SOLUTION: Please help me with this word problem, I'm out of my depth: If 12 g of a radioactive substance are present initially and 4 yr later only 6 g remain, how much of the substance will

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Please help me with this word problem, I'm out of my depth: If 12 g of a radioactive substance are present initially and 4 yr later only 6 g remain, how much of the substance will      Log On


   

Question 703263: Please help me with this word problem, I'm out of my depth:
If 12 g of a radioactive substance are present initially and 4 yr later only 6 g remain, how much of the substance will be present after 7 yr?
Thank you,
Samara
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
Radioactive decay may work like an exponential decay.
Using t for years, A[0] for starting amount, A[t] for amount at time t,
We may use formula, A%5Bt%5D=A%5B0%5De%5E%28k%2At%29

It's not clear if you mean first 4 years and then 3 more years, or are you just saying, "half life is 4 years"+". After 7 years."

Starting with half life, we want to find k.
In a 4 year period, t=4, and A quantity will be reduced to (1/2)*A.
%281%2F2%29=1%2Ae%5E%28k%2A4%29
ln%281%2F2%29=ln%28e%5E%28k%2A4%29%29
ln%281%2F2%29=k%2A4
k=%28ln%281%2F2%29%29%2F4

Computing k gives us k = 0.173.
Our decay formula then is A%5Bt%5D=A%5B0%5D%2Ae%5E%28-0.173%2At%29


NOW, if you want the quantity remaining after 7 years and the starting quantity is 12 grams, then:
A%5B7%5D=12%2Ae%5E%28-0.173%2A7%29
A%5B7%5D=12%2A0.2979
A[7]=3.6 grams