Realizing that the engine exit velocity $u_e$ represents effectively the specific impulse $I_{s p}$ (Sect. 3.4), from Eq. 1.1 derive the following relations. Note that the final mass $m_f$ includes the structural mass $m_s$ comprising engine mass and other accessories, and mass of payload $m_l$.
$$
\begin{aligned}
\frac{m_0}{m_f} & =\frac{m_0}{m_s+m_l}=\exp \left(\frac{\Delta v}{I_{s p}}\right) \\
m_p & =m_0\left[1-\exp \left(\frac{-\Delta v}{I_{s p}}\right)\right] \\
m_p & =m_f\left[\exp \left(\frac{\Delta v}{I_{s p}}\right)-1\right] \\
m_l & =m_0 \exp \left(\frac{-\Delta v}{I_{s p}}\right)-m_s
\end{aligned}
$$