Explanation Given: The provided logistic equation is: y = L 1 + b e − k t Formula used: tanh x = e x − e − x e x + e − x e ln x = x Proof: Consider the target equation, y = 1 2 L [ 1 + tanh ( 1 2 k ( t − ln b k ) ) ] So, let x = 1 2 k ( t − ln b k ) . Thus, the equation becomes as, y = 1 2 L [ 1 + tanh x ] Since this equation contains tan hyperbolic function and it can be expanded as, tanh x = e x − e − x e x + e − x Thus, 1 + tanh x = 1 + e x − e − x e x + e − x = 2 1 + e − 2 x

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ISBN: 9781337275361

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 6, Problem 8PS

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To prove: The logistic equation can be written as y=12L[1+tanh(12k(t−lnbk))].

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Chapter 6, Problem 7PS

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To prove: The logistic equation can be written as y = 1 2 L [ 1 + tanh ( 1 2 k ( t − ln b k ) ) ] .

Solution Summary: The author explains that the provided logistic equation can be written as y=1+be-kt since this equation contains tan hyperbolic function.

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To prove: The logistic equation can be written as y=12L[1+tanh(12k(t−lnbk))].

Want to see the full answer?

Chapter 6 Solutions