= Solution
During a short interval $dt$, let the rocket's mass change by $dm<0$ and its speed by $dv$. The expelled mass $-dm$ has inertial-frame speed $v-u$ to first order. Conservation of <linear momentum>, including the <impulse> $Fdt$, gives
$$
(m+dm)(v+dv)+(-dm)(v-u)-mv=Fdt.
$$
Discarding the second-order product $dm\,dv$ leaves
$$
m\,dv+u\,dm=Fdt,
$$
and hence the <Tsiolkovsky rocket equation>
$$
\boxed{m\frac{dv}{dt}+u\frac{dm}{dt}=F}.
$$
In deep space $F=0$, so $dv=-u\,dm/m$. Burning from initial mass $m_0$ to $m_0/2$ increases the speed by
$$
\boxed{\Delta v=-u\int_{m_0}^{m_0/2}\frac{dm}{m}=u\log2}.
$$
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