Source voltage in discharging capacitor equation
The more general form of the first equation is

$$v_C(t) = v_C(0)\cdot e^{\frac{-t}{RC}}$$

This must be so since, for \$t=0\$

$$e^{\frac{-0}{RC}} = 1$$

So, assuming the capacitor has charged up to the source voltage before you disconnect the source at \$t=0\$, we have

$$v_C(0) = V_S$$

and then

$$t(v_C) = - RC \ln \frac{v_C}{V_S} = RC \ln \frac{V_S}{v_C} $$