The formula for the exponential decay is as follows.
[tex]N(t)=N_0\mleft(\frac{1}{2}\mright)^{\frac{t}{t_{(0.5_)}}}[/tex]where N(t) is the final value, N₀ is the initial value, t is the time that passed, and t₀.₅ is the half-life.
In given, we have the following.
[tex]\begin{gathered} t=24 \\ t_{0.5}=4 \\ N(t)=3 \\ N_0=\text{?} \end{gathered}[/tex]Substituting to the equation, we can solve for N₀.
[tex]\begin{gathered} N(t)=N_0\mleft(\frac{1}{2}\mright)^{\frac{t}{t_{(0.5_)}}} \\ 3=N_0\mleft(\frac{1}{2}\mright)^{\frac{24}{4}} \end{gathered}[/tex]Thus, we have the following.
[tex]\begin{gathered} 3=N_0\mleft(\frac{1}{2}\mright)^6 \\ 3=N_0\mleft(\frac{1}{64}\mright) \\ 192=N_0 \\ N_0=192 \end{gathered}[/tex]Therefore, the initial mass must be 192 mg.
To solve for the mass in 6 weeks, convert the weeks into days. Note that in 1 week, there are 7 days.
[tex]6\text{ weeks}\cdot\frac{7\text{ days}}{1\text{ week}}=42\text{ days}[/tex]Substitute the given values and the 42 days as t into the equation and then solve for N(t).
[tex]\begin{gathered} N(t)=N_0\mleft(\frac{1}{2}\mright)^{\frac{t}{t_{(0.5)}}} \\ N(t)=192\mleft(\frac{1}{2}\mright)^{\frac{42}{4}} \\ =192\mleft(\frac{1}{2}\mright)^{10.5} \\ \approx192(0.000690533966) \\ \approx0.1325825215 \\ \approx0.1326 \end{gathered}[/tex]Therefore, in 6 weeks, the mass must be approximately 0.1326 mg.