{"id":16153,"date":"2020-09-10T11:48:47","date_gmt":"2020-09-10T06:18:47","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=16153"},"modified":"2021-07-02T15:01:53","modified_gmt":"2021-07-02T09:31:53","slug":"kseeb-solutions-for-class-7-maths-chapter-13-ex-13-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-13-ex-13-2\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2"},"content":{"rendered":"

Students can Download Chapter 13 Exponents and Powers Ex 13.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2<\/h2>\n

Question 1.
\nUsing laws of exponents, simplify and write the answer in exponential form :
\n(i) 32<\/sup> \u00d7 34<\/sup> \u00d7 38<\/sup>
\n(ii) 615<\/sup> \u00f7 610<\/sup>
\n(iii) a3<\/sup> \u00d7 a2<\/sup>
\n(iv) 7x<\/sup> \u00d7 72<\/sup>
\n(v) (52<\/sup>)3<\/sup> \u00f7 52<\/sup>
\n(vi) 25<\/sup> \u00d7 55<\/sup>
\n(vii) (a4<\/sup>\u00a0\u00d7 b4<\/sup>)
\n(viii) (34<\/sup>)3<\/sup>
\n(ix) (220<\/sup>\u00a0\u00f7 215<\/sup>) \u00d7 23<\/sup>
\n(x) 8t<\/sup>\u00a0\u00f7 82<\/sup>
\nSolution:
\ni) 32<\/sup> \u00d7 34<\/sup> \u00d7 38<\/sup>
\n= 32 + 4 + 8<\/sup> = 314<\/sup> [\u2235 am<\/sup> \u00d7 an<\/sup> = am + n<\/sup>]<\/p>\n

ii) 615<\/sup> \u00f7 610<\/sup>
\n= 615 – 10<\/sup> = 65<\/sup> [\u2235 am<\/sup> \u00f7 an<\/sup> = am – n<\/sup>]<\/p>\n

iii) a3<\/sup> \u00d7 a2<\/sup>
\n= a3 + 2<\/sup> = a5<\/sup> [\u2235 am<\/sup> \u00d7 an<\/sup> = am + n<\/sup>]<\/p>\n

iv) 7x<\/sup> \u00d7 72<\/sup>
\n= 7x + 2<\/sup> [\u2235 am<\/sup> \u00d7 an<\/sup> = am + n<\/sup>]<\/p>\n

v) (52<\/sup>)3<\/sup> \u00f7 53<\/sup>
\n= 52 \u00d7 3<\/sup> \u00f7 53<\/sup> = 56 – 3<\/sup> [(am<\/sup>)n<\/sup> = amn<\/sup>
\n\u2235 am<\/sup> \u00f7 an<\/sup> = am – n<\/sup>]<\/p>\n

vi) 25<\/sup> \u00d7 55<\/sup>
\n= (2 \u00d7 5)5<\/sup> = 105<\/sup> [\u2235 am<\/sup> \u00d7 bm<\/sup> = (ab)m<\/sup>]<\/p>\n

vii) a4<\/sup> \u00d7 b4<\/sup>
\n= a4<\/sup> \u00d7 b4<\/sup> = (ab)4<\/sup> [\u2235 am<\/sup> \u00d7 bm<\/sup> = (ab)m<\/sup>]<\/p>\n

viii) (34<\/sup>)3<\/sup>
\n= 34 \u00d7 3<\/sup> = 312<\/sup> [(am<\/sup>)n<\/sup> = amn<\/sup>]<\/p>\n

ix) (220<\/sup> \u00f7 215<\/sup>)
\n(220<\/sup> \u00f7 215<\/sup>) \u00d7 23<\/sup> = 220 – 15<\/sup> \u00d7 23<\/sup> [am<\/sup> \u00f7 an<\/sup> = am – n<\/sup>]
\n= 25<\/sup> \u00d7 23<\/sup>
\n= 25 + 3<\/sup> = 28<\/sup> [am<\/sup> \u00d7 an<\/sup> = am + n<\/sup>]<\/p>\n

x) 8t<\/sup> \u00f7 82<\/sup>
\n= 8t – 2<\/sup> [\u2235 am<\/sup> \u00f7 an<\/sup> = am – n<\/sup>]<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nSimplify and express each of the following in exponential form :
\ni)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

ii) ((52<\/sup>)3<\/sup> \u00d7 54<\/sup>) \u00f7 57<\/sup>
\n= (52 \u00d7 3<\/sup> \u00d7 54<\/sup>) \u00f7 57<\/sup> ((am<\/sup>)n<\/sup> = amn<\/sup>)
\n= 56<\/sup> \u00d7 54<\/sup> \u00f7 57<\/sup>
\n= 56 + 4<\/sup> \u00f7 57<\/sup> = 510 – 7<\/sup> = 53<\/sup><\/p>\n

iii) 254<\/sup> \u00f7 53<\/sup>
\n= (52<\/sup>)4<\/sup> \u00f7 53<\/sup> = 52 \u00d7 4<\/sup> \u00f7 53<\/sup>
\n= 58<\/sup> \u00f7 53<\/sup> = 58 – 3<\/sup> = 55<\/sup> [am<\/sup> \u00f7 an<\/sup> = am – n<\/sup><\/p>\n

iv)
\n\"KSEEB
\n\"KSEEB
\n= 72 – 1<\/sup> \u00d7 118 – 3<\/sup> [am<\/sup> \u00f7 an<\/sup> = am – n<\/sup>]
\n= 71<\/sup> \u00d7 115<\/sup> = 7 \u00d7 115<\/sup><\/p>\n

vi) 20<\/sup> + 30<\/sup> + 40<\/sup>
\n= 1 + 1 + 1 = [3] [a0<\/sup> = 1]<\/p>\n

vii) 20<\/sup> \u00d7 30<\/sup> \u00d7 40<\/sup>
\n= 1 \u00d7 1 \u00d7 1 = 1<\/p>\n

viii) (30<\/sup> + 20<\/sup>) \u00d7 50<\/sup>
\n= (1 + 1) \u00d7 1 = 2 \u00d7 1 = 2<\/p>\n

ix)
\n\"KSEEB<\/p>\n

\"KSEEB
\n= (2a)2<\/sup><\/p>\n

x)
\n\"KSEEB
\n= a 5 – 3<\/sup> \u00d7 a8<\/sup>
\n= a2<\/sup> \u00d7 a8<\/sup> a2 + 8<\/sup>
\n\"KSEEB<\/p>\n

xi)
\n\"KSEEB
\n\"KSEEB<\/p>\n

xii) (23<\/sup> \u00d7 2)2<\/sup>
\n= (23<\/sup> \u00d7 21<\/sup>)2<\/sup>
\n= (24<\/sup>)2<\/sup>
\n= 24 \u00d7 2<\/sup> = 28<\/sup><\/p>\n

Question 3.
\nSay true or false and justify your answer:
\ni) 10 \u00d7 1011<\/sup> = 10011<\/sup>
\nL.H.S = 10 \u00d7 1011<\/sup> = 101 + 11<\/sup> = 1012<\/sup> RHS
\n10011<\/sup> = (10 \u00d7 10)11<\/sup>
\n= (102<\/sup>)11<\/sup>
\n= 1o2<\/sup> \u00d7 11 = 1022<\/sup>
\n\u2234 1012<\/sup> \u2260 1022<\/sup>
\nSo 10 \u00d7 1011<\/sup> \u2260 10011<\/sup> \u2234 It is false.<\/p>\n

ii) 23<\/sup> > 52<\/sup>
\n= 2 \u00d7 2 \u00d7 2 > 5 \u00d7 5
\n= 8 > 25
\n\u2234 23<\/sup> > 52<\/sup> \u2234 It is false.<\/p>\n

iii) 23<\/sup> \u00d7 32<\/sup> = 65<\/sup>
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3
\n= 6 \u00d7 6 \u00d7 6 \u00d7 6 \u00d7 6
\n= 72 = 7776
\n\u2234 23<\/sup> \u00d7 32<\/sup> \u2260 65<\/sup> \u2234 It is false.<\/p>\n

iv) 30<\/sup> = (1000)0<\/sup>
\n1 = 1
\n\u2234 30<\/sup> = (1000)0<\/sup> \u2234 It is true<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nExpress each of the following as a product of prime factors only in exponential form :<\/p>\n

i) 108 x 192
\n\"KSEEB
\n= 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3
\n= 28<\/sup> \u00d7 34<\/sup>
\n\"KSEEB
\nIt is the product of prime factor<\/p>\n

ii) 270
\n270 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5
\n= 2 \u00d7 33<\/sup> \u00d7 5
\n\"KSEEB
\nIt is required prime factor.<\/p>\n

iii) 729 \u00d7 64
\n729 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3
\n64 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2
\n\u2234 729 \u00d7 64 = 36<\/sup> \u00d7 26<\/sup>
\n\"KSEEB
\nIt is the required prime factor.<\/p>\n

iv) 768
\n768 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3
\n= 28<\/sup> \u00d7 31<\/sup> or = 28<\/sup> \u00d7 3
\n\"KSEEB
\nIt is a required prime factor.<\/p>\n

Question 5.
\nSimplify :
\ni)
\n\"KSEEB
\nSolution:
\n\"KSEEB<\/p>\n

ii)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

iii)
\n\"KSEEB
\nSolution:
\n\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 13 Exponents and Powers Ex 13.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Question 1. Using laws …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16153"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=16153"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16153\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=16153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=16153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=16153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}