The half-life of radium is 1690 years. If 80 grams are present now, how much will be present in 430 years

Question

I need help with this algebraic word problem, I on't understand what it's asking of me. 

Answer 1

Let A=the remaining amount of radium

Let S=the amount of radium you start with

Let t=the amount of time in years

Let h=the half-life time

 

Then A=S(1/2)^(t/h)

So,

A=80(1/2)^(430/1690)

A=approx. 67g

Answer 2

A(t) = amount remaining in t years

 

       = A₀e^kt, where A₀ is the initial amount and k is a constant to                     be determined.  

 

Since A(1690) = (1/2)A₀ and A₀ = 80, 

 

                 we have 40 ₌ 80e^1690k

 

                 1/2 = e^1690k

 

                 ln(1/2) = 1690k

 

                 k = -0.0004

 

So, A(t) = 80e^-0.0004t 

 

Therefore, A(430) = 80e^-0.0004(430)

 

                            = 80e^-0.172

 

                            ≈ 67.4 g