A weed covers 8 square metres of a pond at the start of a study…

A rate proportional to the amount gives $\dfrac{dN}{dt} = kN$ and $N = N_0 e^{kt}$. The doubling time or the half-life is $\dfrac{\ln 2}{|k|}$. The logistic equation $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{M}\right)$ has the solution $P = \dfrac{M}{1 + Ae^{-kt}}$ with $A = \dfrac{M - P_0}{P_0}$.