The solution would be like this for this specific problem:
T = 12.5%
Initial amount = 1
1*2^(-t/5730) = .125
[tex] \frac{-t}{5730} ln(2) = ln(.125) [/tex]
[tex] \frac{-t}{5730} = -3 [/tex]
[tex] \frac{-t}{5730} = \frac{ln(.125}{ln(2)} [/tex]
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.
