Explanation Given information: P ( Q ) = k Q r for Q ≥ 0 and r ≠ 0 Formula used: Log identities: n log x = log x n log m − log n = log ( m n ) log m x − log n x = log ( m n ) x Calculation: Here, P = k Q r for Q ≥ 0 and r ≠ 0 So, P = k Q r Q r = P k Taking log on both sides: log ( Q r ) = log ( P k ) r log Q =

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Differential Equations

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Chapter 1.5, Problem 8E

To determine

Show that power law relationship P(Q)=kQr for Q≥0 and r≠0 has an inverse that is also a power law, Q(P)=mPs, where m=k−1r and s=1r.

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Chapter 1.5, Problem 7E

Chapter 1.5, Problem 9E

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Chapter 1 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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Prob. 56E Ch. 1.5 - Prob. 57E Ch. 1.5 - Prob. 58E Ch. 1.5 - Prob. 59E Ch. 1.5 - Prob. 60E Ch. 1.6 - Prob. 1PQ Ch. 1.6 - Prob. 2PQ Ch. 1.6 - Prob. 3PQ Ch. 1.6 - Prob. 4PQ Ch. 1.6 - Prob. 5PQ Ch. 1.6 - Prob. 6PQ Ch. 1.6 - Prob. 7PQ Ch. 1.6 - Prob. 8PQ Ch. 1.6 - Prob. 9PQ Ch. 1.6 - Prob. 1E Ch. 1.6 - Prob. 2E Ch. 1.6 - Prob. 3E Ch. 1.6 - Prob. 4E Ch. 1.6 - Prob. 5E Ch. 1.6 - Prob. 6E Ch. 1.6 - Prob. 7E Ch. 1.6 - Prob. 8E Ch. 1.6 - Prob. 9E Ch. 1.6 - Prob. 10E Ch. 1.6 - Prob. 11E Ch. 1.6 - Prob. 12E Ch. 1.6 - Prob. 13E Ch. 1.6 - Prob. 14E Ch. 1.6 - Prob. 15E Ch. 1.6 - Prob. 16E Ch. 1.6 - Prob. 17E Ch. 1.6 - Prob. 18E Ch. 1.6 - Prob. 19E Ch. 1.6 - Prob. 20E Ch. 1.6 - Prob. 21E Ch. 1.6 - Prob. 22E Ch. 1.6 - Prob. 23E Ch. 1.6 - Prob. 24E Ch. 1.6 - Prob. 25E Ch. 1.6 - Prob. 26E Ch. 1.6 - Prob. 27E Ch. 1.6 - Prob. 28E Ch. 1.6 - Prob. 29E Ch. 1.6 - Prob. 30E Ch. 1.6 - Prob. 31E Ch. 1.6 - Prob. 32E Ch. 1.6 - Prob. 33E Ch. 1.6 - Prob. 34E Ch. 1.6 - Prob. 35E Ch. 1.6 - Prob. 36E Ch. 1.6 - Prob. 37E Ch. 1.6 - 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Show that power law relationship P ( Q ) = k Q r for Q ≥ 0 and r ≠ 0 has an inverse that is also a power law, Q ( P ) = m P s , where m = k − 1 r and s = 1 r .

Solution Summary: The author explains that the power law relationship P(Q)=kqr for Qge 0 has an inverse.

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Show that power law relationship P(Q)=kQr for Q≥0 and r≠0 has an inverse that is also a power law, Q(P)=mPs, where m=k−1r and s=1r.

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Chapter 1 Solutions