SOLUTION: A wooden artifact from an ancient tomb contains 35 percent of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? (The half-

Algebra ->  Finance -> SOLUTION: A wooden artifact from an ancient tomb contains 35 percent of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? (The half-      Log On


   

Question 1159088: A wooden artifact from an ancient tomb contains 35 percent of the carbon-14 that is present in living trees.
How long ago, to the nearest year, was the artifact made? (The half-life of carbon-14 is 5730 years.)
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y=pe%5E%28kx%29
-
ln%28y%29=ln%28p%29%2Bln%28e%5Ekx%29
ln%28y%29-ln%28p%29=kx
k=%28ln%28y%29-ln%28p%29%29%2Fx
y=1/2 and p=1 are adequate for half-life; and x is 5730 in this case.
k=ln%281%2F2%29%2F5730
k=-0.00012097
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highlight%28y=pe%5E%28-0.00012097x%29%29

You should be able to do the rest.

Answer by ikleyn(52782) About Me  (Show Source):
You can put this solution on YOUR website!
.

In terms of the half-life, the general formula for radioactive decay of the Carbon-14 is

    C(t) = C%280%29.%281%2F2%29%5E%28t%2F5730%29

where C(t) is the current mass of the carbon-14; C(0) is the initial mass,



Since 335% of the Carbon-14 remained, you have this equation

    0.35*C(0) = C%280%29.%281%2F2%29%5E%28t%2F5730%29,  which reduces to   0.35 = %281%2F2%29%5E%28t%2F5730%29,


To solve it, take logarithm base 10 from both sides. You get an equation 

    log%2810%2C%280.35%29%29 = log%2810%2C%281%2F2%29%5E%28t%2F5730%29%29%29,  or  %28t%2F5730%29%2Alog%2810%2C%280.5%29%29 = log%2810%2C%280.35%29%29.


Therefore,

     t = 5730%2A%28log%2810%2C%280.35%29%29%2Flog%2810%2C%280.5%29%29%29 = 8768 years.


ANSWER.  The piece of wood is about 8768 years old.

Solved.

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The post-solution note: 

    Since the half-life parameter is given, it is NATURALLY to have the ENTIRE SOLUTION in half-life terms.

    The transition to other form (to "ekt-form"), as @josgaritmetic starts his solution, is not necessary.

    It only leads to unnecessary excessive job and creates the room for errors.