Find $t$, as $N=4000e^{\frac{-t}{50}}$

Question

The number of radioactive units in a sample initially containing 4000 units is given by $N=4000e^{\frac{-t}{50}}$. (t in years). Determine how many years have passed when the number of units is decreasing by 15units per annum.

How do you even start? A geometric sequence?? I tried everything and none of them work!

Answer 1

If we are interested to the instantaneous rate then the solution is given by

$$N'(t)=15$$

otherwise, if we are interested to the discrete rate, we have that

$$\Delta N=4000e^{\frac{-t}{50}}-4000e^{\frac{-(t+1)}{50}}=15$$

$$e^{\frac{-t}{50}}\left(1-e^{-\frac{1}{50}}\right)=\frac{15}{4000}$$

$$e^{\frac{-t}{50}}=\frac{\frac{3}{800}}{\left(1-e^{-\frac{1}{50}}\right)}$$

then take $\log$ both sides.

Answer 2

Rate is $dN\over dt$ $=-\frac{4000}{50}e^{-t\over 50}$. We must find $t$ for which this is equal to $15$.

So, we have

$$e^{-t\over 50} = \frac{15}{80}$$

$$\implies-t = 50\log{15\over 80}$$

$$\implies\boxed{t \approx 83.69882\text{ years}}$$