little help findinng t

Question

A wooden artifact from an ancient tomb contains 68 percent of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? (The half-life of carbon-14 is 5730 years.)

Answer 1

Catherine's answer is, of course, correct. We can solve this without first calculating a decay rate.

(1/2)^t/5730 = .68

Taking logs of both sides,

t/5730 ln(1/2) = ln(.68)

t = 5730 ln(.68)/ln(1/2)

t = 3188.13388704473

Answer 2

y = y₀e^kt (y = the amount present at any time, y₀ = the initial amount, k = the constant rate of change, t = time)

First find the constant rate of change by using the half-life information:

1. Let y₀ = 100% = 1, then y = 50% = 0.5 or 1/2
2. Given time for the half-life is t = 5730 years.
3. 1/2 = 1e^k(5730): take the natural log of both sides to get the k to the bottom, which will cancel the e
4. ln(1/2) = 5730k: divide both sides by 5730
5. k = ln(0.5)/5730 = -0.000120968 [we expect this value to be negative bc the artifact is decomposing]

Now use the other given information and the constant rate of change:

1. Let y₀ = 100% = 1, then y = 68% = 0.68 [we expect this time to be less that the time for 0.5]
2. Use the value we found for k = -0.000120968
3. 0.68 = 1e^{-0.000120968t} : take the natural log of both sides again to get the t to the bottom
4. ln(0.68) = -0.000120968t : divide both sides by -0.000120968
5. t = ln(0.68)/-0.000120968 = 3188.13 years