# Ex 12.1, 7 (ii) - Chapter 12 Class 8 Exponents and Powers

Last updated at Sept. 9, 2021 by Teachoo

Last updated at Sept. 9, 2021 by Teachoo

Transcript

Ex 12.1, 7 Simplify. (ii) (3^(−5 ) × 〖10〗^(−5 ) × 125)/(5^(−7 ) × 6^(−5) ) (3^(−5 )× 〖10〗^(−5 ) × 125)/(5^(−7 ) × 6^(−5) ) = 3^(−5)×〖10〗^(−5)×125× 1/5^(−7) ×1/6^(−5) = 1/3^(5 ) ×1/〖10〗^5 ×125×5^7×6^5 = 6^5/3^5 ×1/〖10〗^5 ×125×5^7 = (6/3)^5×1/〖10〗^5 ×125×5^7 = (2)^5×1/〖10〗^5 ×125×5^7 = 2^5×1/〖10〗^5 ×5^3×5^7 = 2^5×1/〖10〗^5 ×5^(3 + 7) = 2^5×1/〖10〗^5 ×5^10 = (2/10)^5×5^10 = (1/5)^5×5^10 = 1^5/5^5 ×5^10 (𝑈𝑠𝑖𝑛𝑔 𝑎^𝑚/𝑏^𝑚 =(𝑎/𝑏)^𝑚 ) (As 125 = 5 × 5 × 5 = 53) (𝑈𝑠𝑖𝑛𝑔 𝑎^𝑚×𝑎^𝑛=(𝑎)^(𝑚 + 𝑛) ) (𝑈𝑠𝑖𝑛𝑔 𝑎^𝑚/𝑏^𝑚 =(𝑎/𝑏)^𝑚 ) = 1/5^5 ×5^10 = 5^10/5^5 = 5^(10 − 5) = 𝟓^𝟓 (𝑈𝑠𝑖𝑛𝑔 𝑎^𝑚/𝑎^𝑛 =𝑎^(𝑚 − 𝑛) )

Ex 12.1

Ex 12.1, 1 (i)
Important

Ex 12.1, 1 (ii)

Ex 12.1, 1 (iii)

Ex 12.1, 2 (i) Important

Ex 12.1, 2 (ii)

Ex 12.1, 2 (iii)

Ex 12.1, 2 (iv)

Ex 12.1, 2 (v)

Ex 12.1, 3 (i) Important

Ex 12.1, 3 (ii)

Ex 12.1, 3 (iii)

Ex 12.1, 3 (iv)

Ex 12.1, 3 (v)

Ex 12.1, 4 (i)

Ex 12.1, 4 (ii)

Ex 12.1, 5

Ex 12.1, 6 (i)

Ex 12.1, 6 (ii)

Ex 12.1, 7 (i) Important

Ex 12.1, 7 (ii) You are here

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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.