Question 960955
In carbon dating scientists use the fact that carbon-14 has a half-life of
5750 years to determine the age of organic material.
 If a particular fossil is discovered and it contains only 1% (so has lost 99%) of it's carbon-14,
roughly how old is the fossil?
:
The radioactive decay formula for any substance, we can use:
A = Ao*2^(-t/h), where
A = amt remaining after t time
Ao = Initial amt (t=0)
t = time of decay
h = half-life of the substance
:
In this problem
A = .01
Ao = 1
t = time of decay (age)
h = 5750 yrs
:
1 * 2(-t/5750) = .01
or just
 2(-t/5750) = .01
using the nat logs we can write it
 ln(2(-t/5750)) = ln(.01)
the log equiv of exponents
{{{-t/5750}}}*ln(2) = ln(.01)
{{{-t/5750}}} = {{{ln(.01)/ln(2)}}}
use your calc
{{{-t/5750}}} = -6.643856
t = -6.643856 * -5750
t = +38,202 yrs old (roughly 38 thousand years old)