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 =

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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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

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

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

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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