Answer:
About 0.6548 grams will be remaining.
Step-by-step explanation:
We can write an exponential function to model the situation. The standard exponential function is:
[tex]f(t)=a(r)^t[/tex]
The original sample contained 510 grams. So, a = 510.
Each half-life, the amount decreases by half. So, r = 1/2.
For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.
Therefore, our function is:
[tex]\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}[/tex]
One year has 365 days.
Therefore, the amount remaining after one year will be:
[tex]\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548[/tex]
About 0.6548 grams will be remaining.
Alternatively, we can use the standard exponential growth/decay function modeled by:
[tex]f(t)=Ce^{kt}[/tex]
The starting sample is 510. So, C = 510.
After one half-life (38 days), the remaining amount will be 255. Therefore:
[tex]255=510e^{38k}[/tex]
Solving for k:
[tex]\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)[/tex]
Thus, our function is:
[tex]f(t)=510e^{t\ln(.5)/38}[/tex]
Then after one year or 365 days, the amount remaining will be about:
[tex]f(365)=510e^{365\ln(.5)/38}\approx 0.6548[/tex]
