Question 1159088
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In terms of the half-life, the general formula for radioactive decay of the Carbon-14 is

    C(t) = {{{C(0)}}}.{{{(1/2)^(t/5730)}}}

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(0)}}}.{{{(1/2)^(t/5730)}}},  which reduces to   0.35 = {{{(1/2)^(t/5730)}}},


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

    {{{log(10,(0.35))}}} = {{{log(10,(1/2)^(t/5730)))}}},  or  {{{(t/5730)*log(10,(0.5))}}} = {{{log(10,(0.35))}}}.


Therefore,

     t = {{{5730*(log(10,(0.35))/log(10,(0.5)))}}} = 8768 years.


<U>ANSWER</U>.  The piece of wood is about 8768 years old.
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Solved.


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The post-solution note:


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    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.
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