A radioactive material is reduced to 1/16 of its original amount in four days. How much material should one begin with so that 4 * 10^-3 kg of the material is left after six days?

Radioactivity Decay Formula is given as
$$\begin{gathered} N=N_0{e}^{-\lambda t} ---\left(1\right) \\ Where, \\ \lambda =Radioactivity Decay cons\tan t \\ {N}_0= Initial Amount of material \\ N=Material left after time t \end{gathered}$$

Here, Given Data is
$N=\frac{1}{16}{N}_0 and t=4 days$
Putting the values in Eq. (1) we get
$16=e^{4\lambda }\Rightarrow \lambda =\frac{\ln 16}{4}=\ln 2 ---\left(2\right)$
Now in second case
Let the initial amount of material be M₀ Kg
Given, N = 4 x 10^-3 Kg
  t = 6 days
Putting these values in Eq. (1), we get
$$\begin{gathered} 4\times {10}^{-3}=M_0{e}^{-6\lambda } \\ Putting the value of \lambda from Eq. \left(2\right), we get \\ 4\times {10}^{-3}=M_0{e}^{-6\ln 2} \\ \Rightarrow 4\times {10}^{-3}=M_0{e}^{-\ln 2^6} \\ \Rightarrow 4\times {10}^{-3}=M_0{e}^{-\ln 64}=\frac{M_0}{64} \\ \Rightarrow M \end{gathered}$$₀ = 256 X 10^-3 kg
Therefore, initial amount of material required is 256 x 10^-3 Kg