SOLUTION: 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 i

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 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 i      Log On


   

Question 181467: 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?
I am having a terrible time trying to figure out this question and I need to show my work. Can you please help me, Heather
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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 formula: A = Ao%2A%282%5E%28-t%2Fh%29%29
Where
A = final amt
Ao = initial amt
t = time years
h = half-life
:
In this problem, we assume the initial amt is 1 and final amt .086 (8.6%),
time and half-life in millions of years.
Ao%2A%282%5E%28-t%2Fh%29%29 = A
1%2A%282%5E%28-t%2F2%29%29 = .086
we can ignore the 1
%282%5E%28-t%2F2%29%29 = .086
using logs here
log%282%5E%28-t%2F2%29%29 = log(.086)
log equiv of exponents
%28-t%2F2%29*log(2) = log(.086)
Find the logs
.301*-t%2F2 = -1.0655
Multiply both sides by 2
-.301t = -2.131
t = %28-2.131%29%2F%28-.301%29
t = 7.08 million years to decay to 8.6%
:
:
Check on a calc: enter: 2^(-7.08/2) = .086 ~ 8.6%
:
did this help you? any questions