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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!

Noa Even
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CountDOOKU
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2 Answers2

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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.

user
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  • for the first line $15=-\frac{4000}{50}e^{-t\over 50}$ which is $e^{-t\over 50} = 15 * \frac{-50}{4000}$ isnt it? and this wont work because natural log cannot take negative values? Am i missing something? Thanks for the help! – CountDOOKU Oct 20 '19 at 01:32
  • No it isn’t. The first term is greater than the second one because N is decreasing. Try with a calculator. – user Oct 20 '19 at 07:07
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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

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

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

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

Certainly not a dog
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  • wouldnt be $e^{-t\over 50} = 15 * \frac{-50}{4000}$ where did the negative sign go? I reached this line before but i couldnt figure out t since loge cannot take negative values please help? – CountDOOKU Oct 20 '19 at 01:22
  • @FredWeasley Log can absolutely take negative values. I assume you mean you cannot input negative values to log, which is true. We are not doing that here, as you can see from my working. Note that $dN\over dt$ is the rate of growth which is negative (since we are observing decay), and $-\times - = +$. This automatically comes from my method, which was to use differentiation – Certainly not a dog Oct 28 '19 at 06:27