Here is the problem I am struggling with. I can get the answer to part (a), but I couldn’t get the answer to part (b). I have attached an image showing the question and my working for part (b). If anyone could help with showing me where I went wrong and what the correct method is, that would be great, thanks.
As a side question, part (b) measures the activity in grams, is this just a mistake, or am I missing something?
On
Instead of solving for $k$, you can just infer its value very simply: $$ \begin{split} A(t) & = A_0 e^{-kt} \\ & = A_0 2^{-t/T_{1/2}} \\ & = A_0 e^{- \frac{\ln{2}}{T_{1/2}} t} \end{split} $$ where $T_{1/2}$ is the half-life. This derivation is very natural when you think about what a half-life is: It's the time in which some quantity gets halved. Thus the equation above.
Now we have the simple relation $$ k = \frac{\ln{2}}{T_{1/2}} \approx 0.0693 $$ The first line in your solution of the b-part is correct. You just have to plug in $k$: $$ A_0 = 200 e^{5k} = 200 e^{5\times 0.0693} \approx 282.82 $$
Recall that $\ln\left(\frac{e^{5t}}4\right)=5t-\ln4$, since $\log ab=\log|a|+\log|b|$. This gives you $k=\frac{\ln2}{10}$. Note that if you wanted to solve for $k$, you could have just taken logarithm on both sides in the second line, i.e. $\frac{A_0}4=A_0e^{-20k}\implies e^{20k}=4$ and $k=\frac{\ln4}{20}$.