SOLUTION: Wood deposits recovered from an archaeological site contain 13% of the C-14 they originally contained. How long ago did the tree from which the wood was obtained die? (Assume the h

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Question 1192687: Wood deposits recovered from an archaeological site contain 13% of the C-14 they originally contained. How long ago did the tree from which the wood was obtained die? (Assume the half life of carbon-14 is 5,730 years. Round your answer to the nearest whole number.)
Found 3 solutions by Boreal, ikleyn, josgarithmetic:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Nf=Noe^(-kt)
(1/2)=e^(-kt) for half life
ln(1/2)=(-5730k)
-0.693=-5730k
k=0.0001210
now Nf/No=0.13
so 0.13=e^(-0.0001210t)
ln both sides to remove e
-2.040=-0.0001210t
t=16,861 years.
13% is not quite 3 half lives (12.5% is), which is 17,190 years, so the answer makes sense.

Answer by ikleyn(52818) About Me  (Show Source):
You can put this solution on YOUR website!
.

If you don't care about quality of the solution,  you may accept it from the other tutors posts.

But if you will come to a respectful company for an interview and if they will give you a similar problem,

then,  if you solve it in this way,  they will fail you.


Because  NEXT  TO  YOU  will be another person who  highlight%28highlight%28KNOWS%29%29  how to do it right.



Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
y=Ae%5E%28-kx%29
-
If using the description for this model, then A=100, y=13 but first need a value for k.
Half-life information means if A=100 and y=50, then 50=100e%5E%28-5730%2Ak%29,
1%2F2=e%5E%28-5730k%29
ln%281%2F2%29=-5730k
ln%282%29=5730k
k=ln%282%29%2F5730
k=0.00012097

highlight_green%28y=Ae%5E%28-0.00012097x%29%29

How long ago the tree died?
ln%28y%29=ln%28A%29-0.00012097x
ln%28y%29-ln%28A%29=-0.00012097x
ln%28y%2FA%29=-0.00012097x
highlight_green%28x=%28ln%28y%2FA%29%29%2F%28-0.00012097%29%29
x=-ln%280.13%29%2F0.00012097
highlight%28x=16900%29