Question 273388
The half-life of carbon-14 is 5730 years.
 A bone is discovered which has 30 percent of the carbon-14 found in the bones of other living animals. 
How old is the bone?
:
The half-life formula
A = Ao*2^(-t/h)
where
A = amt after t yrs
Ao = initial amt
t = time 
h = half-life of substance
:
Set the initial amt at 1, then the resulting amt = .3
:
2^(-t/5730) = .3
using nat logs
ln(2^(-t/5730)) = ln(.3)
log equiv of exponents
{{{-t/5730}}}ln(2)= ln(.3)
{{{-t/5730}}}= {{{ln(.3)/ln(2)}}}
{{{-t/5730}}}= -1.737
t = -1.737 * -5730
t = 9,953 yrs
:
:
Check solution on a calc: enter 2^(-9953/5730) results: .29999 ~ .3 (30%)