Question 242017
Carbon dating: The amount of carbon-14 present in animal bones after t years is given by {{{P(t)=P(0)e^(-0.00012t)}}}, a bone has lost 18% of it's carbon-14, How old are the bones?
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If the bone has lost 18% of its carbon in t years, 
then 82% of it remains after t years.

Since the amount of carbon was {{{P(0)}}} when it was fresh, after t
years {{{P(t)}}} is 82% of {{{P(0)}}} or {{{P(t)=0.82P(0)}}}.
 So we substitute {{{ 0.82P(0) }}} for {{{P(t)}}} in

{{{P(t)=P(0)e^(-0.00012t)}}}

{{{0.82P(0)=P(0)e^(-0.00012t)}}}.

Divide both sides by {{{P(0)}}}

{{{0.82=e^(-0.00012t)}}}

Take natural logs of both sides:

{{{ln(0.82)=-0.00012t}}}

Divide both sides by {{{-0.00012}}}

{{{(ln(0.82))/(-0.00012)=t}}}

{{{1653.757823=t}}}

Answer: about 1650 years old.

Edwin</pre>