"Note that $T(x; a) = e^aT(x - a; 0)$ achieves its minimum where $T(x - a; 0)$
does, namely at the unique $x$ value with $x - a = t_0$. Thus $t_a = a + t_0$ and so $$ T(t_a; a) = e^aT(t_a -a;0) = e^{t_a-t_0} T(t_0; 0) = [e^{-t_0}T(t_0; 0)]e^{t_a}$$ which is exponential in $t_a$, as desired."