Question 1157177
Carbon-14 has a half-life of 5,730 years.
 A fossil is found that has 26% of the carbon-14 found in a living sample.
 How old is the fossil? (Round your answer to the nearest year.)
:
The radioactive decay formula: A = Ao*2^(-t/h) where:
A = amt after t time
Ao = initial amt
t = time of decay
h = half-life of substance
:
let initial amt = 1, resulting amt = .26
1*2^(t/5730) = .26
use natural logs
ln(2^(-t/5730)) = ln(.26)
log equiv of exponents
{{{-t/5730}}}ln(2) = ln(.26)
:
{{{-t/5730}}}= {{{ln(.26)/ln(2)}}}
using your calc
{{{-t/5730}}} = -1.94341
t = -5730 * -1.94341
t = 11,136 years