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

 

$\frac{-t}{5730} ln(2) = ln(.125)$


$\frac{-t}{5730} = -3$


$\frac{-t}{5730} = \frac{ln(.125}{ln(2)}$

 

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.