The first step to answer this problem is to determine the decay constant "k" :

`N = N_oe^(-kT)` divide both sides by `N_o` and applying the symmetric property of equality gives

`e^(-kT) = N/N_o` we can then take the natural log of both sides

`ln(e^(-kT)) = ln(N/N_o)` which produces `-kT = ln(N/N_o)` therefore

`k = -(ln(N/N_o))/T` we can now substitute the information for two years and solve for k

`k = -(ln(80/100)/2) = 0.113`

This gives the general equation for N of

`N = 100 e^(-0.113T)` . Now substituting 5 years for T we get

`N = 100e^(-0.565) = 56.8 grams`

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