Question 154765
{{{A(t)=A[0]e^(kt)}}} Start with the general exponential formula



Since we're talking about the half life, this means that when {{{t=5730}}}, then {{{A(t)=(1/2)A[0]}}}. In other words, when the time is 5,730 years, the amount left over will be half of the original amount.



{{{(1/2)A[0]=A[0]e^(k*5730)}}} Plug in {{{A(t)=(1/2)A[0]}}} and {{{t=5730}}}. This equation represents the half life. We can now solve for the growth/decay constant "k"




{{{1/2=e^(5730k)}}} Divide both sides by {{{A[0]}}}.




{{{ln(1/2)=ln(e^(5730k))}}} Take the natural log of both sides.



{{{ln(1/2)=5730k*ln(e)}}} Rewrite the right side using the identity  {{{ln(x^y)=y*ln(x))}}}



{{{-0.693147=5730k(1)}}} Take the natural log of {{{1/2}}} to get approximately -0.693147 and take the natural log of "e" to get 1




{{{-0.693147=5730k}}} Multiply



{{{-0.693147/5730=k}}} Divide both sides by 5,730 to isolate k.



{{{-0.000121=k}}} Divide



So the growth/decay constant is {{{k=-0.000121}}}



So the equation is {{{A(t)=A[0]e^(-0.000121t)}}} 



Now I'm assuming that we're using the time frame from 1632 to 1675 (and ignoring the current year 2008). So this would mean that the time span is 1675-1632=43 years. So the claim is that the painting is 43 years old (from the viewpoint of someone living in 1675).


So if the claim is correct, then if {{{t=43}}} is plugged into the equation {{{A(t)=A[0]e^(-0.000121t)}}}, then the resulting amount of carbon should be 99.5% of the original amount. Let's see if this is true:



{{{A(t)=A[0]e^(-0.000121t)}}} Start with the given equation.



{{{A(43)=A[0]e^(-0.000121*43)}}} Plug in {{{t=43}}} 



{{{A(43)=A[0]e^(-0.005203)}}} Multiply



{{{A(43)=A[0](0.994811)}}} Raise "e" to the -0.005203th power to get 0.994811



{{{A(43)=0.994811A[0]}}} Rearrange the terms



If we round to the nearest thousandth, we get


{{{A(43)=0.995A[0]}}} 



Since {{{0.995A[0]}}} is 99.5% of the original amount {{{A[0]}}}, this means that the amount left over is about 99.5% of the original amount. So this shows us that the painting is approximately 43 years old, which means that the painting was most likely created in 1632 (from the viewpoint of 1675). So there is a small chance that this picture is a fake.



Note: we introduced rounding errors due to the evaluation of logarithms and  division. So our answer is not accurate.