Question 199966
The level of thorium in a sample decreases by a factor of one-half every 2 million years. 
A meteorite is discovered to have only 8.6% of its original thorium remaining.
 How old is the meteorite?
:
The half life equation
 Ao*(2^(-t/h)) = A
Where
A = resulting amt 
Ao - initial amt
t = time 
h = half-life of the substance
:
Time in this problems is millions of yrs:
Let Ao = 100
then
A = 8.6
h = 2 (million yrs)
Find t in millions of yrs
:
100(2^(-t/2)) = 8.6
2^(-t/2) = {{{8.6/100}}}
2^(-t/2) = .086
:
log(2^(-t/2)) = log(.086)
;
log equiv of exponents
{{{-t/2}}}log(2) = log(.086)
:
find the logs
{{{(-.301t)/2}}} = -1.0655
Multiply equation by 2
-.301t = 2(-1.0655)
:
-.301t = -2.131
t = {{{(-2.131)/(-.301)}}}
t = 7.078 million years
:
:
Check solution on calc: enter: 2^(-7.078/2)*100 = 8.6%