Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.5 Exponential and Logarithmic Equations - 4.5 Exercises - Page 470: 92

$ b=-\frac{1}{x}ln(\frac{\frac{K}{y}-1}{a})$

$y=\frac{K}{1+ae^{-bx}} \longrightarrow 1+ae^{-bx}=\frac{K}{y} \longrightarrow e^{-bx}=\frac{\frac{K}{y}-1}{a} \longrightarrow -bx=ln(\frac{\frac{K}{y}-1}{a})\longrightarrow b=-\frac{1}{x}ln(\frac{\frac{K}{y}-1}{a})$