Question

How old is a bone that has 12.5 percent of the original amount of radioactive carbon

Answer 1

The solution would be like this for this specific problem:

T = 12.5%

Initial amount = 1

1*2^(-t/5730) = .125

`(-t)/(5730) ln(2) = ln(.125)`



`(-t)/(5730) = -3`



`(-t)/(5730) = (ln(.125)/(ln(2))`

t = -3 * -5730
t = 17,190 yrs

So, a bone that has 12.5 percent of the original amount of radioactive carbon is 17,190 yrs old.