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Find (2/3)^2 ×(2/5)^(−3) × (3/5)^2 (2/3)^2 ×(2/5)^(−3) × (3/5)^2 = (2/3)^2 ×(5/2)^3 × (3/5)^2 = 2^2/3^2 × 5^3/2^3 × 3^2/5^2 = 𝟓/𝟐 Evaluate (i) ∛((343)^(−2) ) ∛((343)^(−2) ) = [(343)^(−2) ]^(1/3) = (343)^(−2 × 1/3) = (7^3 )^( (−2)/3) = 7^[3 × ((−2)/3)] = 7^(−2) = 1/7^2 = 1/49 (ii) √(5&(32)^(−3) ) √(5&(32)^(−3) ) = [(32)^(−3) ]^(1/5) = (32)^(−3 × 1/5) = (2^5 )^( (−3)/5) = 2^[5 × ((−3)/5)] = 2^(−3) = 1/2^3 = 1/8 Which of the following is not equal to [(5/6)^(1/5) ]^((−1)/6) (a) 1/[(5/6)^(1/5) ]^(1/6) (b) (6/5)^(1/30) (c) (5/6)^((−1)/30) (d) (5/6)^(1/5 − 1/6) Value of (256)0.16 × (256)0.09 is (A) 4 (B) 16 (C) 64 (D) 256.25 (256)0.16 × (256)0.09 = (256)0.16 + 0.09 = (256)0.25 = (256)^(1/4) = (2^8 )^(1/4) Which of the following is equal to x (a) 𝑥^(12/7)−𝑥^(5/7) (b) √(12&(𝑥^4 )^(1/3) ) (c) (√(𝑥^3 ))^(2/3) (d) 𝑥^(12/7) × 𝑥^(5/7) Find value of x, if 5x−3 × 32x−8 = 225 Given 5x − 3 × 32x − 8 = 225 5x − 3 × 32x − 8 = 25 × 9 5x − 3 × 32x − 8 = 52 × 32 Comparing powers of 5 and 3 If 〖25〗^(𝑥 − 1)=5^(2𝑥 − 1)−100, find the value of x Given 〖25〗^(𝑥 − 1)=5^(2𝑥 − 1)−100 〖(5^2)〗^(𝑥 − 1)=5^(2𝑥 − 1)−100 5^(2𝑥 − 2)=5^(2𝑥 − 1)−100 5^(2𝑥 − 2)− 5^(2𝑥 − 1)=−100 5^2𝑥 × 5^(−2)− 5^2𝑥 × 5^(−1)=−100 5^2𝑥/5^2 −5^2𝑥/5 =−100 5^2𝑥 (1/5^2 −1/5) =−100 5^2𝑥 ((1 − 5)/5^2 ) =−100 5^2𝑥 ((−4)/5^2 ) =−100 5^2𝑥 =100 ×5^2/4 5^2𝑥 =25 × 25 5^2𝑥 =5^2 × 5^2 Prove that (2^30 + 2^29+ 2^28)/(2^31 + 2^30 − 2^29 ) = 7/10 We have (2^30 + 2^29+ 2^28)/(2^31 + 2^30 − 2^29 ) = (2^28 (2^2 + 2 + 1))/(2^29 (2^2 + 2 − 1)) = (2^28 (4 + 2 + 1))/(2^29 (4 + 2 − 1)) = 2^((28 − 29))× 7/5 Simplify: [5 (8^(1/3)+〖27〗^(1/3) )^3 ]^(1/4) [5 (8^(1/3)+〖27〗^(1/3) )^3 ]^(1/4) = [5 (〖(2^3)〗^(1/3)+〖(3^3)〗^(1/3) )^3 ]^(1/4) = [5 (2+3)^3 ]^(1/4) = [5 (5)^3 ]^(1/4) = [5^4 ]^(1/4) Simplify (81/16)^(−3/4)×[(25/9)^((−3)/2)÷(5/2)^(−3) ] (81/16)^(−3/4)×[(25/9)^((−3)/2)÷(5/2)^(−3) ] = [(3/2)^4 ]^(−3/4)×[[(5/3)^2 ]^((−3)/2)÷(5/2)^(−3) ] = (3/2)^(−3)×[(5/3)^(−3)÷(5/2)^(−3) ] =(2/3)^3×[(3/5)^3÷(2/5)^3 ] = 2^3/3^3 ×[3^3/5^3 ÷2^3/5^3 ] = 2^3/3^3 ×[3^3/5^3 ×5^3/2^3 ] = 2^3/3^3 ×3^3/2^3 = 1

  1. Chapter 1 Class 9 Number Systems
  2. Concept wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo