Use half-life data in a signed exponential decay model

$dx/dt=-kx$ with $k>0$. For a nonzero amount, $\ln|x|=-kt+C$. The teacher highlights that the rate must carry a negative coefficient when k is defined positive.

Method 1 uses x>0: $\ln x=-kt+C$, hence $x=Ae^{-kt}$ with $A=e^C>0$. At t=0, x=x0 gives A=x0. Method 2 exponentiates the modulus first: $x=\pm e^Ce^{-kt}=Ae^{-kt}$, then the initial positive amount again selects A=x0. The zero-initial-amount equilibrium is obtained directly rather than through its logarithm.

$x(8.02)=x_0/2$ gives $1/2=e^{-8.02k}$, so $k=\ln2/8.02$. Therefore $x=x_0e^{-(\ln2)t/8.02}=x_0\,2^{-t/8.02}$, with t in days. For positive x0, it never reaches exactly zero at finite time in this continuous model.