Transforming a Nonlinear Model to a Linear One

The following is a comprehensive list of transforming a nonlinear model to a linear one.

Nonlinear Linear Relationships

$$y = \frac{\textstyle ax}{\textstyle 1+bx} \underbrace{\frac{\textstyle 1}{\textstyle y}}_Y = \underbrace{\frac{\textstyle... ...xtstyle 1}{\textstyle x} + \underbrace{\frac{\textstyle b}{\textstyle a}}_\beta a = \frac{\textstyle 1}{\textstyle \alpha} b = \frac{\textstyle \beta}{\textstyle \alpha}$$

$$y = \frac{\textstyle a}{\textstyle x+b} \underbrace{\frac{\textstyle 1}{\textstyle y}}_Y = \underbrace{\frac{\textstyle... ...\textstyle a}}_\alpha x + \underbrace{\frac{\textstyle b}{\textstyle a}}_\beta a = \frac{\textstyle 1}{\textstyle \alpha} b = \frac{\textstyle \beta}{\textstyle \alpha}$$

$$y = \frac{\textstyle ax}{\textstyle b^2+x^2} \underbrace{\frac{\textstyle 1}{\textstyle y}}_Y = \underbrace{\frac{\textstyle... ...ce{\frac{\textstyle b^2}{\textstyle a}}_\beta \frac{\textstyle 1}{\textstyle x} a = \frac{\textstyle 1}{\textstyle \alpha} b^2=\frac{\textstyle \beta}{\textstyle \alpha}$$

$$\underbrace{\ln y}_Y = \underbrace{b}_\alpha \ln x + \underbrace{\ln a}_\beta a = e^\beta b=\alpha$$ y = ax^b

$$y = \frac{\textstyle 1}{\textstyle 1+ax^b} \underbrace{\ln \left(\frac{\textstyle 1-y}{\textstyle y}\right)}_Y = \underbrace{b}_\alpha \ln x + \underbrace{\ln a}_\beta a = e^\beta b=\alpha$$

$$y = \frac{\textstyle 1}{\textstyle 1+\exp \left( \frac{\textstyle ax}{\textstyle b+x} \right)} \underbrace{\left[\ln \left(\frac{\textstyle 1-y}{\textstyle y}\right)\right]^{... ...tstyle 1}{\textstyle x} + \underbrace{\frac{\textstyle 1}{\textstyle a}}_\beta a = \frac{\textstyle 1}{\textstyle \beta} b=\frac{\textstyle \alpha}{\textstyle \beta}$$

$$y = \ln a + x - \ln (e^x+b) \underbrace{e^{-y}}_Y = \underbrace{\frac{\textstyle b}{\textstyle a}}_\alpha e^{-x} + \underbrace{\frac{\textstyle 1}{\textstyle a}}_\beta a = \frac{\textstyle 1}{\textstyle \beta} b=\frac{\textstyle \alpha}{\textstyle \beta}$$

$$\frac{\textstyle x^2}{\textstyle a^2} + \frac{\textstyle y^2}{\textstyle b^2} = 1 \underbrace{y^2}_Y = \underbrace{-\frac{\textstyle b^2}{\textstyle a^2}}_\alpha x^2 + \underbrace{b^2}_\beta a^2 = -\frac{\textstyle \beta}{\textstyle \alpha} b^2 = \beta$$

$$y = a \exp \left\{-\left(\frac{\textstyle x-c}{\textstyle b} \right)^2\right\} \underbrace{\ln y}_Y = \underbrace{-\frac{\textstyle 1}{\textstyle b^2}}_\alph... ...}}_\beta x + \underbrace{\ln a - \frac{\textstyle c^2}{\textstyle b^2}}_\gamma a=\exp \left(\gamma-\frac{\textstyle \beta^2}{\textstyle 4}\right) b=\pm \sqrt{-\frac{\textstyle 1}{\textstyle \alpha}} c=-\frac{\textstyle \beta}{\textstyle 2 \alpha}$$

$$y = \frac{\textstyle a}{\textstyle \sqrt{(1+bx^2)^2+c}} \frac{\textstyle 1}{\textstyle y^2}= \underbrace{\frac{\textstyle b^2}{\textsty... ...tyle a^2}}_\beta x^2+ \underbrace{\frac{\textstyle c+1}{\textstyle a^2}}_\gamma a = \pm \frac{\textstyle \sqrt{4\alpha}}{\textstyle \beta} b = \frac{\textstyle 2 \alpha}{\textstyle \beta} c = \frac{\textstyle 4 \alpha \gamma}{\textstyle \beta^2}-1$$