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  1. Chapter 13 Class 7 Exponents and Powers
  2. Serial order wise

Transcript

Example 12 Simplify and write the answer in the exponential form (i) (〖12〗^4 × 〖 9〗^3 × 4)/(6^3 ×〖 8〗^2 × 27) (〖12〗^4 × 9^3 × 4)/(6^3 × 8^2 × 27) = ((2 × 2 × 3)^4 × (3 × 3)^3 × (2 × 2))/((2 × 3)^3 × 〖(2 × 2 × 2)〗^2 × (3 × 3 × 3)) = ((2^2 × 3)^4 × (3^2 )^3 × (2^2 ))/((2 × 3)^3 × 〖(2^3)〗^2 × (3^3)) = ((2^2 )^4 × (3)^4 × (3^2 )^3 × (2^2))/(2^3 × 3^3 × (2^3 )^2×〖 3〗^3 ) = (2^(2 × 4) × 3^4 × 3^(2 × 3) ×〖 2〗^2)/(2^3 × 3^3 × 2^(3 × 2) × 3^3 ) [(a × b)m = am × bm] [(am)n = am × n] = (2^8 × 3^4 × 3^6 ×〖 2〗^2)/(2^3 × 3^3 × 2^6 × 3^3 ) = ((2^8 ×〖 2〗^2 ) × (3^4 × 3^6))/((2^3 × 2^6 ) × (3^3 × 3^3)) = (2^(8 + 2) × 3^(4 + 6))/(2^(3 + 6) ×〖 3〗^(3 + 3) ) = 2^10/2^9 × 3^10/3^6 = 210 − 9 × 310 − 6 = 21 × 34 = 2 × 81 = 162 (Using am × an = am + n) (Using 𝑎^𝑚/𝑎^𝑛 = am − n) Example 12 Simplify and write the answer in the exponential form (ii) 23 × a3 × 5a4 2^3 × 𝑎^3 × 〖5𝑎〗^4 = 2^3 × 𝑎^3 × 5 × 𝑎^4 = (2^3×5)×(𝑎^3×𝑎^4) = (8 × 5) × 𝑎^(3 + 4) = 〖𝟒𝟎𝒂〗^𝟕 (am × an = am + n) Example 12 Simplify and write the answer in the exponential form (iii) (2 × 3^4 × 2^5)/(9 ×〖 4〗^2 ) (2 × 3^4 × 2^5)/(9 × 4^2 ) = (2 × 3^4 × 25)/(9 ×〖 (2^2 )〗^2 ) = ((2 × 2^5 ) × 3^4)/(9 × 2^4 ) = (2^(1 + 5) × 3^4)/(3^2 × 2^4 ) = 2^6/2^4 × 3^4/3^2 = 2^(6 − 4) × 3^(4 − 2) ((am)n = am × n) ((am)n = am × n) [𝑎^𝑚/𝑎^𝑛 " = " 𝑎^(𝑚 − 𝑛) ] = 2^2 × 3^2 = 〖(2×3)〗^2 = 6^2 = 36 (𝑎^𝑚×𝑏^𝑚 " = " (𝑎×𝑏)^𝑚 )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.