can you help with this exponential decay question?

If $A_t$ is the amount left at time $t$, $A_0$ is the initial amount and $k$ is the rate constant for the first order reaction, we know for the radioactive decay (which is first order reaction) -

$displaystyle A_t = A_0e^{-kt} ,$ ...(i)

or $ , displaystyle ln (frac{A_t}{A_0}) = -kt$

As $A_0 = 100, A_t = 80, t = 20, -> , displaystyle ln (frac{80}{100}) = -20k$

$displaystyle k = -frac{ln 0.8}{20}$ ...(ii)

At half life, $displaystyle ln (frac{50}{100}) = -kt_{1/2}$ or $displaystyle t_{1/2} = - frac {ln (0.5)}{k} = 20 times frac {ln (0.5)} {ln(0.8)} approx 62.1$ years.

Now, find $k$ using equation $(ii)$ and plug $k$ and $A_0 = 100$ into equation $(i)$ and that is your answer to $(b)$.

Hint: You have the radioactive decay law: $$N(t)=N_0e^{-lambda t}$$ You have $N=80$ the time $t$ and $N_0=100$ deduce $lambda$. Then half life $T$ is: $$N=dfrac {N_0}2$$ $$dfrac {N_0}2=N_0e^{-lambda T}$$ $$implies dfrac 12=e^{-lambda T}$$ $$ T=dfrac {ln 2}{lambda}$$