2<\/sup><\/td>\n8<\/td>\n | 2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Note: 1. xxxxxxx = x4is read as \u2018x raised to the power 4\u2019 or \u20184th<\/sup> power of x\u2019. \n2. x2<\/sup>y5<\/sup> is read as \u2018x squared into y raised to power 5\u2019. \n3. p6<\/sup>q3<\/sup> is reas as \u2018p raised to the power 6 into q cubed\u2019.<\/p>\n<\/p>\n NCERT In-text Question Page No. 251<\/span><\/p>\nQuestion 1. \nExpress: \n(i) 729 as a power of 3 \n(ii) 128 as a power of 2 \n(iii) 343 as a power of 7 \nAnswer: \n(i) 729 \nWe have: 729 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 = 36<\/sup> \nThus, 729 = 36<\/sup> \n<\/p>\n(ii) 128 \nWe have: 128 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 = 27<\/sup> \nThus, 128 = 27<\/sup> \n<\/p>\n(iii) 343 \nWe have: 343 = 7 \u00d7 7 \u00d7 7 \u00d7 7= 73<\/sup> \nThus, 343 = 73<\/sup> \n<\/p>\nNCERT In-text Question Page No. 254<\/span><\/p>\nQuestion 1. \nSimplify and write in exponential form: \n(i) 25<\/sup> \u00d7 23<\/sup> \n(ii) p3<\/sup> \u00d7 p2<\/sup> \n(iii) 43<\/sup> \u00d7 42<\/sup> \n(iv) a3<\/sup> \u00d7 a2<\/sup> \u00d7 a7<\/sup> \n(v) 53<\/sup> \u00d7 57<\/sup> \u00d7 512<\/sup> \n(vi) (-4)100<\/sup> \u00d7 (-4)20<\/sup> \nAnswer: \n(i) 25<\/sup> \u00d7 23<\/sup> \nWe have: 25<\/sup> \u00d7 23<\/sup> = 25 + 3<\/sup> = 28<\/p>\n<\/p>\n (ii) p3<\/sup> \u00d7 p2<\/sup> \nWe have: p3<\/sup> \u00d7 p2<\/sup> = p3 + 2<\/sup> = p5<\/sup><\/p>\n(iii) 43<\/sup> \u00d7 42<\/sup> \nWe have: 43<\/sup> \u00d7 42<.sup> = 43+2<\/sup> = p5<\/sup><\/sup><\/p>\n(iv) a3<\/sup> \u00d7 a2<\/sup> \u00d7 a7<\/sup> \nWe have: a3<\/sup> \u00d7 a2<\/sup> \u00d7 a7<\/sup> = a3 + 2 + 7<\/sup> = a12<\/sup><\/p>\n(v) 53<\/sup> \u00d7 57<\/sup> \u00d7 512<\/sup> \nWe have: 53<\/sup> \u00d7 57<\/sup> \u00d7 512<\/sup> = 53 + 7 + 12<\/sup> = 522<\/sup><\/p>\n(vi) (-4)100<\/sup> \u00d7 (-4)20<\/sup> \nWe have: (-4)100<\/sup> \u00d7 (-4)20<\/sup> = (-4)100+2<\/sup> = (4)120<\/sup><\/p>\nNote: The above rule is possible only for same bases. It is not true for different bases. Thus, 23<\/sup> x 32<\/sup> will no obev this rule.<\/p>\nNCERT In-text Question Page No. 255<\/span><\/p>\nQuestion 1. \nSimplify and write in exponential form: (eg., 116<\/sup> \u00f7 112<\/sup> = 114<\/sup>) \n(i) 29<\/sup> \u00f7 23<\/sup> \n(ii) 108<\/sup> \u00f7 104<\/sup> \n(iii) 911<\/sup> \u00f7 97<\/sup> \n(iv) 2015<\/sup> \u00f7 2013<\/sup> \n(v) 713<\/sup> \u00f7 710<\/sup> \nAnswer: \nSince am<\/sup> an<\/sup> = am-n<\/sup>, therefore; \n(i) 29<\/sup> \u00f7 23<\/sup> \nWe have: 29<\/sup> \u00f7 23<\/sup> = 29-3<\/sup> = 26<\/sup><\/p>\n<\/p>\n (ii) 108<\/sup> \u00f7 104<\/sup> \nWe have: 108<\/sup> \u00f7 104<\/sup> = 108-4<\/sup> = 104<\/sup><\/p>\n(iii) 911<\/sup> \u00f7 97<\/sup> \nWe have: 911<\/sup> \u00f7 97<\/sup> = 911-7<\/sup>= 94<\/sup><\/p>\n(iv) 2015<\/sup> \u00f7 2013<\/sup> \nWe have: 2015<\/sup> \u00f7 2013<\/sup> = 2015 -13<\/sup> = 202<\/sup><\/p>\n(v) 713<\/sup> \u00f7 710<\/sup> \nWe have: 713<\/sup> \u00f7 710<\/sup> = 713-10<\/sup> = 73<\/sup><\/p>\nQuestion 15. \nSimplify and write the answer in exponential form: \n(i) (62<\/sup>)4<\/sup> \n(ii) (22<\/sup>)100<\/sup> \n(iii) (750<\/sup>)2<\/sup> \n(iv) ( 53<\/sup>)7<\/sup> \nAnswer: \nSince (am<\/sup>)n<\/sup> = amxn<\/sup> = amn<\/sup>, therefore; \n(i) (62<\/sup>)4<\/sup> \nWe have: (62<\/sup>)4<\/sup> = 62 \u00d7 4<\/sup> = 68<\/sup><\/p>\n(ii) (22<\/sup>)100<\/sup> \nWe have: ( 22<\/sup>)100<\/sup> = 22 \u00d7 100<\/sup> = 2200<\/sup><\/p>\n(iii) (750<\/sup>)2<\/sup> \nWe have: (750<\/sup> )2<\/sup> = 750 \u00d7 2<\/sup> = 7100<\/sup><\/p>\n(iv) ( 53<\/sup>)7<\/sup> \nWe have: (53<\/sup>)7<\/sup> = 53 \u00d7 7<\/sup> = 521<\/sup><\/p>\n<\/p>\n NCERT In-text Question Page No. 256<\/span><\/p>\nQuestion 1. \nPut into another form using am<\/sup> \u00d7 bm<\/sup>= (ab)m<\/sup> \n(i) 43<\/sup> \u00d7 23<\/sup> \n(ii) 25<\/sup> \u00d7 b5<\/sup> \n(iii) a2<\/sup> \u00d7 t2<\/sup> \n(iv) 56<\/sup> \u00d7 (-2)6<\/sup> \n(v) (-2)4<\/sup> \u00d7 (-3)4<\/sup> \nAnswer: \nSince am<\/sup> \u00f7 an<\/sup> = am-n<\/sup>, therefore; \n(i) 43<\/sup> \u00d7 23<\/sup> \nWe have: 43<\/sup> \u00d7 23<\/sup> = (4 \u00d7 2)3<\/sup> = 83<\/sup><\/p>\n(ii) 25<\/sup> x b5<\/sup> \nWe have: 25<\/sup> \u00d7 b5<\/sup> = (2 \u00d7 b)5<\/sup>= (2b)5<\/sup><\/p>\n(iii) a2<\/sup> \u00d7 t2<\/sup> \nWe have: a2<\/sup> x t2<\/sup> = (a \u00d7 t)2<\/sup> = (at)2<\/sup><\/p>\n(iv) 56<\/sup> x (-2)6<\/sup> \nWe have: 56<\/sup> \u00d7 (-2)6<\/sup> = [5 \u00d7 (-2)]6<\/sup> = (-10)6<\/sup><\/p>\n(v) (-2)4<\/sup> x (-3)4<\/sup> \nWe have: (-2)4<\/sup> \u00d7 (-3)4<\/sup> = [(-2) \u00d7 (-3)]4<\/sup> = (6)4<\/sup><\/p>\nNCERT In-text Question Page No. 257<\/span><\/p>\nQuestion 2. \nPut into another form using am<\/sup> \u00f7 bm<\/sup> \n(i) 45<\/sup> \u00f7 3 |