求解 t_{2}=frac{ln(frac{x_{m}}{x_{0}}-1)+(.31)}{gamma_{0}}

$$t _ { 2 } = \frac { \ln ( \frac { x _ { m } } { x _ { 0 } } - 1 ) + ( .31 ) } { \gamma _ { 0 } }$$

求解 t_2 的值 (复数求解)

$$t_{2}=\frac{\ln(\frac{x_{m}-x_{0}}{x_{0}})+0.31}{\gamma _{0}}$$

$$\gamma _{0}\neq 0\text{ and }x_{m}\neq x_{0}\text{ and }x_{0}\neq 0$$

求解 x_0 的值 (复数求解)

$$\left\{\begin{matrix}\\x_{0}\neq 0\text{, }&\text{unconditionally}\\x_{0}=\frac{x_{m}}{e^{t_{2}\gamma _{0}-0.31}+1}\text{, }&x_{m}\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }t_{2}=\frac{2\pi n_{1}i}{\gamma _{0}}+\frac{0.31+\pi i}{\gamma _{0}}\text{ and }Im(\ln(e^{t_{2}\gamma _{0}-0.31}))-Re(t_{2})Im(\gamma _{0})-Re(\gamma _{0})Im(t_{2})=0\text{ and }\gamma _{0}\neq 0\end{matrix}\right.$$

求解 t_2 的值

$$t_{2}=\frac{\ln(\frac{x_{m}-x_{0}}{x_{0}})+0.31}{\gamma _{0}}$$

$$\left(\gamma _{0}\neq 0\text{ and }x_{m}<x_{0}\text{ and }x_{0}<0\right)\text{ or }\left(\gamma _{0}\neq 0\text{ and }x_{m}>x_{0}\text{ and }x_{0}>0\right)$$

求解 x_0 的值

$$x_{0}=\frac{x_{m}}{e^{t_{2}\gamma _{0}-0.31}+1}$$

$$\gamma _{0}\neq 0\text{ and }x_{m}\neq 0$$