}}^{\text{5}}\text{ × }{\left(\text{5 × 2}\right)}^{\text{5}}{\text{ × 5}}^{\text{2}}}{{\text{5}}^{\text{7}}\text{ × }{\left(\text{3 × 2}\right)}^{\text{5}}}\\ \text{= }\frac{{\text{3}}^{\text{5}}{\text{ × 5}}^{\text{5}}{\text{ × 2}}^{\text{5}}{\text{ × 5}}^{\text{2}}}{{\text{5}}^{\text{7}}{\text{ × 3}}^{\text{5}}{\text{ × 2}}^{\text{5}}}\\ \text{= }\frac{{\text{3}}^{\text{5}}{\text{ × 5}}^{\text{5+2}}{\text{ × 2}}^{\text{5}}}{{\text{5}}^{\text{7}}{\text{ × 3}}^{\text{5}}{\text{ × 2}}^{\text{5}}}\left({\text{Using a}}^{\text{m}}{\text{ × a}}^{\text{n}}{\text{ = a}}^{\text{m+n}}\right)\\ \text{= }\frac{{\text{3}}^{\text{5}}{\text{ × 5}}^{\text{7}}{\text{ × 2}}^{\text{5}}}{{\text{5}}^{\text{7}}{\text{ × 3}}^{\text{5}}{\text{ × 2}}^{\text{5}}}\\ {\text{= 5}}^{\text{7}-\text{7}}{\text{ × 2}}^{\text{5}-\text{5}}{\text{ × 3}}^{\text{5}-\text{5}}\\ \text{= 1 × 1 × 1 = 1}\end{array}$