Given the equation $${V_m} = u(\ln {m_0} - \ln {m_8}) - g{t_f}$$ I need to solve for ${m_0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way
1st attempt
$$\eqalign{ & {V_m} = u(\ln {m_0} - \ln {m_8}) - g{t_f} \cr & {V_m} = u(\ln {m_0}) - u(\ln {m_8}) - g{t_f} \cr & {V_m} = u(\ln {m_0}) - u(\ln {m_8}) - g{t_f} \cr & {V_m} = u(\ln {m_0}) - u(\ln {m_8}) - g{t_f} \cr & u(\ln {m_0}) - u(\ln {m_8}) - g{t_f} - {V_m} = 0 \cr & u(\ln {m_0}) - u(\ln {m_8}) - {V_m} = g{t_f} \cr & u(\ln {m_0}) = g{t_f} + u(\ln {m_8}) + {V_m} \cr & \ln {m_0} = (g{t_f} + u(\ln {m_8}) + {V_m}) \div u \cr & {e^{(g{t_f} + u(\ln {m_8}) + {V_m}) \div u}} = {m_0} \cr} $$
2nd attempt - think this looks a little better but still not there yet
$$\eqalign{ & {V_m} = u(\ln {{{m_0}} \over {{m_8}}}) - g{t_f} \cr & {V_m} + g{t_f} = u(\ln {{{m_0}} \over {{m_8}}}) \cr & {{{V_m} + g{t_f}} \over u} = \ln {{{m_0}} \over {{m_8}}} \cr & {e^{{{{V_m} + g{t_f}} \over u}}} = {{{m_0}} \over {{m_8}}} \cr & {m_8}{e^{{{{V_m} + g{t_f}} \over u}}} = {m_0} \cr} $$
Hint:
$$\ln(a) - \ln(b) = \ln \left(\frac{a}{b} \right)$$