Practice Questions on Laws of Exponents

Last updated at Dec. 16, 2024 by

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Chapter 1 Class 9 Chapter Summary - Part 8 Laws of Exponents.pdf
 
 

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Practice Questions on Laws of Exponents - Part 2

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Practice Questions on Laws of Exponents - Part 3

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Practice Questions on Laws of Exponents - Part 4

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Practice Questions on Laws of Exponents - Part 5

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Practice Questions on Laws of Exponents - Part 6

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Practice Questions on Laws of Exponents - Part 7

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Practice Questions on Laws of Exponents - Part 8 Practice Questions on Laws of Exponents - Part 9 Practice Questions on Laws of Exponents - Part 10

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Practice Questions on Laws of Exponents - Part 11

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Practice Questions on Laws of Exponents - Part 12

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Practice Questions on Laws of Exponents - Part 13 Practice Questions on Laws of Exponents - Part 14   

 

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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

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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