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Ex 13.2, 2 (ix) - Chapter 13 Class 7 Exponents and Powers
Last updated at Sept. 7, 2021 by Teachoo
Transcript
Ex 13.2, 2 Simplify and express each of the following in exponential form: (ix) (2^8 รใ ๐ใ^5)/(4^3 รใ ๐ใ^3 ) (2^8 ร ๐^5)/(4^3 ร ๐^3 ) = (2^8 รใ ๐ใ^5)/((2 ร 2)^3 รใ ๐ใ^3 ) = 2^8/(2^2 )3 ร ๐^5/๐^3 Using, (am)n = ๐^(๐ ร ๐) = 2^8/2^6 ร ๐^5/๐^3 Using, ๐^๐/๐^๐ = ๐^(๐ โ ๐) = 28 โ 6 ร a5 โ 3 = 22 ร a2 Using, am ร bm = (a ร b)m + n = (2 ร a)2 = (2a)2 = (2 ร a)2 = (2a)2
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Ex 13.2
Ex 13.2, 1 (i)
Ex 13.2, 1 (ii)
Ex 13.2, 1 (iii)
Ex 13.2, 1 (iv)
Ex 13.2, 1 (v)
Ex 13.2, 1 (vi)
Ex 13.2, 1 (vii)
Ex 13.2, 1 (viii)
Ex 13.2, 1 (ix)
Ex 13.2, 1 (x)
Ex 13.2, 2 (i) Important
Ex 13.2, 2 (ii)
Ex 13.2, 2 (iii)
Ex 13.2, 2 (iv)
Ex 13.2, 2 (v)
Ex 13.2, 2 (vi)
Ex 13.2, 2 (vii)
Ex 13.2, 2 (viii)
Ex 13.2, 2 (ix) You are here
Ex 13.2, 2 (x)
Ex 13.2, 2 (xi)
Ex 13.2, 2 (xii)
Ex 13.2, 3 (i)
Ex 13.2, 3 (ii)
Ex 13.2, 3 (iii)
Ex 13.2, 3 (iv)
Ex 13.2, 4 (i)
Ex 13.2, 4 (ii)
Ex 13.2, 4 (iii)
Ex 13.2, 4 (iv)
Ex 13.2, 5 (i) Important
Ex 13.2, 5 (ii)
Ex 13.2, 5 (iii)
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.