becky0307 said:
Use the formula N = Ie^kt, where N is the number of items in terms of the initial population I, at time t, and k is the growth constant equal to the percent of growth per unit of time. An artifact is discovered at a certain site. If it has 65% of the carbon-14 it originally contained, what is the approximate age of the artifact? (carbon-14 decays at the rate of 0.0125% annually.)
I am lost at how to set this problem up.
I think K represents; 0.0125% annually but then I don't have a clue or best idea how to proceed. would t be 65%. If so, how do you represent that in an exponent?
So we have the equation:
\(\displaystyle N = Ie^{kt}\)
We know that the
initial "population" of carbon is "I" and the "population" of carbon at time t (in years) is N.
We are told that it has 65% of it's original value so we know that the amount we have is .65 of the original, or the fraction...
\(\displaystyle \frac{N}{I} = .65\)
Another way to explain this is that if the INITIAL carbon is \(\displaystyle I\) then the FINAL carbon is \(\displaystyle N = .65I\), therefore if we divide...
\(\displaystyle \frac{N}{I} = e^{kt}\) we get that \(\displaystyle \frac{N}{I}=\frac{.65I}{I} = .65\)
The next thing that is given is the value of k which is \(\displaystyle .0125%\), this means that the RATE is actually \(\displaystyle k = .000125\), since to convert a percent into a fraction we divide by 100. (EX 100% = 1 in fraction form, or 50% would be .5)
So now we have everything we need to solve... and we are
looking for the value of \(\displaystyle t\) for which all of this given information is true.
\(\displaystyle .65 = e^{-.000125*t}\) , note that k is negative since this is exponential
decay, which means that it will get SMALLER as time increases, therefore requiring a negative exponent.
Now we have only one variable and it should be a simple matter of algebra to solve for t from here!
Hope this helps
