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8 December, 00:10

A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. when did the tree die from which the charcoal came? use 5600 years as the half-life of carbon 14.

A (t) = Aoe^kt k<0

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  1. 8 December, 00:35
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    Using the equation of half-life, the given is 0.5 = e^ (5600k)

    in order to solve for k

    we then get the log for both sides

    log (0.5) = log (e^ (5600k))

    then we simplify

    log (0.5) = 5600k * log (e)

    k = log (0.5) / (5600*log (e))

    k = - 0.301029996 / (5600 * (.434294482)

    k = - 0.000123776

    Back to the original equation, we have

    0.3 = 1*e^ (k*t)

    where we substitute for k

    0.3 = e^ (-.000123776*t)

    and then get the log for each

    log (0.3) = log (e^ (-.000123776*t))

    and then simplify

    log (0.3) = -.000123776*t*log (e)

    and then solve for t

    t = log (0.3) / (-.000123776*log (e))

    t = 9727 years

    That is when the tree died.
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